Count of pairs of elements with Sum X and multiplication Y in given Array
Last Updated :
23 Jul, 2025
Given array A[] (-109 <= A[i] <= 109) of size N (2 <= N <= 107) along with integers X (-2*109 <= X <= 2*109) and Y (-1018 <= Y <= 1018), the task for this problem is to find the count of pairs (A[i], A[j]) such that X = A[i] + A[j] and Y = A[i] * A[j] also i < j. Also A[i], X, Y != 0
Examples:
Input: A[] = {1, 4, -2, 3, 3, 3}, X = 7, Y = 12
Output: 3
Explanation: pairs (A[2], A[4]) = (4, 3), (A[2], A[5]) = (4, 3), and (A[2], A[6]) = (4, 3) are the pairs which satisfy above conditions.
Input: A[] = {1, 3, 2}, X = 5, Y = 6
Output: 1
Explanation: There is only one pair (A[2], A[3]) = (3, 2) which satisfy above conditions (X = A[2] + A[3] = 5 and Y = A[2] * A[3] = 6).
Naïve Approach: The basic way to solve the problem is as follows:
Simply iterating with two loops and maintaining a counter of pairs which satisfy above conditions.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To solve the problem follow the below idea:
If X = A[i] + A[j] and Y = A[i] * A[j] then a2 - Xa + Y = 0 which is Quadratic equation in terms of sum and product of roots.
Here roots are A[i] and A[j]. so we just have to find the count of roots of quadratic equation in array A[] their multiplication will be answer. If the roots are integers then only its possible to find them in array otherwise answer will be zero. Here root will be (X + sqrt(Discriminant)) / 2 and (X - sqrt(Discriminant)) / 2 where Discriminant = X * X - 4 * Y.
Below are the steps for the above approach:
- Creating Variable for discriminant and calculate it.
- Check condition whether roots are integers or not. if not integer return 0.
- Find both the roots by above formula.
- If Discriminant is zero then there will be only one unique root find its count as cntRoot in array A[] and return answer: cntRoot * (cntRoot - 1) / 2.
- If Discriminant is not zero then find count of both the roots as firstRootCnt and secondRootCnt then return answer as : firstRootCnt * secondRootCnt.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function will tell whether given
// number is perfect square or not
bool isPerfectSquare(int number)
{
return sqrt(number) * sqrt(number) == number;
}
// Function to find count of pairs of integers
// with Sum X and multiplication Y
int findPairCount(int A[], int N, int X, int Y)
{
// Creating Variable for discriminant
int discriminant = X * X - 4 * Y;
// Check if both the roots are
// integers or not
if (isPerfectSquare(discriminant)
and (X + (int)sqrt(discriminant)) % 2 == 0
and (X - (int)sqrt(discriminant)) % 2 == 0) {
// First and second root
int firstRoot = (X + sqrt(discriminant)) / 2;
int secondRoot = (X - sqrt(discriminant)) / 2;
// If both roots are same that is
// discriminant is zero
if (discriminant == 0) {
// Variable that track count
// of cntRoot
int cntRoot = 0;
// Find count of given root
for (int i = 0; i < N; i++) {
if (A[i] == firstRoot)
cntRoot++;
}
// Answer in this case will be number
// of ways of choosing 2 elements
// from cntRoot
return cntRoot * (cntRoot - 1) / 2;
}
else {
// Find count of given roots
int firstRootCnt = 0, secondRootCnt = 0;
// Find count of given roots
for (int i = 0; i < N; i++) {
if (firstRoot == A[i])
firstRootCnt++;
else if (secondRoot == A[i])
secondRootCnt++;
}
// answer will be number of ways of
// choosing one element rom
// firstRootCnt and one element
// from second RootCnt;
return firstRootCnt * secondRootCnt;
}
}
// If roots are not interger then return 0
else {
return 0;
}
}
// Driver Code
int32_t main()
{
// Input 1
int X = 7, Y = 12;
int A[] = { 1, 4, -2, 3, 3, 3 }, N = 6;
// Function Call
