id stringlengths 17 53 | domain stringclasses 5
values | category stringclasses 24
values | difficulty stringclasses 3
values | language stringclasses 10
values | problem stringlengths 29 149 | thinking_1 stringlengths 236 493 | solution_1 stringlengths 166 541 | thinking_2 stringlengths 302 467 | solution_2 stringlengths 184 565 | thinking_3 stringlengths 302 463 | solution_3 stringlengths 169 1.14k | metadata dict |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
max_subarray_python_10_64c0b219 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_javascript_10_637c5c0a | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_java_10_c1ff4385 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_cpp_10_a93b2767 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Si... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eit... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_go_10_3a6b24eb | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step ... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Sin... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eith... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_rust_10_925c2015 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_typescript_10_11502f47 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_csharp_10_8a791ca4 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_ruby_10_c921ecd1 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_swift_10_9cf62e2e | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Return empty result for invalid input. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_python_11_0d11e404 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_javascript_11_f3bf92a9 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_java_11_0c579f45 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_cpp_11_3205772b | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_go_11_2b761524 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_rust_11_4fdbe89f | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_typescript_11_d8853c05 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_csharp_11_ba552448 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_ruby_11_297dca2e | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_swift_11_7fe90b4d | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Minimize memory allocations. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_python_12_13ce533e | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_javascript_12_4d74ebc4 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_java_12_7dc02744 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_cpp_12_8bf8cef9 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Si... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eit... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_go_12_6880f876 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step ... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Sin... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eith... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_rust_12_1ababe2b | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_typescript_12_0ec85e58 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_csharp_12_307a046e | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_ruby_12_c98b5fd3 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_swift_12_ba781b20 | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Support generic types where applicable. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_python_13_940f088e | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_javascript_13_8880106a | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_java_13_a57c85e5 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_cpp_13_29b8e320 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_go_13_eaab26df | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_rust_13_544b24ae | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_typescript_13_ff6e4ebd | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_csharp_13_32af6163 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_ruby_13_23178f76 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_swift_13_68f73ae3 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Document time and space complexity. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_python_14_b566dff0 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_javascript_14_3a4a7583 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_java_14_17867ff8 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_cpp_14_901516eb | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Si... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eit... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_go_14_c749cd73 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step ... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Sin... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eith... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_rust_14_ea396596 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_typescript_14_e669b967 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_csharp_14_82446dae | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_ruby_14_c9322dcf | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_swift_14_4dde7d50 | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Include unit tests for verification. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_python_15_a0d6dff1 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_javascript_15_c521a367 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_java_15_90873b52 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_cpp_15_73856832 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_go_15_eec05306 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_rust_15_e27b90a0 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_typescript_15_cbf2c863 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_csharp_15_eae83c2e | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_ruby_15_8dce5d21 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_swift_15_1011f6fc | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Handle overflow for large values. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_python_16_504f4c22 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_javascript_16_0d97607a | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_java_16_7d91f90d | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_cpp_16_205e8e04 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Si... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eit... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_go_16_64e1b399 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step ... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Sin... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eith... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_rust_16_46dad607 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_typescript_16_49767c8b | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_csharp_16_110d8aff | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_ruby_16_70bc22f7 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_swift_16_ecb52071 | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Use tail recursion where possible. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_python_17_191f321f | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_javascript_17_3812fdbc | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_java_17_d1482b8d | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_cpp_17_37ab8116 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_go_17_7d7c5796 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_rust_17_6e73f672 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_typescript_17_fa98caaa | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_csharp_17_d9250daa | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_ruby_17_3aa4e39f | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_swift_17_f8b1cebc | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Leverage immutable data structures. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_python_18_8ea516ee | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_javascript_18_8adcf143 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_java_18_ad1246ea | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_cpp_18_a5fb6ef9 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Si... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eit... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_go_18_bb23f063 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Step ... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: Sin... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution eith... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_rust_18_2d139021 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_typescript_18_9e10bcb4 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spac... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Ste... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solut... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_csharp_18_f7a2df14 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_ruby_18_e95c2213 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_swift_18_03e97faf | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Apply early termination for sorted input. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: easy
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"max_subarray"
],
... |
max_subarray_python_19_966bbfc9 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | def max_subarray_sum(nums):
"""
Brute Force: Check all possible subarrays.
