Count of pairs in an array whose sum is a perfect square
Last Updated :
28 Mar, 2023
Given an array arr of distinct elements, the task is to find the total number of two element pairs from the array whose sum is a perfect square.
Examples:
Input: arr[] = {2, 3, 6, 9, 10, 20}
Output: 2 Only possible pairs are (3, 6) and (6, 10)
Input: arr[] = {9, 2, 5, 1}
Output: 0
Naive Approach: Use nested loops and check every possible pair for whether their sum is a perfect square or not. This technique is not effective when the length of the array is very large.
Efficient Approach:
- Store all the elements of the array in a HashSet named nums and save the sum of the maximum two elements in a variable named max.
- It is clear that the sum of any two elements from the array will not exceed max. So, find all the perfect squares which are ? max and save it in an ArrayList named perfectSquares.
- Now for every element in the array say arr[i] and for every perfect square saved in perfectSquares, check whether perfectSquares.get(i) - arr[i] exists in nums or not i.e. if there is any element in the original array that when added with the currently chosen element gives any perfect square from the list.
- If the above condition is satisfied, increment the count variable.
- Print the value of count in the end.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return an ArrayList containing
// all the perfect squares upto n
vector<int> getPerfectSquares(int n)
{
vector<int> perfectSquares;
int current = 1, i = 1;
// while current perfect square is less than or equal to n
while (current <= n)
{
perfectSquares.push_back(current);
current = static_cast<int>(pow(++i, 2));
}
return perfectSquares;
}
// Function to print the sum of maximum
// two elements from the array
int maxPairSum(vector<int> &arr)
{
int n = arr.size();
int max, secondMax;
if (arr[0] > arr[1])
{
max = arr[0];
secondMax = arr[1];
}
else
{
max = arr[1];
secondMax = arr[0];
}
for (int i = 2; i < n; i++)
{
if (arr[i] > max)
{
secondMax = max;
max = arr[i];
}
else if (arr[i] > secondMax)
{
secondMax = arr[i];
}
}
return (max + secondMax);
}
// Function to return the count of numbers that
// when added with n give a perfect square
int countPairsWith(int n, vector<int> &perfectSquares, unordered_set<int> &nums)
{
int count = 0;
for (int i = 0; i < perfectSquares.size(); i++)
{
int temp = perfectSquares[i] - n;
// temp > n is checked so that pairs
// (x, y) and (y, x) don't get counted twice
if (temp > n && find(nums.begin(), nums.end(), temp) != nums.end())
{
count++;
}
}
return count;
}
// Function to count the pairs whose sum is a perfect square
int countPairs(vector<int> &arr)
{
int i, n = arr.size();
// Sum of the maximum two elements from the array
int max = maxPairSum(arr);
// List of perfect squares upto max
vector<int> perfectSquares = getPerfectSquares(max);
// Contains all the array elements
unordered_set<int> nums;
for (i = 0; i < n; i++)
{
nums.insert(arr[i]);
}
int count = 0;
for (i = 0; i < n; i++)
{
// Add count of the elements that when
// added with arr[i] give a perfect square
count += countPairsWith(arr[i], perfectSquares, nums);
}
return count;
}
// Driver code
int main()
{
vector<int> arr = {2, 3, 6, 9, 10, 20};
cout << countPairs(arr) << endl;
return 0;
}
// This code is contributed by mits
// Java implementation of the approach
import java.util.*;
public class GFG {
// Function to return an ArrayList containing
// all the perfect squares upto n
public static ArrayList<Integer> getPerfectSquares(int n)
{
ArrayList<Integer> perfectSquares = new ArrayList<>();
int current = 1, i = 1;
// while current perfect square is less than or equal to n
while (current <= n) {
perfectSquares.add(current);
current = (int)Math.pow(++i, 2);
}
return perfectSquares;
}
// Function to print the sum of maximum
// two elements from the array
public static int maxPairSum(int arr[])
{
int n = arr.length;
int max, secondMax;
if (arr[0] > arr[1]) {
max = arr[0];
secondMax = arr[1];
}
else {
max = arr[1];
secondMax = arr[0];
}
for (int i = 2; i < n; i++) {
if (arr[i] > max) {
secondMax = max;
max = arr[i];
}
else if (arr[i] > secondMax) {
secondMax = arr[i];
}
}
return (max + secondMax);
}
// Function to return the count of numbers that
// when added with n give a perfect square
public static int countPairsWith(
