16 months ago

PRIYANSHU GUPTA • 10

Mumbai

solution for problem 1 :

   let the given vector and x be:

    vector<int> v={3,7,27,23};

    int x=10;

    map<int,int> m;

    int cnt=0;

    int ans=0;

    for(int i=0;i<v.size();i++){

        if(v[i]%x==0){

            cnt++;

        }

        else{

            int rem=v[i]%x;

            int check=x-rem;

            if(m[check]){

                ans+=m[check];

            }

        }

        m[v[i]%x]++;

    }

    int val=((cnt*(cnt-1))/2);

    cout<<val+ans<<endl;

12 months ago

suryansh jaiswal • 370

Lucknow

Question 1 solution

// C++ Program to count pairs
// whose sum divisible by 'K'
#include <bits/stdc++.h>
using namespace std;

// Program to count pairs whose sum divisible
// by 'K'
int countKdivPairs(int A[], int n, int K)
{
	// Create a frequency array to count
	// occurrences of all remainders when
	// divided by K
	int freq[K] = { 0 };

	// Count occurrences of all remainders
	for (int i = 0; i < n; i++)
		++freq[A[i] % K];

	// If both pairs are divisible by 'K'
	int sum = freq[0] * (freq[0] - 1) / 2;

	// count for all i and (k-i)
	// freq pairs
	for (int i = 1; i <= K / 2 && i != (K - i); i++)
		sum += freq[i] * freq[K - i];
	// If K is even
	if (K % 2 == 0)
		sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
	return sum;
}

// Driver code
int main()
{

	int A[] = { 2, 2, 1, 7, 5, 3 };
	int n = sizeof(A) / sizeof(A[0]);
	int K = 4;
	cout << countKdivPairs(A, n, K);

	return 0;
}

13 months ago

rahul • 0

Prayagraj

Q2->

int avg(vector<int> &arr,int k){
    int n=arr.size();
   int result = 0, curSum = 0;
    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++) {
        curSum += (arr[i] - k);
        if (curSum == 0)result++;
          
        if (mp.find(curSum) != mp.end())result += mp[curSum];
        mp[curSum]++;
    }
    return result;
}
vector<int> solve(vector<int> &arr){
    vector<int> res;
    for(int i=1;i<=arr.size();i++){
        res.push_back(avg(arr,i));
    }
    return res;
}