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974. Subarray Sums Divisible by K

Description

Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k.

A subarray is a contiguous part of an array.

 

Example 1:

Input: nums = [4,5,0,-2,-3,1], k = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by k = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Example 2:

Input: nums = [5], k = 9
Output: 0

 

Constraints:

  • 1 <= nums.length <= 3 * 104
  • -104 <= nums[i] <= 104
  • 2 <= k <= 104

Solutions

Solution 1: Hash Table + Prefix Sum

  1. Key Insight:

    • If there exist indices $i$ and $j$ such that $i \leq j$, and the sum of the subarray $nums[i, ..., j]$ is divisible by $k$, then $(s_j - s_i) \bmod k = 0$, this implies: $s_j \bmod k = s_i \bmod k$
    • We can use a hash table to count the occurrences of prefix sums modulo $k$ to efficiently check for subarrays satisfying the condition.

  2. Prefix Sum Modulo:

    • Use a hash table $cnt$ to count occurrences of each prefix sum modulo $k$.
    • $cnt[i]$ represents the number of prefix sums with modulo $k$ equal to $i$.
    • Initialize $cnt[0] = 1$ to account for subarrays directly divisible by $k$.

  3. Algorithm:

    • Let a variable $s$ represent the running prefix sum, starting with $s = 0$.
    • Traverse the array $nums$ from left to right.
      • For each element $x$:
        • Compute $s = (s + x) \bmod k$.
        • Update the result: $ans += cnt[s]$.
        • Increment $cnt[s]$ by $1$.
    • Return the result $ans$.

Note: if $s$ is negative, adjust it to be non-negative by adding $k$ and taking modulo $k$ again.

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

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

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

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

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

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function subarraysDivByK(nums: number[], k: number): number {
    const counter = new Map();
    counter.set(0, 1);
    let s = 0,
        ans = 0;
    for (const num of nums) {
        s += num;
        const t = ((s % k) + k) % k;
        ans += counter.get(t) || 0;
        counter.set(t, (counter.get(t) || 0) + 1);
    }
    return ans;
}

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