README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1500-1599/1546.Maximum%20Number%20of%20Non-Overlapping%20Subarrays%20With%20Sum%20Equals%20Target/README_EN.md	1855	Weekly Contest 201 Q3	Greedy Array Hash Table Prefix Sum

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

中文文档

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

Solution 1: Greedy + Prefix Sum + Hash Table

We traverse the array $nums$, using the method of prefix sum + hash table, to find subarrays with a sum of $target$. If found, we increment the answer by one, then we set the prefix sum to $0$ and continue to traverse the array $nums$ until the entire array is traversed.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Python3

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

Java

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

C++

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

Go

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

TypeScript

function maxNonOverlapping(nums: number[], target: number): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        let s = 0;
        const vis: Set<number> = new Set();
        vis.add(0);
        for (; i < n; ++i) {
            s += nums[i];
            if (vis.has(s - target)) {
                ++ans;
                break;
            }
            vis.add(s);
        }
    }
    return ans;
}