# 53. Maximum Subarry

## The Problem

Link to original problem on Leetcode.

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Examples

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.


Example 2:

Input: nums = [1]
Output: 1


Example 3:

Input: nums = [0]
Output: 0


Example 4:

Input: nums = [-1]
Output: -1


Example 5:

Input: nums = [-100000]
Output: -100000

Constraints
• 1 <= nums.length <= 3 * 104
• -105 <= nums[i] <= 105

Follow up: If you have figured out the $O(n)$ solution, try coding another solution using the divide and conquer approach, which is more subtle.

## My Solution

### Naïve Approach

The worst thing I can think of would be to compute the sum of every conceivable subarray. This would be $O(n{^3})$ time complexity and $O(1)$ space complexity.

// Bad, don't do thisconst maxSubArray = (nums) => {    let sum = nums[0];    for (let i = 0; i < nums.length; i++) {        for (let j = i; j < nums.length; j++) {            // I'm using slice and reduce to get the subarray            // sum instead of writing a third for loop,            // because why not?            const newSum = nums                .slice(i, j + 1)                .reduce((acc, curr) => {                    return acc + curr;                }, 0);            sum = Math.max(sum, newSum);        }    }    return sum;};

This code passed Leetcode's example test cases, but times out when submitted. No surprises there!

It can be improved to $O(n{^2})$ time by noticing that we don't need to compute each piece of each subarry. For example, with an array [2, 5, -3, 4], I would start with 2, 2+5, then 2+5-3, then 2+5-3+4 for the first loop of i. See how I recompute every value every time? Instead, I could do it as 2, 2+5, 7-3, 4+4. Here's what that would look like:

// Better but still bad, don't do this eitherconst maxSubArray = (nums) => {    let sum = nums[0];    for (let i = 0; i < nums.length; i++) {        let leftSideSum = 0;        for (let j = i; j < nums.length; j++) {            leftSideSum += nums[j];            sum = Math.max(sum, leftSideSum);        }    }    return sum;};

I did some research and found Kadane's algorithm for solving this problem in $O(n)$ time. It breaks the problem down into the question: would I get a higher sum by continuing the largest subarry ending at index i - 1, or just starting a new subarry at i? To do this, it keeps track of the best sum we've seen so far and the best subarray sum ending at i. No lie, this took some time to wrap my head around. Here's an implementation.

// Much improved, could do thisconst maxSubArray = (nums) => {    let current = -Infinity, best = -Infinity;    for (let i = 0; i < nums.length; i++) {        current = Math.max(nums[i], current + nums[i]);        best = Math.max(best, current);    }    return best;};