int n, ArrayList<Integer> perfectSquares,
HashSet<Integer> nums)
{
int count = 0;
for (int i = 0; i < perfectSquares.size(); i++) {
int temp = perfectSquares.get(i) - n;
// temp > n is checked so that pairs
// (x, y) and (y, x) don't get counted twice
if (temp > n && nums.contains(temp))
count++;
}
return count;
}
// Function to count the pairs whose sum is a perfect square
public static int countPairs(int arr[])
{
int i, n = arr.length;
// Sum of the maximum two elements from the array
int max = maxPairSum(arr);
// List of perfect squares upto max
ArrayList<Integer> perfectSquares =
getPerfectSquares(max);
// Contains all the array elements
HashSet<Integer> nums = new HashSet<>();
for (i = 0; i < n; i++)
nums.add(arr[i]);
int count = 0;
for (i = 0; i < n; i++) {
// Add count of the elements that when
// added with arr[i] give a perfect square
count += countPairsWith(arr[i], perfectSquares, nums);
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 6, 9, 10, 20 };
System.out.println(countPairs(arr));
}
}
# Python3 implementation of the approach
# Function to return an ArrayList containing
# all the perfect squares upto n
def getPerfectSquares(n):
perfectSquares = [];
current = 1;
i = 1;
# while current perfect square is
# less than or equal to n
while (current <= n):
perfectSquares.append(current);
i += 1;
current = int(pow(i, 2));
return perfectSquares;
# Function to print the sum of maximum
# two elements from the array
def maxPairSum(arr):
n = len(arr);
max = 0;
secondMax = 0;
if (arr[0] > arr[1]):
max = arr[0];
secondMax = arr[1];
else:
max = arr[1];
secondMax = arr[0];
for i in range(2, n):
if (arr[i] > max):
secondMax = max;
max = arr[i];
elif (arr[i] > secondMax):
secondMax = arr[i];
return (max + secondMax);
# Function to return the count of numbers that
# when added with n give a perfect square
def countPairsWith(n, perfectSquares, nums):
count = 0;
for i in range(len(perfectSquares)):
temp = perfectSquares[i] - n;
# temp > n is checked so that pairs
# (x, y) and (y, x) don't get counted twice
if (temp > n and (temp in nums)):
count += 1;
return count;
# Function to count the pairs whose
# sum is a perfect square
def countPairs(arr):
n = len(arr);
# Sum of the maximum two elements
# from the array
max = maxPairSum(arr);
# List of perfect squares upto max
perfectSquares = getPerfectSquares(max);
# Contains all the array elements
nums = [];
for i in range(n):
nums.append(arr[i]);
count = 0;
for i in range(n):
# Add count of the elements that when
# added with arr[i] give a perfect square
count += countPairsWith(arr[i],
perfectSquares, nums);
return count;
# Driver code
arr = [ 2, 3, 6, 9, 10, 20 ];
print(countPairs(arr));
# This code is contributed by mits
// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
public class GFG {
// Function to return an ArrayList containing
// all the perfect squares upto n
public static ArrayList getPerfectSquares(int n)
{
ArrayList perfectSquares = new ArrayList();
int current = 1, i = 1;
// while current perfect square is less than or equal to n
while (current <= n) {
perfectSquares.Add(current);
current = (int)Math.Pow(++i, 2);
}
return perfectSquares;
}
// Function to print the sum of maximum
// two elements from the array
public static int maxPairSum(int[] arr)
{
int n = arr.Length;
int max, secondMax;
if (arr[0] > arr[1]) {
max = arr[0];
secondMax = arr[1];
}
else {
max = arr[1];
secondMax = arr[0];
}
for (int i = 2; i < n; i++) {
if (arr[i] > max) {
secondMax = max;
max = arr[i];
}
else if (arr[i] > secondMax) {
secondMax = arr[i];
}
}
return (max + secondMax);
}
// Function to return the count of numbers that
// when added with n give a perfect square
public static int countPairsWith(
int n, ArrayList perfectSquares, ArrayList nums)
{
int count = 0;
for (int i = 0; i < perfectSquares.Count; i++) {
int temp = (int)perfectSquares[i] - n;
// temp > n is checked so that pairs
// (x, y) and (y, x) don't get counted twice
if (temp > n && nums.Contains(temp))
count++;
}
return count;
}
// Function to count the pairs whose sum is a perfect square
public static int countPairs(int[] arr)
{
int i, n = arr.Length;
// Sum of the maximum two elements from the array
int max = maxPairSum(arr);
// List of perfect squares upto max
ArrayList perfectSquares = getPerfectSquares(max);
// Contains all the array elements
ArrayList nums = new ArrayList();
