README_EN.md

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

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Description

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.

 

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104
• 0 <= target <= 106

Solutions

Python3

class Solution:
    def maxNonOverlapping(self, nums: List[int], target: int) -> int:
        i, n = 0, len(nums)
        ans = 0
        while i < n:
            s = 0
            seen = {0}
            while i < n:
                s += nums[i]
                if s - target in seen:
                    ans += 1
                    break
                i += 1
                seen.add(s)
            i += 1
        return ans

Java

class Solution {
    public int maxNonOverlapping(int[] nums, int target) {
        int i = 0, n = nums.length;
        int ans = 0;
        while (i < n) {
            int s = 0;
            Set<Integer> seen = new HashSet<>();
            seen.add(0);
            while (i < n) {
                s += nums[i];
                if (seen.contains(s - target)) {
                    ++ans;
                    break;
                }
                ++i;
                seen.add(s);
            }
            ++i;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxNonOverlapping(vector<int>& nums, int target) {
        int i = 0, n = nums.size();
        int ans = 0;
        while (i < n) {
            int s = 0;
            unordered_set<int> seen;
            seen.insert(0);
            while (i < n) {
                s += nums[i];
                if (seen.count(s - target)) {
                    ++ans;
                    break;
                }
                ++i;
                seen.insert(s);
            }
            ++i;
        }
        return ans;
    }
};

Go

func maxNonOverlapping(nums []int, target int) int {
	i, n, ans := 0, len(nums), 0
	for i < n {
		s := 0
		seen := map[int]bool{0: true}
		for i < n {
			s += nums[i]
			if seen[s-target] {
				ans++
				break
			}
			seen[s] = true
			i++
		}
		i++
	}
	return ans
}

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