1546. Maximum Number of Non-Overlapping Subarrays With Sum Equals Target

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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

Solution

Using a map to denote the sum of elements (from 0 -> i) and the accepted count from (0->i)

Like what we do in 560. Subarray Sum Equals K, we can find if there is a subarray's sum is K, and add 1 to its value (accepted count).

Keep traversing all elements in the array

 public int maxNonOverlapping(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>(); //sum, count
        map.put(0,0); //inital status
        int cur = 0;
        int count = 0;
        for(int i = 0; i < nums.length; i++){
            cur += nums[i];
            if(map.containsKey(cur-target)){
                count = Math.max(count, map.get(cur-target) + 1);
            }
            map.put(cur, count);
        }
        return count;
    }