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:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
• 0 <= target <= 10^6

Accepted Answer:

class Solution {
    public int maxNonOverlapping(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        if (nums.length == 1) {
            return nums[0] == target ? 1: 0;
        }
        // to check if subarray sum is Target
        // core logic is F(i) - F(y) == Target(i is current index, and y is index 
        // somewhere before it)
        // which means there exists 1 path from y to x,
        // F(i) is sum from index -1 to i, F(-1) = 0
        //
        // F(i-1) = F(i-2) + V(i-1)
        Set<Integer> set = new HashSet<>();
        set.add(0);
        int res = 0, cur = 0;
        for (int i = 0; i < nums.length; i++) {
            cur += nums[i];
            if (set.contains(cur - target)) {
                res++;
                cur = 0;
                set = new HashSet<>();
                set.add(0);
            } else {
                set.add(cur);
            }
        }
        return res;
}