LC - 1590. Make Sum Divisible by P

class Solution {
public:
    int minSubarray(vector<int>& nums, int p) 
    {
        long sum = 0;
        sum = accumulate(nums.begin(), nums.end(), sum);

        // The valid subarrays have the sum of the form
        // sumToFind, sumToFind + p, sumToFind + 2p, ...
        int sumToFind = sum % p;
        if(sumToFind < 0) sumToFind += p;
        if(sumToFind == 0) return 0; // the whole sum is already divisible by P

        // Keep track of the latest (remainder % p)  of the prefix sums
        unordered_map<int, int> lastRem; // <rem, lastIndex>
        int prefixSum = 0;
        lastRem[0] = -1;
        int result = nums.size();
        for(int i = 0; i < nums.size(); ++i) 
        {
            prefixSum = (prefixSum + nums[i])%p;
            if(prefixSum < 0) prefixSum += p;                     

            // when we have prefixSum%p, we need to
            // find the latest occurrence of prefix sum mod p 
            // equals to (prefixSum - sumToFind) % p.
            int counterpart = (prefixSum - sumToFind + p)%p;
            if(lastRem.count(counterpart)) 
            {
                result = min(result, i - lastRem[counterpart]);
            }
            lastRem[prefixSum] = i;
        }
        return (result == nums.size()) ? -1 : result;
        
    }
};