cout << findPairCount(A, N, X, Y) << endl;
// Input 2
int X1 = 5, Y1 = 6;
int A1[] = { 1, 3, 2 }, N1 = 3;
// Function Call
cout << findPairCount(A1, N1, X1, Y1) << endl;
return 0;
}
import java.util.Arrays;
public class Solution {
// Function will tell whether given number is a perfect square or not
static boolean isPerfectSquare(int number) {
int sqrt = (int) Math.sqrt(number);
return sqrt * sqrt == number;
}
// Function to find the count of pairs of integers with Sum X and multiplication Y
static int findPairCount(int[] A, int N, int X, int Y) {
// Creating Variable for discriminant
int discriminant = X * X - 4 * Y;
// Check if both the roots are integers or not
if (isPerfectSquare(discriminant)
&& (X + (int) Math.sqrt(discriminant)) % 2 == 0
&& (X - (int) Math.sqrt(discriminant)) % 2 == 0) {
// First and second root
int firstRoot = (X + (int) Math.sqrt(discriminant)) / 2;
int secondRoot = (X - (int) Math.sqrt(discriminant)) / 2;
// If both roots are the same, that is discriminant is zero
if (discriminant == 0) {
// Variable that tracks the count of cntRoot
int cntRoot = 0;
// Find count of given root
for (int i = 0; i < N; i++) {
if (A[i] == firstRoot)
cntRoot++;
}
// Answer in this case will be the number of ways of choosing 2 elements from cntRoot
return cntRoot * (cntRoot - 1) / 2;
} else {
// Find count of given roots
int firstRootCnt = 0, secondRootCnt = 0;
// Find count of given roots
for (int i = 0; i < N; i++) {
if (firstRoot == A[i])
firstRootCnt++;
else if (secondRoot == A[i])
secondRootCnt++;
}
// Answer will be the number of ways of choosing one element from firstRootCnt and one element from secondRootCnt;
return firstRootCnt * secondRootCnt;
}
}
// If roots are not integers, then return 0
else {
return 0;
}
}
// Driver Code
public static void main(String[] args) {
// Input 1
int X = 7, Y = 12;
int[] A = {1, 4, -2, 3, 3, 3};
int N = 6;
// Function Call
System.out.println(findPairCount(A, N, X, Y));
// Input 2
int X1 = 5, Y1 = 6;
int[] A1 = {1, 3, 2};
int N1 = 3;
// Function Call
System.out.println(findPairCount(A1, N1, X1, Y1));
}
}
# Function to check if a number is a perfect square
def isPerfectSquare(number):
root = int(number ** 0.5)
return root * root == number
# Function to find the count of pairs of integers
# with Sum X and multiplication Y
def findPairCount(A, N, X, Y):
# Creating a variable for the discriminant
discriminant = X * X - 4 * Y
# Check if both the roots are integers or not
if isPerfectSquare(discriminant):
root = int(discriminant ** 0.5)
if (X + root) % 2 == 0 and (X - root) % 2 == 0:
# First and second root
firstRoot = (X + root) // 2
secondRoot = (X - root) // 2
# If both roots are the same (discriminant is zero)
if discriminant == 0:
# Variable that tracks the count of cntRoot
cntRoot = 0
# Find the count of the given root
for i in range(N):
if A[i] == firstRoot:
cntRoot += 1
# Answer in this case will be the number
# of ways of choosing 2 elements from cntRoot
return cntRoot * (cntRoot - 1) // 2
else:
# Find the count of given roots
firstRootCnt = 0
secondRootCnt = 0
# Find count of given roots
for i in range(N):
if firstRoot == A[i]:
firstRootCnt += 1
elif secondRoot == A[i]:
secondRootCnt += 1
# Answer will be the number of ways of
# choosing one element from
# firstRootCnt and one element
# from secondRootCnt;
return firstRootCnt * secondRootCnt
# If roots are not integers, then return 0
return 0
# Driver Code
def main():
# Input 1
X = 7
Y = 12
A = [1, 4, -2, 3, 3, 3]
N = 6
# Function Call
print(findPairCount(A, N, X, Y))
# Input 2
X1 = 5
Y1 = 6
A1 = [1, 3, 2]
N1 = 3
# Function Call
print(findPairCount(A1, N1, X1, Y1))
if __name__ == "__main__":
main()
