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_0_90579e5d | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. | 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_0_b268da7a | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. | 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_0_231d376f | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. | 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_0_793ed0d5 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. | 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_0_6758aee4 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. | 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_0_8bea64fe | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. | 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_0_215f3b64 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. | 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_0_d24a056f | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. | 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_0_7aa157e6 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. | 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_0_8613de8e | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. | 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_1_8aeef2e1 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_8f04cd11 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_dee09732 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_8674fb5b | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_d1fabf13 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_7030e336 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_e689f9c0 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_066d8270 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_55db8259 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_1_1db59ba7 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Optimize for large input sizes up to 10^6. | 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_2_c5c25023 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_292c6a3e | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_3253e4f0 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_08a59e9e | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_09795791 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_e309bd44 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_16275f11 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_a2d7e603 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_55c7eafe | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_2_5fc3ae83 | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Use iterative approach to avoid recursion limits. | 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_3_5b480d87 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_516b1d12 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_3de711e3 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_b00e2773 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_2108bbeb | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_46d482f2 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_ab3647f4 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_fad506db | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_1ef3c1e5 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_3_5a27cab3 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Consider memory constraints for streaming input. | 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_4_88886a67 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_b57bab2b | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_49621984 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_77251ec7 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_b9ebfc81 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_54bd65e9 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_f7c02d35 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_c1dd16ba | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_e47c51fb | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_4_a400e62f | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Support both ascending and descending order. | 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_5_55aec332 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_cea153b5 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_16b57c92 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_481dd854 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_80bd90f4 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_d2336914 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_4c51c5be | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_88a43dd6 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_42092f71 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_5_9aabf0e4 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Ensure thread-safety for concurrent access. | 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_6_c272d36c | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_8c509202 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_d1219087 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_823850c2 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_c279cff5 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_78f9c2b1 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_fa2ab076 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_9fa777bf | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_bf8d96c4 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_6_85cc089f | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Add comprehensive input validation. | 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_7_68188819 | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_79bec5e1 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_41b96249 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_2497f91e | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_5213cbc8 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_b5cfeb75 | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_972729b9 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_057573e3 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_8bb7c2d2 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_7_4108dee6 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Optimize for cache efficiency. | 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_8_bd1d9b34 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_73b38e4c | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_3bfff54e | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_2ba67a88 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_e277e17a | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_94aeae2b | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_bfd545ac | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_75841453 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_9cecaa26 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_8_30b553eb | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Use zero-based indexing throughout. | 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_9_68ccb84b | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_309fc873 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_7edce153 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_58466c60 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_ea3ffcbf | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_db7b2c8b | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_22e93920 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_9c436cb2 | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_22e01a01 | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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_9_b111fd98 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Handle duplicate elements correctly. | 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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