for (i = 0; i < n; i++)
nums.Add(arr[i]);
int count = 0;
for (i = 0; i < n; i++) {
// Add count of the elements that when
// added with arr[i] give a perfect square
count += countPairsWith(arr[i], perfectSquares, nums);
}
return count;
}
// Driver code
public static void Main()
{
int[] arr = { 2, 3, 6, 9, 10, 20 };
Console.WriteLine(countPairs(arr));
}
}
// This code is contributed by mits
<?php
// PHP implementation of the approach
// Function to return an ArrayList containing
// all the perfect squares upto n
function getPerfectSquares($n)
{
$perfectSquares = array();
$current = 1;
$i = 1;
// while current perfect square
// is less than or equal to n
while ($current <= $n)
{
array_push($perfectSquares, $current);
$current = (int)pow(++$i, 2);
}
return $perfectSquares;
}
// Function to print the sum of maximum
// two elements from the array
function maxPairSum($arr)
{
$n = count($arr);
$max;
$secondMax;
if ($arr[0] > $arr[1])
{
$max = $arr[0];
$secondMax = $arr[1];
}
else
{
$max = $arr[1];
$secondMax = $arr[0];
}
for ($i = 2; $i < $n; $i++)
{
if ($arr[$i] > $max)
{
$secondMax = $max;
$max = $arr[$i];
}
else if ($arr[$i] > $secondMax)
{
$secondMax = $arr[$i];
}
}
return ($max + $secondMax);
}
// Function to return the count of numbers that
// when added with n give a perfect square
function countPairsWith($n, $perfectSquares, $nums)
{
$count = 0;
for ($i = 0; $i < count($perfectSquares); $i++)
{
$temp = $perfectSquares[$i] - $n;
// temp > n is checked so that pairs
// (x, y) and (y, x) don't get counted twice
if ($temp > $n && in_array($temp, $nums))
$count++;
}
return $count;
}
// Function to count the pairs whose
// sum is a perfect square
function countPairs($arr)
{
$n = count($arr);
// Sum of the maximum two elements
// from the array
$max = maxPairSum($arr);
// List of perfect squares upto max
$perfectSquares = getPerfectSquares($max);
// Contains all the array elements
$nums = array();
for ($i = 0; $i < $n; $i++)
array_push($nums, $arr[$i]);
$count = 0;
for ($i = 0; $i < $n; $i++)
{
// Add count of the elements that when
// added with arr[i] give a perfect square
$count += countPairsWith($arr[$i],
$perfectSquares, $nums);
}
return $count;
}
// Driver code
$arr = array( 2, 3, 6, 9, 10, 20 );
echo countPairs($arr);
// This code is contributed by mits
?>
<script>
// Javascript implementation of the approach
// Function to return an ArrayList containing
// all the perfect squares upto n
function getPerfectSquares(n)
{
let perfectSquares = [];
let current = 1;
let i = 1;
// while current perfect square
// is less than or equal to n
while (current <= n)
{
perfectSquares.push(current);
current = Math.pow(++i, 2);
}
return perfectSquares;
}
// Function to print the sum of maximum
// two elements from the array
function maxPairSum(arr)
{
let n = arr.length
let max;
let secondMax;
if (arr[0] > arr[1])
{
max = arr[0];
secondMax = arr[1];
}
else
{
max = arr[1];
secondMax = arr[0];
}
for (let i = 2; i < n; i++)
{
if (arr[i] > max)
{
secondMax = max;
max = arr[i];
}
else if (arr[i] > secondMax)
{
secondMax = arr[$i];
}
}
return (max + secondMax);
}
// Function to return the count of numbers that
// when added with n give a perfect square
function countPairsWith(n, perfectSquares, nums)
{
let count = 0;
for (let i = 0; i < perfectSquares.length; i++)
{
let temp = perfectSquares[i] - n;
// temp > n is checked so that pairs
// (x, y) and (y, x) don't get counted twice
if (temp > n && nums.includes(temp))
count++;
}
return count;
}
// Function to count the pairs whose
// sum is a perfect square
function countPairs(arr)
{
let n = arr.length
// Sum of the maximum two elements
// from the array
let max = maxPairSum(arr);
// List of perfect squares upto max
let perfectSquares = getPerfectSquares(max);
// Contains all the array elements
let nums = [];
for (let i = 0; i < n; i++)
nums.push(arr[i]);
let count = 0;
for (let i = 0; i < n; i++)
{
// Add count of the elements that when
// added with arr[i] give a perfect square
count += countPairsWith(arr[i], perfectSquares, nums);
}
return count;
}
// Driver code
let arr = new Array( 2, 3, 6, 9, 10, 20 );
document.write(countPairs(arr));
// This code is contributed by Saurabh jaiswal
</script>
Time Complexity: O(n*sqrt(max))
Auxiliary Space: O(n + sqrt(max))
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