Maximum Number of Non-Overlapping Subarrays With Sum Equals Target

Given an array nums and an integer target.

Return the maximum number of non-empty non-overlapping subarrays such that the sum of values in each subarray is equal to target.

 

Example 1:

Input: nums = [1,1,1,1,1], target = 2
Output: 2
Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to target(2).
Example 2:

Input: nums = [-1,3,5,1,4,2,-9], target = 6
Output: 2
Explanation: There are 3 subarrays with sum equal to 6.
([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping.
Example 3:

Input: nums = [-2,6,6,3,5,4,1,2,8], target = 10
Output: 3
Example 4:

Input: nums = [0,0,0], target = 0
Output: 3
 

Constraints:


Solution:

class Solution {
    public int maxNonOverlapping(int[] nums, int target) {
        int count = 0;
        Map<Integer, Integer> sumToNextStartIndex = new HashMap();
        int preSum = 0;
        int start = -1;
        sumToNextStartIndex.put(0, -1);
        for (int i = 0; i < nums.length; i ++) {
            preSum += nums[i];
            // from start to i are indices that are available
            if (sumToNextStartIndex.containsKey(preSum - target) && sumToNextStartIndex.get(preSum - target) >= start) {
                count ++;
                start = i;
            }
            sumToNextStartIndex.put(preSum, i);
        }
        return count;
    }
}