using System;
public class GFG
{
// Function will tell whether given number is a perfect square or not
static bool IsPerfectSquare(int number)
{
int sqrt = (int)Math.Sqrt(number);
return sqrt * sqrt == number;
}
// Function to find the count of pairs of integers with Sum X and multiplication Y
static int FindPairCount(int[] A, int N, int X, int Y)
{
// Creating Variable for discriminant
int discriminant = X * X - 4 * Y;
// Check if both the roots are integers or not
if (IsPerfectSquare(discriminant)
&& (X + (int)Math.Sqrt(discriminant)) % 2 == 0
&& (X - (int)Math.Sqrt(discriminant)) % 2 == 0)
{
// First and second root
int firstRoot = (X + (int)Math.Sqrt(discriminant)) / 2;
int secondRoot = (X - (int)Math.Sqrt(discriminant)) / 2;
// If both roots are the same, that is discriminant is zero
if (discriminant == 0)
{
// Variable that tracks the count of cntRoot
int cntRoot = 0;
// Find count of given root
for (int i = 0; i < N; i++)
{
if (A[i] == firstRoot)
cntRoot++;
}
// Answer in this case will be the number of ways of choosing 2 elements from cntRoot
return cntRoot * (cntRoot - 1) / 2;
}
else
{
// Find count of given roots
int firstRootCnt = 0, secondRootCnt = 0;
// Find count of given roots
for (int i = 0; i < N; i++)
{
if (firstRoot == A[i])
firstRootCnt++;
else if (secondRoot == A[i])
secondRootCnt++;
}
// Answer will be the number of ways of choosing one element from firstRootCnt and one element from secondRootCnt;
return firstRootCnt * secondRootCnt;
}
}
// If roots are not integers, then return 0
else
{
return 0;
}
}
// Driver Code
public static void Main(string[] args)
{
// Input 1
int X = 7, Y = 12;
int[] A = { 1, 4, -2, 3, 3, 3 };
int N = 6;
// Function Call
Console.WriteLine(FindPairCount(A, N, X, Y));
// Input 2
int X1 = 5, Y1 = 6;
int[] A1 = { 1, 3, 2 };
int N1 = 3;
// Function Call
Console.WriteLine(FindPairCount(A1, N1, X1, Y1));
}
}
function GFG(number) {
const sqrt = Math.sqrt(number);
return Math.floor(sqrt) * Math.floor(sqrt) === number;
}
// Function to find the count of pairs of integers with
// Sum X and multiplication Y
function Count(A, N, X, Y) {
// Creating a variable for the discriminant
const discriminant = X * X - 4 * Y;
// Check if both the roots are integers or not
if (GFG(discriminant)
&& (X + Math.sqrt(discriminant)) % 2 === 0
&& (X - Math.sqrt(discriminant)) % 2 === 0) {
// First and second root
const firstRoot = (X + Math.sqrt(discriminant)) / 2;
const secondRoot = (X - Math.sqrt(discriminant)) / 2;
// If both roots are the same
// that is discriminant is zero
if (discriminant === 0) {
// Variable that tracks the count of cntRoot
let cntRoot = 0;
// Find count of given root
for (let i = 0; i < N; i++) {
if (A[i] === firstRoot)
cntRoot++;
}
// Answer in this case will be the number of ways of
// choosing 2 elements from cntRoot
return (cntRoot * (cntRoot - 1)) / 2;
} else {
// Find count of given roots
let firstRootCnt = 0;
let secondRootCnt = 0;
// Find count of given roots
for (let i = 0; i < N; i++) {
if (firstRoot === A[i])
firstRootCnt++;
else if (secondRoot === A[i])
secondRootCnt++;
}
// Answer will be the number of ways of choosing one element from
// firstRootCnt and one element from secondRootCnt;
return firstRootCnt * secondRootCnt;
}
}
// If roots are not integers, then return 0
else {
return 0;
}
}
// Driver Code
// Input 1
const X = 7;
const Y = 12;
const A = [1, 4, -2, 3, 3, 3];
const N = 6;
// Function Call
console.log(Count(A, N, X, Y));
// Input 2
const X1 = 5;
const Y1 = 6;
const A1 = [1, 3, 2];
const N1 = 3;
// Function Call
console.log(Count(A1, N1, X1, Y1));
Time Complexity: O(N)
Auxiliary Space: O(1)
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