Sum of elements whose square root is present in the array

02 Jun, 2021
Given an array arr[], the task is to find the sum of all those elements from the given array whose square root is present in the same array.
Examples: 
Input: arr[] = {1, 2, 3, 4, 6, 9, 10} Output: 13 4 and 9 are the only numbers whose square roots 2 and 3 are present in the array
Input: arr[] = {4, 2, 36, 6, 10, 100} Output: 140  
Naive Approach: To find the sum of elements whose square root is present in the given array, check for square root of every element by iterating from arr[0] to arr[n] which will do the job but in O(n*n) complexity.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
PHP
Javascript
// CPP program to find the sum of all the elements// from the array whose square root is present// in the same array #include<bits/stdc++.h>using namespace std;     // Function to return the required sumint getSum(int arr[], int n)    {         int sum = 0;         for (int i = 0; i < n; i++) {            double sqrtCurrent = sqrt(arr[i]);             for (int j = 0; j < n; j++) {                double x = arr[j];                 // If sqrtCurrent is present in array                if (x == sqrtCurrent) {                    sum += (sqrtCurrent * sqrtCurrent);                    break;                }            }        }         return sum;    }     // Driver code    int main()    {        int arr[] = { 2, 4, 5, 6, 7, 8, 9, 3 };        int n = sizeof(arr)/sizeof(arr[0]);        cout<<(getSum(arr, n));    }// This code is contributed by// Surendra_Gangwar
// Java program to find the sum of all the elements// from the array whose square root is present// in the same arraypublic class GFG {    // Function to return the required sum    public static int getSum(int arr[], int n)    {         int sum = 0;         for (int i = 0; i < n; i++) {            double sqrtCurrent = Math.sqrt(arr[i]);             for (int j = 0; j < n; j++) {                double x = arr[j];                 // If sqrtCurrent is present in array                if (x == sqrtCurrent) {                    sum += (sqrtCurrent * sqrtCurrent);                    break;                }            }        }         return sum;    }     // Driver code    public static void main(String args[])    {        int arr[] = { 2, 4, 5, 6, 7, 8, 9, 3 };        int n = arr.length;        System.out.println(getSum(arr, n));    }}
# Python3 program to find the sum of# all the elements from the array# whose square root is present in# the same arrayimport math # Function to return the required sumdef getSum(arr, n):    sum = 0    for i in range(0, n):        sqrtCurrent = math.sqrt(arr[i])        for j in range(0, n):            x = arr[j]                         # If sqrtCurrent is present in array            if (x == sqrtCurrent):                sum += (sqrtCurrent *                        sqrtCurrent)                break    return int(sum) # Driver codeif __name__ == '__main__':     arr = [ 2, 4, 5, 6, 7, 8, 9, 3]    n = len(arr)    print(getSum(arr, n))     # This code is contributed# by 29AjayKumar
// C# program to find the sum of all the elements// from the array whose square root is present// in the same arrayusing System ; public class GFG {         // Function to return the required sum    public static float getSum(int []arr, int n)    {         float sum = 0;         for (int i = 0; i < n; i++) {            float sqrtCurrent = (float)Math.Sqrt(arr[i]);             for (int j = 0; j < n; j++) {                float x = (float)arr[j];                 // If sqrtCurrent is present in array                if (x == sqrtCurrent) {                    sum += (sqrtCurrent * sqrtCurrent);                    break;                }            }        }         return sum;    }     // Driver code    public static void Main()    {        int []arr = { 2, 4, 5, 6, 7, 8, 9, 3 };        int n = arr.Length;        Console.WriteLine(getSum(arr, n));    }    // This code is contributed by Ryuga}
<?php// PHP program to find the sum of all// the elements from the array whose// square root is present in the same array // Function to return the required sumfunction getSum(&$arr, $n){    $sum = 0;     for ($i = 0; $i < $n; $i++)    {        $sqrtCurrent = sqrt($arr[$i]);         for ($j = 0; $j < $n; $j++)        {            $x = $arr[$j];             // If sqrtCurrent is present in array            if ($x == $sqrtCurrent)            {                $sum += ($sqrtCurrent * $sqrtCurrent);                break;            }        }    }     return $sum;} // Driver code$arr = array(2, 4, 5, 6, 7, 8, 9, 3);$n = sizeof($arr);echo (getSum($arr, $n));     // This code is contributed// by Shivi_Aggarwal?>
<script>    // Javascript program to find the sum of all the elements    // from the array whose square root is present    // in the same array         // Function to return the required sum    function getSum(arr, n)    {          let sum = 0;          for (let i = 0; i < n; i++) {            let sqrtCurrent = Math.sqrt(arr[i]);              for (let j = 0; j < n; j++) {                let x = arr[j];                  // If sqrtCurrent is present in array                if (x == sqrtCurrent) {                    sum += (sqrtCurrent * sqrtCurrent);                    break;                }            }        }          return sum;    }         let arr = [ 2, 4, 5, 6, 7, 8, 9, 3 ];    let n = arr.length;    document.write(getSum(arr, n));         // This code is contributed by suresh07.</script>
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Efficient Approach: We can create a HashSet of all the elements present in the array and than check for the square root of each element of the array in O(n) time.
