Hi Geeks! Welcome to 100 Days of Leetcode Challenge.

In this article, we going to see about Maximum Subarray.

**Day 29 **

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:6Explanation:[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.

## Approach

### Using DP

public class Solution { public int MaxSubArray(int[] nums) { int[] dp = new int[nums.Length]; dp[0] = nums[0]; int max = nums[0]; for(int i=1;i<nums.Length;i++) { if(dp[i-1] <= 0) dp[i] = nums[i]; else if(dp[i-1] > 0){ if(nums[i] + dp[i-1] <= 0) dp[i] = nums[i]; else dp[i] = nums[i] + dp[i-1]; } max = Math.Max(dp[i],max); } return max; } }

It takes, O(n) time complexity and O(n) Space Complexity.

How can we reduce space complexity to O(1)? Can we able to solve that?

Yes, we can, by introducing two variables.

public class Solution { public int MaxSubArray(int[] nums) { int prev = nums[0]; int current = 0; int max = nums[0]; for(int i=1;i<nums.Length;i++) { if(prev <= 0) current = nums[i]; else{ if(nums[i] + prev <= 0) current = nums[i]; else current = nums[i] + prev; } max = Math.Max(current,max); prev = current; } return max; } }