Count of pairs in an array whose sum is a perfect square

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++

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// 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

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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     
?>

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Output:

2


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Second year Department of Information Technology Jadavpur University

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