Maximum Subarray

Maximum Subarray - Problem

Given an integer array nums, find the subarray with the largest sum, and return its sum.

A subarray is a contiguous non-empty sequence of elements within an array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [-2,1,-3,4,-1,2,1,-5,4]

› Output: 6

💡 Note: The subarray [4,-1,2,1] has the largest sum = 6

Example 2 — Single Element

$ Input: nums = [1]

› Output: 1

💡 Note: Array has only one element, so maximum subarray sum is 1

Example 3 — All Negative

$ Input: nums = [-2,-1]

› Output: -1

💡 Note: The subarray [-1] has the largest sum = -1

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32 🍎 Apple 28

The key insight is that any subarray with a negative sum should be discarded since it would only reduce the sum of any larger subarray containing it. Kadane's Algorithm efficiently tracks the running sum and resets when it becomes negative, achieving optimal performance. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate every possible subarray by trying all start and end positions. Calculate the sum of each subarray and keep track of the maximum.

Kadane's Algorithm

⏱️ Time: O(n) Space: O(1)

Maintain current sum as we iterate. If current sum becomes negative, reset to 0 since any subarray starting from that point would be better off without the negative prefix.

Prefix Sum

⏱️ Time: O(n²) Space: O(n)

Build prefix sum array where prefix[i] contains sum from index 0 to i. Any subarray sum from i to j can be calculated as prefix[j] - prefix[i-1].

Brute Force — Algorithm Steps

For each starting position i
For each ending position j ≥ i
Calculate sum from i to j
Update maximum sum found

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Try all start and end positions

Calculate Sums

Sum each subarray individually

Find Maximum

Track the largest sum found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int maxSum = nums[0];
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            int currentSum = 0;
            for (int k = i; k <= j; k++) {
                currentSum += nums[k];
            }
            if (currentSum > maxSum) {
                maxSum = currentSum;
            }
        }
    }
    
    return maxSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: outer for start, middle for end, inner for sum calculation

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses variables to track maximum sum and current sum

⚡ Linearithmic Space

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Very High Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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