53. Maximum Subarray

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

Example:

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

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.

class Solution {
    public int maxSubArray(int[] nums) {
        int maxSum = Integer.MIN_VALUE;
        int curSum = 0;
        int left = 0, right = 0;
        while (right < nums.length) {
            curSum += nums[right];
            maxSum = Math.max(maxSum, curSum);
            right++;
            if (curSum < 0) {
                curSum = 0;
                left = right;
            }
        }
        return maxSum;
    }
}

public class Solution {
   public int maxSubArray(int[] nums) {
        int currSum = nums[0], maxSum = nums[0];
        
        for(int i = 1; i < nums.length; i++) {
            
            currSum = (currSum < 0) ? nums[i] : (currSum + nums[i]);
            
            if (currSum > maxSum) {
                maxSum = currSum;
            }
        }
        
        return maxSum;
    }
}

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        res = nums[0]
        cur = nums[0]
        for i in range(1, len(nums)):
            cur += nums[i]
            if cur < nums[i]:
                cur = nums[i]
            res = max(res, cur)            
        return res