Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

Constraints:

1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9
0 <= sum(nums[i]) <= 2^31 - 1
1 <= k <= 2^31 - 1

Approach 1: Hash + Prefix Sum

A continuous subarray sum can be represented as $sum_j - sum_i$ where $sum_j$ is the prefix sum till index $j$ and $sum_i$ is the prefix sum till index $i$ . We are looking for the subarray sum which can be divisible by $k$ . That means $(sum_j - sum_i) \% k == 0$ . We can further rewrite as $sum_j \% k - sum_i \% k== 0$ and got $sum_j \% k == sum_i \% k$ .

Therefore, we can calculate the prefix sum and store its remainder to a hash map. If we see the same remainder in the hash map, we need to make sure that the length is at least $2$ . If so, we can return true.

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        if (n == 1) return false;
        // hash map to store the remainders
        unordered_map<int, int> mod {{0, -1}};
        int sum = 0, remainder;
        for (int i = 0; i < n; i++) {
            // prefix sum
            sum += nums[i];
            // calculate the remainder
            remainder = sum % k;
            if (mod.count(remainder)) {
                // if remainder exists in hash map
                // check the length
                if (i - mod[remainder] >= 2) return true;
            } else {
                // mark the current index to hash map
                mod[remainder] = i;
            }
        }
        return false;
    }
};

0523 - Continuous Subarray Sum (Medium)

Problem Link

Problem Statement

Approach 1: Hash + Prefix Sum

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