Time: O(N^2), Space: O(1)
"""
n = len(nums)
max_sum = float('-inf')
for i in range(n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, curren... | Approach: Kadane for Max Subarray
Language: Python | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | def max_subarray_sum(nums):
"""
Kadane's Algorithm: Track maximum subarray ending at each position.
Time: O(N), Space: O(1)
"""
max_so_far = current_max = nums[0]
for num in nums[1:]:
current_max = max(num, current_max + num)
max_so_far = max(max_so_far, current_max)
return m... | Approach: Divide Conquer for Max Subarray
Language: Python | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | def max_subarray_sum(nums):
"""
Divide and Conquer: Max subarray crosses midpoint or lies in one half.
Time: O(N log N), Space: O(log N)
"""
def max_crossing_sum(arr, left, mid, right):
sm = 0
left_sum = float('-inf')
for i in range(mid, left - 1, -1):
sm += arr[i... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_javascript_19_2105865b | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums) {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
maxSum = M... | Approach: Kadane for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums) {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
maxSoFar =... | Approach: Divide Conquer for Max Subarray
Language: Javascript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums) {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr, left, mid, right) {
let sm = 0, leftSum = -Infinity;
for (let i = mid; i >= left; i--) {
sm += arr[i];
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_java_19_08c313af | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int currentSum = 0;
for (int j = i; j < nums.length; j++) {
... | Approach: Kadane for Max Subarray
Language: Java | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i],... | Approach: Divide Conquer for Max Subarray
Language: Java | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int maxSubArray(int[] nums) {
return maxSubarrayHelper(nums, 0, nums.length - 1);
}
private int maxCrossingSum(int[] arr, int left, int mid, int right) {
... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_cpp_19_269358bd | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
St... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSum = INT_MIN;
for (size_t i = 0; i < nums.size(); i++) {
... | Approach: Kadane for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: ... | #include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (size_t i... | Approach: Divide Conquer for Max Subarray
Language: Cpp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution e... | #include <vector>
#include <algorithm>
#include <climits>
using namespace std;
class Solution {
public:
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
int maxSubArray(vector<int>& nums) {
return maxSubarrayHelper(nums, 0, nums.size() -... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_go_19_318ee736 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
Ste... | package main
import "math"
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
func maxSubArray(nums []int) int {
maxSum := math.MinInt64
for i := 0; i < len(nums); i++ {
currentSum := 0
for j := i; j < len(nums); j++ {
currentSum += nums[j]
if cur... | Approach: Kadane for Max Subarray
Language: Go | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4: S... | package main
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
func maxSubArray(nums []int) int {
maxSoFar := nums[0]
currentMax := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] > currentMax+nums[i] {
currentMax = nums[i]
} e... | Approach: Divide Conquer for Max Subarray
Language: Go | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ei... | package main
import "math"
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxSubArray(nums []int) int {
return maxSubarrayHelper(nums, 0, len(nums)-1)
}
func maxCrossingSum(arr []int, left, mid, right int) int {
sm := 0
leftSum := math.Mi... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_rust_19_55fb5156 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let mut max_sum = i32::MIN;
for i in 0..nums.len() {
let mut current_sum = 0;
for j in i..nums.len() {
current_sum += nums[j];
if current_sum >... | Approach: Kadane for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let mut max_so_far = nums[0];
let mut current_max = nums[0];
for i in 1..nums.len() {
current_max = nums[i].max(current_max + nums[i]);
max_... | Approach: Divide Conquer for Max Subarray
Language: Rust | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | pub fn max_sub_array(nums: Vec<i32>) -> i32 {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
fn max_crossing_sum(arr: &[i32], left: usize, mid: usize, right: usize) -> i32 {
let mut sm = 0;
let mut left_sum = i32::MIN;
f... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_typescript_19_0540c781 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | function maxSubArraySum(nums: number[]): number {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
