Count of pairs whose sum of pairwise product with X and Y is K

Given an array arr[] of size N and three integers X, Y and K, the task is to count the number of pairs (i, j) where i < j such that (arr[i] * X + arr[j] * Y) = K.

Examples:

Input: arr[] = {3, 1, 2, 3}, X = 4, Y = 2, K = 14
Output: 2
Explanation: The possible pairs are: (1, 2), (3, 4). 
For i = 1, j = 2, Value of the expression = 4 * 3 + 2 * 1 = 14.
For i = 3, j = 4, Value of the expression = 4 * 2 + 2 * 3 = 14.

Input: arr[] = [1, 3, 2], X = 1, Y = 3, K = 7
Output: 1
Explanation: The possible pairs are: (1, 2).
For i = 1, j = 2, Value of the expression = 1 * 1 + 2 * 3 = 7.

Input: N = 2, arr[] = [1, 2], X =1, Y = 1, target = 2
Output: 0
Explanation: No pair satisfy the above condition.
For i = 1, j = 2, Value of the expression = 1 * 1 + 1 * 2 = 3.

Naive Approach: The basic idea to solve the problem is as follows:

Find all possible pairs of (i, j) and for each pair check if (arr[i]*X + arr[j]*Y = K). If so then increment the count of pairs.

Follow the below steps to implement the idea:

• Initialize, an integer variable (say cnt as 0) to store the count of the pairs.
• Loop through the array from i = 0 to N – 2 :
  • Loop for all the second elements from j = i + 1 to N-1:
    • Check if the sum of arr[i] * X and arr[j] * Y is equal to K.
    • Then increment the value of cnt by 1.
• Return the value of cnt as the final answer.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the number of pairs
int countPairs(int arr[], int n, int x,
               int y, int k)
{
    // Store the count of the pairs.
    int cnt = 0;

    // Loop for the first element.
    for (int i = 0; i < n - 1; ++i) {

        // Loop for the second element.
        for (int j = i + 1; j < n; ++j) {

            // Calculate the sum
            int cur_sum = arr[i] * x
                          + arr[j] * y;

            // Increment 'cnt' if current
            // sum is equal to required target
            if (cur_sum == k) {
                cnt++;
            }
        }
    }

    // Return the count of pairs.
    return cnt;
}

// Driver Code
int main()
{
    int X = 4, Y = 2, K = 14;
    int arr[] = { 3, 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << countPairs(arr, N, X, Y, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

public class GFG
{

// Function to count the number of pairs
static int countPairs(int[] arr, int n, int x,
               int y, int k)
{
    // Store the count of the pairs.
    int cnt = 0;

    // Loop for the first element.
    for (int i = 0; i < n - 1; ++i) {

        // Loop for the second element.
        for (int j = i + 1; j < n; ++j) {

            // Calculate the sum
            int cur_sum = arr[i] * x
                          + arr[j] * y;

            // Increment 'cnt' if current
            // sum is equal to required target
            if (cur_sum == k) {
                cnt++;
            }
        }
    }

    // Return the count of pairs.
    return cnt;
}

// Driver Code
public static void main(String[] args)
{
    int X = 4, Y = 2, K = 14;
    int[] arr = { 3, 1, 2, 3 };
    int N = arr.length;

    // Function call
   System.out.println( countPairs(arr, N, X, Y, K));
}
}

// This code is contributed by shinjanpatra.

Python3

# Python code for the above approach

# Function to count the number of pairs
def countPairs(arr, n, x,
               y, k) :

    # Store the count of the pairs.
    cnt = 0

    # Loop for the first element.
    for i in range(n-1) :

        # Loop for the second element.
        for j in range(i+1, n) :

            # Calculate the sum
            cur_sum = arr[i] * x + arr[j] * y

            # Increment 'cnt' if current
            # sum is equal to required target
            if (cur_sum == k) :
                cnt += 1




    # Return the count of pairs.
    return cnt

# Driver Code
if __name__ == "__main__":

    X = 4
    Y = 2
    K = 14
    arr = [ 3, 1, 2, 3 ]
    N = len(arr)

    # Function call
    print(countPairs(arr, N, X, Y, K))

     # This code is contributed by code_hunt.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to count the number of pairs
static int countPairs(int[] arr, int n, int x,
               int y, int k)
{
    // Store the count of the pairs.
    int cnt = 0;

    // Loop for the first element.
    for (int i = 0; i < n - 1; ++i) {

        // Loop for the second element.
        for (int j = i + 1; j < n; ++j) {

            // Calculate the sum
            int cur_sum = arr[i] * x
                          + arr[j] * y;

            // Increment 'cnt' if current
            // sum is equal to required target
            if (cur_sum == k) {
                cnt++;
            }
        }
    }

    // Return the count of pairs.
    return cnt;
}

// Driver Code
public static void Main(String[] args)
{
    int X = 4, Y = 2, K = 14;
    int[] arr = { 3, 1, 2, 3 };
    int N = arr.Length;

    // Function call
    Console.WriteLine( countPairs(arr, N, X, Y, K));
}
}

// This code is contributed by avijitmondal1998.

