# Largest sum subarray

## Problem

Given an integer array find the subarray which has the largest sum.

## Solution

We will keep two pointers at the start of the array. We will keep on incrementing the end pointer and calculate the current sum. If at any index current sum is larger than the maximum sum we will update the maximum sum and the corresponding indexes. If the current sum is zero or negative at any point, that means current subarray will not be a part of the maximum sum subset. In this scenario we move both the pointer to the next index and continue the process. The complexity of this algorithm is O(n) because the while loop can be looped at most 2*n times

### Code

public class LargestSumSubArray { public static void main(String[] args) { int[] arr = { 4, 3, -5, 0, 6, -8, 12, 3, -9, 2, 5, 8, -3, 4, 8, 0, 3, -3, -5, -9, 4, 2 }; maxSumSubArray(arr); } private static void maxSumSubArray(int[] arr) { int currentStart = 0; int currentEnd = 0; int currentSum = 0; int maxStart = 0; int maxEnd = 0; int maxSum = 0; while (currentEnd != arr.length) { currentSum += arr[currentEnd]; if (currentSum > maxSum) { maxSum = currentSum; maxStart = currentStart; maxEnd = currentEnd; } if (currentSum <= 0) { currentStart = currentEnd + 1; currentSum=0; } currentEnd++; } System.out.println("Maximum sum = " + maxSum); System.out.println("Indexes (" + maxStart + "," + maxEnd + ")"); } }