1590. Make Sum Divisible by P

typedef long long ll;
class Solution {
public:
    int minSubarray(vector<int>& nums, int p) {
        /*
          3 1 4 2 totalsum=10
          remove 3 or 31 or 314 
          remove 1 or 14 or 142
          remove 4 or 42
          remove 2
             0 3 1 4 2
          ps[0 3 4 8 10]
         brute force O(N*2);
         how to do it nlogn or n 
         if on subtracting from sum a number which has same remainder when%p
         thats the ans
         now question becomes finding subarrya with sum as sum%p
        */
        int n=nums.size();
        ll totalsum=0;
        for(auto& i:nums)totalsum+=i;
        if(totalsum%p==0)return 0;
        ll rem=totalsum%p;
        unordered_map<ll,int>mp;
        ll sum=0;int ans=n;
        mp[0]=-1;
        for(int i=0;i<n;i++){
            sum = (sum+nums[i])%p;
            ll remsum=(sum-rem+p)%p;
            if(mp.find(remsum)!=mp.end())ans=min(ans,i-mp[remsum]);
            mp[sum]=i;
        }
        if(ans==n)return -1;
        return ans;
    }
};