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 arrayInput: 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 the 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:
// 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 sum
int 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 array public 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 array import math
# Function to return the required sum def 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 code if __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 array using 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 sum function 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> |
13
Complexity Analysis:
- Time Complexity: O(n2)
- Auxiliary Space: O(1)
Efficient Approach: We can create a HashSet of all the elements present in the array and then check for the square root of each element of the array in O(n) time.
Below is the implementation of the above approach:
// 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 sum int 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 code int 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 array using 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 array import math
# Function to return the required sum def 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 code arr = [ 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 sum function 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 sum function 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 code let 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> |
13
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)