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560. Subarray Sum Equals K

DifficultyMedium

Description

Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [1,1,1], k = 2
Output: 2

Example 2:

Input: nums = [1,2,3], k = 3
Output: 2

 

Constraints:

  • 1 <= nums.length <= 2 * 104
  • -1000 <= nums[i] <= 1000
  • -107 <= k <= 107

Solutions

Solution 1: Hash Table + Prefix Sum

Thinking

Count subarrays whose sum is \(k\). Pairwise endpoints are \(O(n^2)\), tight for \(n \le 2 \times 10^4\).

A prefix \(s\) needs earlier prefixes equal to \(s-k\). A hash map stores prefix frequencies, seeded with \(\textit{cnt}[0]=1\) for the empty prefix. Query first, then insert, so the current index is not reused.

We define a hash table cnt to store the number of times the prefix sum of the array nums appears. Initially, we set the value of cnt[0] to 1, indicating that the prefix sum 0 appears once.

We traverse the array nums, calculate the prefix sum s, then add the value of cnt[s - k] to the answer, and increase the value of cnt[s] by 1.

After the traversal, we return the answer.

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

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class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        cnt = Counter({0: 1})
        ans = s = 0
        for x in nums:
            s += x
            ans += cnt[s - k]
            cnt[s] += 1
        return ans

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class Solution {
    public int subarraySum(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        cnt.put(0, 1);
        int ans = 0, s = 0;
        for (int x : nums) {
            s += x;
            ans += cnt.getOrDefault(s - k, 0);
            cnt.merge(s, 1, Integer::sum);
        }
        return ans;
    }
}

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class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> cnt{{0, 1}};
        int ans = 0, s = 0;
        for (int x : nums) {
            s += x;
            ans += cnt[s - k];
            ++cnt[s];
        }
        return ans;
    }
};

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func subarraySum(nums []int, k int) (ans int) {
    cnt := map[int]int{0: 1}
    s := 0
    for _, x := range nums {
        s += x
        ans += cnt[s-k]
        cnt[s]++
    }
    return
}

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function subarraySum(nums: number[], k: number): number {
    const cnt: Map<number, number> = new Map();
    cnt.set(0, 1);
    let [ans, s] = [0, 0];
    for (const x of nums) {
        s += x;
        ans += cnt.get(s - k) || 0;
        cnt.set(s, (cnt.get(s) || 0) + 1);
    }
    return ans;
}

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use std::collections::HashMap;

impl Solution {
    pub fn subarray_sum(nums: Vec<i32>, k: i32) -> i32 {
        let mut cnt = HashMap::new();
        cnt.insert(0, 1);
        let mut ans = 0;
        let mut s = 0;
        for &x in &nums {
            s += x;
            if let Some(&v) = cnt.get(&(s - k)) {
                ans += v;
            }
            *cnt.entry(s).or_insert(0) += 1;
        }
        ans
    }
}

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