7. Continuous Subarray Sum

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

Solution: (Prefix Sum of Remainder)

class Solution
{
public:
    bool checkSubarraySum(vector<int> &nums, int k)
    {

        unordered_map<int, int> m;
        int ps = 0;
        m.insert({0, -1});

        for (int i = 0; i < nums.size(); i++)
        {
            ps += nums[i];
            int rem = ps % k;

            if (m.find(rem) != m.end())
            {
                int dif = i - m[rem];
                if (dif >= 2)
                {
                    return true;
                }
            }
            m.insert({rem, i});
        }

        return false;
    }
};

Time Complexity: O(n)