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_20_69b9d4b0 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_52fb8d96 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_3eb3ceec | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_a1747fb3 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_91313a6f | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_4099195c | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_e2834432 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_0cd305e6 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_897e6d97 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_20_39c1430c | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Implement using functional programming style. | 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_21_d1b1ea7c | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_c05f6506 | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_0fc89549 | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_11959de3 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_809b02b4 | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_c7f4c9ee | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_80ac96aa | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_6e47b13b | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_ddfbf7cf | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_21_4e512459 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Add memoization to avoid recomputation. | 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_22_3c244c53 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_a0cfe8c7 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_c5f6dc15 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_c62f2504 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_622251df | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_ead4f041 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_45844f75 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_a98ebf84 | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_9977810c | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_22_a5c8df96 | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Use sentinel values for boundary conditions. | 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_23_e1b895eb | algorithms | arrays | medium | python | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_b8ea80bd | algorithms | arrays | medium | javascript | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_fb26a08d | algorithms | arrays | medium | java | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_b01ab954 | algorithms | arrays | medium | cpp | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_15e3e61a | algorithms | arrays | medium | go | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_a869024c | algorithms | arrays | medium | rust | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_feaf65c9 | algorithms | arrays | medium | typescript | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_8281eb2b | algorithms | arrays | medium | csharp | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_f4bceb8f | algorithms | arrays | medium | ruby | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_23_3cf78625 | algorithms | arrays | medium | swift | Find the contiguous subarray with the largest sum in the given array. Apply divide and conquer for parallelization. | 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_24_a16fd289 | algorithms | arrays | easy | python | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_af7837d4 | algorithms | arrays | easy | javascript | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_1f830763 | algorithms | arrays | easy | java | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_b128c987 | algorithms | arrays | easy | cpp | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_591701f9 | algorithms | arrays | easy | go | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_04057ae0 | algorithms | arrays | easy | rust | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_9c29a537 | algorithms | arrays | easy | typescript | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_86ee7e6e | algorithms | arrays | easy | csharp | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_a501ace7 | algorithms | arrays | easy | ruby | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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_24_78cbc8fa | algorithms | arrays | easy | swift | Find the contiguous subarray with the largest sum in the given array. Handle circular references correctly. | 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"
],
... |
merge_intervals_python_0_95edca6b | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_0_c3412e6a | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
return resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_0_8f0dedd0 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math.max(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_0_adc0be26 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_0_48ef40d6 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_0_444e1b1b | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_0_a7cbd14e | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_0_5df14a5e | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_0_b0da8a5b | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_0_fad87e31 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_1_7f50c163 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_1_ed1845b5 | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
return resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_1_98495bdb | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math.max(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_1_1f2384dd | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_1_301a8d1d | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_1_b47facd3 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_1_e1167e8c | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_1_807bb77b | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_1_5c837a33 | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_1_7aedc954 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_2_03bade28 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_2_c578910a | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
return resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_2_1d591e2d | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math.max(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_2_491cc21c | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_2_9c7f48a4 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_2_f5a3db75 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_2_4567330a | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_2_35eff41f | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_2_d52865bf | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_2_509bc79f | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_3_9b9db3ba | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_3_cf1ddcbf | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
return resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_3_b6e37bdd | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math.max(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_3_a6d78724 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_3_f2a3bcd3 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_3_8edf8b4c | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_3_fc0209d6 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_3_fd598ee1 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_3_a691916d | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_3_70746b12 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Consider memory constraints for streaming input. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_4_fd762976 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_4_32ae9d77 | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
return resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_4_d3f323c8 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math.max(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_4_1c2b0605 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_4_201c0c9c | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_4_b0d9e6f5 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_4_2f66eb28 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_4_971e2c95 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_4_043c6542 | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_4_6c0d7d54 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Support both ascending and descending order. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
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