325. Maximum Size Subarray Sum Equals k

Description

Given an integer array nums and an integer k, return the maximum length of a subarray that sums to k. If there is not one, return 0 instead.

 

Example 1:

Input: nums = [1,-1,5,-2,3], k = 3
Output: 4
Explanation: The subarray [1, -1, 5, -2] sums to 3 and is the longest.

Example 2:

Input: nums = [-2,-1,2,1], k = 1
Output: 2
Explanation: The subarray [-1, 2] sums to 1 and is the longest.

 

Constraints:

  • 1 <= nums.length <= 2 * 105
  • -104 <= nums[i] <= 104
  • -109 <= k <= 109

Solutions

Solution 1

Python Code
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class Solution:
    def maxSubArrayLen(self, nums: List[int], k: int) -> int:
        d = {0: -1}
        ans = s = 0
        for i, x in enumerate(nums):
            s += x
            if s - k in d:
                ans = max(ans, i - d[s - k])
            if s not in d:
                d[s] = i
        return ans

Java Code
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class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        Map<Long, Integer> d = new HashMap<>();
        d.put(0L, -1);
        int ans = 0;
        long s = 0;
        for (int i = 0; i < nums.length; ++i) {
            s += nums[i];
            ans = Math.max(ans, i - d.getOrDefault(s - k, i));
            d.putIfAbsent(s, i);
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int maxSubArrayLen(vector<int>& nums, int k) {
        unordered_map<long long, int> d{{0, -1}};
        int ans = 0;
        long long s = 0;
        for (int i = 0; i < nums.size(); ++i) {
            s += nums[i];
            if (d.count(s - k)) {
                ans = max(ans, i - d[s - k]);
            }
            if (!d.count(s)) {
                d[s] = i;
            }
        }
        return ans;
    }
};

Go Code
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func maxSubArrayLen(nums []int, k int) (ans int) {
	d := map[int]int{0: -1}
	s := 0
	for i, x := range nums {
		s += x
		if j, ok := d[s-k]; ok && ans < i-j {
			ans = i - j
		}
		if _, ok := d[s]; !ok {
			d[s] = i
		}
	}
	return
}

TypeScript Code
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function maxSubArrayLen(nums: number[], k: number): number {
    const d: Map<number, number> = new Map();
    d.set(0, -1);
    let ans = 0;
    let s = 0;
    for (let i = 0; i < nums.length; ++i) {
        s += nums[i];
        if (d.has(s - k)) {
            ans = Math.max(ans, i - d.get(s - k)!);
        }
        if (!d.has(s)) {
            d.set(s, i);
        }
    }
    return ans;
}