Below is the implementation of the above approach:  
C++
Java
C#
Python3
PHP
Javascript
// C++ program to find the sum of all the elements// from the array whose square root is present// in the same array#include<bits/stdc++.h>using namespace std; // Function to return the required sumint getSum(int arr[], int n){     int i, sum = 0;     // Initialization of hash map    set<int> hashSet;     // Store each element in the hash map    for (i = 0; i < n; i++)        hashSet.insert(arr[i]);     for (i = 0; i < n; i++)    {        double sqrtCurrent = sqrt(arr[i]);         // If sqrtCurrent is a decimal number        if (floor(sqrtCurrent) != ceil(sqrtCurrent))            continue;         // If hash set contains sqrtCurrent        if (hashSet.find((int)sqrtCurrent) !=            hashSet.end())        {            sum += (sqrtCurrent * sqrtCurrent);        }    }     return sum;} // Driver codeint main(){    int arr[] = { 2, 4, 5, 6, 7, 8, 9, 3 };    int n = sizeof(arr)/sizeof(arr[0]);    cout << (getSum(arr, n));    return 0;} // This code is contributed by Rajput-Ji
// Java program to find the sum of all the elements// from the array whose square root is present// in the same array import java.util.*;public class GFG {     // Function to return the required sum    public static int getSum(int arr[], int n)    {         int i, sum = 0;         // Initialization of hash map        Set<Integer> hashSet = new HashSet<>();         // Store each element in the hash map        for (i = 0; i < n; i++)            hashSet.add(arr[i]);         for (i = 0; i < n; i++) {            double sqrtCurrent = Math.sqrt(arr[i]);             // If sqrtCurrent is a decimal number            if (Math.floor(sqrtCurrent) != Math.ceil(sqrtCurrent))                continue;             // If hash set contains sqrtCurrent            if (hashSet.contains((int)sqrtCurrent)) {                sum += (sqrtCurrent * sqrtCurrent);            }        }         return sum;    }     // Driver code    public static void main(String args[])    {        int arr[] = { 2, 4, 5, 6, 7, 8, 9, 3 };        int n = arr.length;        System.out.println(getSum(arr, n));    }}
// C# program to find the sum of all the elements// from the array whose square root is present// in the same arrayusing System;using System.Collections.Generic; class GFG{     // Function to return the required sum    public static int getSum(int []arr, int n)    {         int i, sum = 0;         // Initialization of hash map        HashSet<int> hashSet = new HashSet<int>();         // Store each element in the hash map        for (i = 0; i < n; i++)            hashSet.Add(arr[i]);         for (i = 0; i < n; i++)        {            double sqrtCurrent = Math.Sqrt(arr[i]);             // If sqrtCurrent is a decimal number            if (Math.Floor(sqrtCurrent) !=                Math.Ceiling(sqrtCurrent))                continue;             // If hash set contains sqrtCurrent            if (hashSet.Contains((int)sqrtCurrent))            {                sum += (int)(sqrtCurrent * sqrtCurrent);            }        }         return sum;    }     // Driver code    public static void Main(String []args)    {        int []arr = { 2, 4, 5, 6, 7, 8, 9, 3 };        int n = arr.Length;        Console.WriteLine(getSum(arr, n));    }} // This code contributed by Rajput-Ji
# Python3 program to find the sum of all the# elements from the array whose square# root is present in the same arrayimport math # Function to return the required sumdef getSum(arr, n):    sum = 0;     # Initialization of hash map    hashSet = set();     # Store each element in the hash map    for i in range(n):        hashSet.add(arr[i]);         for i in range(n):             sqrtCurrent = math.sqrt(arr[i]);         # If sqrtCurrent is a decimal number        if (math.floor(sqrtCurrent) != math.ceil(sqrtCurrent)):            continue;         # If hash set contains sqrtCurrent        if (int(sqrtCurrent) in hashSet):            sum += int(sqrtCurrent * sqrtCurrent);     return sum; # Driver codearr = [ 2, 4, 5, 6, 7, 8, 9, 3 ];n = len(arr);print(getSum(arr, n)); # This code is contributed by mits
<?php// PHP program to find the sum of all the// elements from the array whose square// root is present in the same array // Function to return the required sumfunction getSum($arr, $n){    $sum = 0;     // Initialization of hash map    $hashSet = array();     // Store each element in the hash map    for ($i = 0; $i < $n; $i++)        array_push($hashSet, $arr[$i]);             $hashSet = array_unique($hashSet);         for ($i = 0; $i < $n; $i++)    {        $sqrtCurrent = sqrt($arr[$i]);         // If sqrtCurrent is a decimal number        if (floor($sqrtCurrent) != ceil($sqrtCurrent))            continue;         // If hash set contains sqrtCurrent        if (in_array((int)$sqrtCurrent, $hashSet))        {            $sum += ($sqrtCurrent * $sqrtCurrent);        }    }     return $sum;} // Driver code$arr = array( 2, 4, 5, 6, 7, 8, 9, 3 );$n = count($arr);print(getSum($arr, $n)); // This code is contributed by mits?>
<script> // Javascript program to find the sum of// all the elements from the array whose// square root is present in the same array     // Function to return the required sumfunction getSum(arr, n){    let i, sum = 0;         // Initialization of hash map    let hashSet = new Set();         // Store each element in the hash map    for(i = 0; i < n; i++)        hashSet.add(arr[i]);         for(i = 0; i < n; i++)    {        let sqrtCurrent = Math.sqrt(arr[i]);             // If sqrtCurrent is a decimal number        if (Math.floor(sqrtCurrent) !=            Math.ceil(sqrtCurrent))            continue;             // If hash set contains sqrtCurrent        if (hashSet.has(sqrtCurrent))        {            sum += (sqrtCurrent * sqrtCurrent);        }    }    return sum;} // Driver codelet arr = [ 2, 4, 5, 6, 7, 8, 9, 3 ];let n = arr.length; document.write(getSum(arr, n));     // This code is contributed by unknown2108 </script>
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SURENDRA_GANGWAR
ankthon
Shivi_Aggarwal
29AjayKumar
Rajput-Ji
Mithun Kumar
unknown2108
suresh07
Hash
HashSet
Arrays
Arrays
Hash
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