Javascript

<script>
// JavaScript code to implement the approach

// Function to count the number of pairs
function countPairs(arr, n, x, y, k)
{

    // Store the count of the pairs.
    var cnt = 0;

    // Loop for the first element.
    for (var i = 0; i < n - 1; ++i) {

        // Loop for the second element.
        for (var j = i + 1; j < n; ++j) {

            // Calculate the sum
            var cur_sum = arr[i] * x + arr[j] * y;

            // Increment 'cnt' if current
            // sum is equal to required target
            if (cur_sum == k) {
                cnt++;
            }
        }
    }

    // Return the count of pairs.
    return cnt;
}

// Driver Code
var X = 4;
var Y = 2;
var K = 14;
var arr = [ 3, 1, 2, 3 ];
var N = arr.length;

// Function call
document.write(countPairs(arr, N, X, Y, K));

// this code is contributed by phasing17.
</script>

Output

2

Time Complexity: O(N²)
Space Complexity: O(1)

Efficient Approach: The problem can be solved efficiently based on the following mathematical observation:

arr[i]*X + arr[j]*Y = K
arr[i]*X = K – arr[j]*Y
So (K – arr[j]*Y) must be divisible by X and if arr[j] is found then the value of arr[i] must be (K – arr[j]*Y)/X.

So to solve the problem based on the above observation, consider each value as arr[j] and try to find the presence of arr[i] using the above relation where i is less than j.

The above observation can be implemented using frequency counting. Follow the below steps to solve this problem :

• Initialize a variable (say cnt = 0) to store the count of pairs.
• Create a frequency array freq[] to store the frequency of the array elements.
• Loop through all the elements from j = 0 to N-1:
  • Find the value of required arr[i] as shown in the above observation.
  • If that element is present then increase the cnt by the frequency of that element.
  • Increase the frequency of arr[j].
• Return the value of cnt as the final answer.

Below is the implementation of the above approach :

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the count of pairs
int countPairs(int arr[], int n, int x,
               int y, int k)
{
    // Variable to store count of the pairs
    int cnt = 0;

    // Map for storing the frequency
    unordered_map<int, int> freq;

    // Loop for finding the count of pairs
    for (int i = 0; i < n; ++i) {

        // RHS Value of the equation given
        // in approach.
        int rhs = k - arr[i] * y;

        // If RHS value is not divisible by
        // x or is negative, there is no pair
        // possible for 'arr[i]' as second
        // element.
        if (rhs % x || rhs <= 0) {

            // Update the frequency of arr[i]
            freq[arr[i]]++;
            continue;
        }
        rhs /= x;

        // Adding frequency of rhs in array
        cnt += freq[rhs];

        // Update the frequency of arr[i]
        freq[arr[i]]++;
    }

    // Return the count of pairs
    return cnt;
}

// Driver Code
int main()
{
    int X = 4, Y = 2, K = 14;
    int arr[] = { 3, 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << countPairs(arr, N, X, Y, K);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

public class GFG
{

// Function to find the count of pairs
static int countPairs(int[] arr, int n, int x,
               int y, int k)
{
    // Variable to store count of the pairs
    int cnt = 0;

    // Map for storing the frequency
    Map<Integer,Integer> freq = new HashMap<Integer,Integer>();

    for (int i = 0; i < n; i++) {

        if(freq.containsKey(arr[i])){
            freq.put(arr[i], freq.get(arr[i]) + 0);
        }else{
            freq.put(arr[i], 0);
        }
    }

    // Loop for finding the count of pairs
    for (int i = 0; i < n; ++i) {

        // RHS Value of the equation given
        // in approach.
        int rhs = k - arr[i] * y;

        // If RHS value is not divisible by
        // x or is negative, there is no pair
        // possible for 'arr[i]' as second
        // element.
        if ((rhs % x) != 0 || rhs <= 0) {

            // Update the frequency of arr[i]
            freq.put(arr[i], freq.get(arr[i])+ 1);
            continue;
        }
        rhs /= x;

        // Adding frequency of rhs in array
        cnt += freq.get(rhs);

        // Update the frequency of arr[i]
        freq.put(arr[i], freq.get(arr[i])+ 1);