let currentSum = 0;
for (let j = i; j < nums.length; j++) {
currentSum += nums[j];
... | Approach: Kadane for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
S... | function maxSubArraySum(nums: number[]): number {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
let maxSoFar = nums[0];
let currentMax = nums[0];
for (let i = 1; i < nums.length; i++) {
currentMax = Math.max(nums[i], currentMax + nums[i]);
... | Approach: Divide Conquer for Max Subarray
Language: Typescript | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The sol... | function maxSubArraySum(nums: number[]): number {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
let sm = 0, leftSum = -Infinity;
for (le... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_csharp_19_d18d0ccb | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.... | using System;
public class Solution {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSum = int.MinValue;
for (int i = 0; i < nums.Length; i++) {
int currentSum = 0;
for (int j = i; j < nums.Lengt... | Approach: Kadane for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step ... | using System;
public class Solution {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
public int MaxSubArray(int[] nums) {
int maxSoFar = nums[0];
int currentMax = nums[0];
for (int i = 1; i < nums.Length; i++) {
currentM... | Approach: Divide Conquer for Max Subarray
Language: Csharp | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solutio... | using System;
public class Solution {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
public int MaxSubArray(int[] nums) {
return MaxSubarrayHelper(nums, 0, nums.Length - 1);
}
private int MaxCrossingSum(int[] arr, int left, int... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_ruby_19_661f776d | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
S... | def max_sub_array_sum(nums)
# Brute Force: Check all possible subarrays.
# Time: O(N^2), Space: O(1)
max_sum = -Float::INFINITY
nums.each_index do |i|
current_sum = 0
(i...nums.length).each do |j|
current_sum += nums[j]
max_sum = [max_sum, current_sum].max
end
end
max_sum
end | Approach: Kadane for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4:... | def max_sub_array_sum(nums)
# Kadane's Algorithm: Track maximum subarray ending at each position.
# Time: O(N), Space: O(1)
max_so_far = current_max = nums[0]
nums[1..-1].each do |num|
current_max = [num, current_max + num].max
max_so_far = [max_so_far, current_max].max
end
max_so_far
end | Approach: Divide Conquer for Max Subarray
Language: Ruby | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution ... | def max_sub_array_sum(nums)
# Divide and Conquer: Max subarray crosses midpoint or lies in one half.
# Time: O(N log N), Space: O(log N)
def max_crossing_sum(arr, left, mid, right)
sm = 0
left_sum = -Float::INFINITY
mid.downto(left) do |i|
sm += arr[i]
left_sum = [left_sum, sm].max
end... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
max_subarray_swift_19_146870a0 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Use bit manipulation for space optimization. | Approach: Brute Force for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra space.
... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Brute Force: Check all possible subarrays.
// Time: O(N^2), Space: O(1)
var maxSum = Int.min
for i in 0..<nums.count {
var currentSum = 0
for j in i..<nums.count {
currentSum += nums[j]
maxSum = max(maxSum, currentSum... | Approach: Kadane for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Kadane's Algorithm: track the maximum subarray ending at each position.
Step 2: The max ending at position i is either element i or element i plus previous max.
Step 3: Maintain current_max (ending here) and global_max (best so far).
Step 4... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Kadane's Algorithm: Track maximum subarray ending at each position.
// Time: O(N), Space: O(1)
var maxSoFar = nums[0]
var currentMax = nums[0]
for i in 1..<nums.count {
currentMax = max(nums[i], currentMax + nums[i])
maxSoFar = max(maxSo... | Approach: Divide Conquer for Max Subarray
Language: Swift | Difficulty: medium
Step 1: Split problem into smaller subproblems, solve recursively, combine results.
Step 2: Identify how to split: midpoint, pivot, or natural partition.
Step 3: Define the combine step that merges subproblem solutions.
Step 4: The solution... | func maxSubArraySum(_ nums: [Int]) -> Int {
// Divide and Conquer: Max subarray crosses midpoint or lies in one half.
// Time: O(N log N), Space: O(log N)
func maxCrossingSum(_ arr: [Int], _ left: Int, _ mid: Int, _ right: Int) -> Int {
var sm = 0, leftSum = Int.min
for i in stride(from: mid... | {
"approach_1_type": "brute_force",
"approach_2_type": "dp_optimized",
"approach_3_type": "divide_conquer",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N log N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"max_subarray"
],
... |
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