    }

    // Return the count of pairs
    return cnt;
}

// Driver Code
public static void main(String []args)
{
    int X = 4, Y = 2, K = 14;
    int[] arr = { 3, 1, 2, 3 };
    int N = arr.length;

    // Function call
    System.out.print(countPairs(arr, N, X, Y, K));
}
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to find the count of pairs
def countPairs(arr, n, x, y, k):

    # Variable to store count of the pairs
    cnt = 0

    # Dictionary for storing the frequency
    freq = dict()

    # Loop for finding the count of pairs
    for i in range(n):

        # RHS Value of the equation given
        # in approach.
        rhs = k - arr[i] * y;

        # If RHS value is not divisible by
        # x or is negative, there is no pair
        # possible for 'arr[i]' as second
        # element.
        if (rhs % x or rhs <= 0) :

            # Update the frequency of arr[i]
            if arr[i] in freq:
                freq[arr[i]] += 1
            else:
                freq[arr[i]] = 1
            continue

        rhs = rhs // x

        # Adding frequency of rhs in array
        if rhs in freq:
            cnt += freq[rhs]

        # Update the frequency of arr[i]
        if arr[i] in freq:
            freq[arr[i]] += 1
        else:
            freq[arr[i]] = 1

    # Return the count of pairs
    return cnt;

# Driver Code
X = 4
Y = 2
K = 14;
arr = [3, 1, 2, 3]
N = len(arr)

# Function Call
print(countPairs(arr, N, X, Y, K))

# This code is contributed by phasing17

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to find the count of pairs
static int countPairs(int[] arr, int n, int x,
               int y, int k)
{
    // Variable to store count of the pairs
    int cnt = 0;

    // Map for storing the frequency
    Dictionary<int,int> freq = new Dictionary<int,int>();

    for (int i = 0; i < n; i++) {

        if(freq.ContainsKey(arr[i])){
            freq[arr[i]] = freq[arr[i]] + 0;
        }else{
            freq.Add(arr[i], 0);
        }
    }

    // Loop for finding the count of pairs
    for (int i = 0; i < n; ++i) {

        // RHS Value of the equation given
        // in approach.
        int rhs = k - arr[i] * y;

        // If RHS value is not divisible by
        // x or is negative, there is no pair
        // possible for 'arr[i]' as second
        // element.
        if ((rhs % x) != 0 || rhs <= 0) {

            // Update the frequency of arr[i]
             // Update the frequency of arr[i]
            freq[arr[i]]++;
            continue;
        }
        rhs /= x;

        // Adding frequency of rhs in array
        cnt += freq[rhs];

        // Update the frequency of arr[i]
        freq[arr[i]]++;

    }

    // Return the count of pairs
    return cnt;
}

// Driver Code
public static void Main()
{
    int X = 4, Y = 2, K = 14;
    int[] arr = { 3, 1, 2, 3 };
    int N = arr.Length;

    // Function call
    Console.Write(countPairs(arr, N, X, Y, K));
}
}

Javascript

<script>
       // JavaScript program for the above approach

       // Function to find the count of pairs
       function countPairs(arr, n, x, y, k)
       {

           // Variable to store count of the pairs
           let cnt = 0;

           // Map for storing the frequency
           let freq = new Map();

           // Loop for finding the count of pairs
           for (let i = 0; i < n; ++i) {

               // RHS Value of the equation given
               // in approach.
               let rhs = k - arr[i] * y;

               // If RHS value is not divisible by
               // x or is negative, there is no pair
               // possible for 'arr[i]' as second
               // element.
               if (rhs % x || rhs <= 0) {

                   // Update the frequency of arr[i]
                   if (freq.has(arr[i])) {
                       freq.set(arr[i], freq.get(arr[i]) + 1)
                   }
                   else {
                       freq.set(arr[i], 1)
                   }
                   continue;
               }
               rhs = Math.floor(rhs / x);

               // Adding frequency of rhs in array
               if (freq.has(arr[i]))
                   cnt += freq.get(arr[i]) + 1;

               // Update the frequency of arr[i]
               if (freq.has(arr[i]))
                   freq.set(arr[i], freq.get(arr[i]) + 1)
               else
                   freq.set(arr[i], 1)
           }

           // Return the count of pairs
           return cnt;
       }

       // Driver Code
       let X = 4, Y = 2, K = 14;
       let arr = [3, 1, 2, 3];
       let N = arr.length;

       // Function call
       document.write(countPairs(arr, N, X, Y, K));

   // This code is contributed by Potta Lokesh
   </script>

Output

2

Time Complexity: O(N)
Auxiliary Space: O(N)