 Open in App
Not now

# Sum of elements whose square root is present in the array

• Last Updated : 12 Sep, 2022

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 the square root of every element by iterating from arr to arr[n] which will do the job but in O(n*n) complexity.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the sum of all the elements` `// from the array whose square root is present` `// in the same array`   `#include` `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);` `        ``cout<<(getSum(arr, n));` `    ``}` `// This code is contributed by ` `// Surendra_Gangwar`

## Java

 `// 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

 `# 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#

 `// 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

 ``

## Javascript

 ``

Output

`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++

 `// C++ program to find the sum of all the elements` `// from the array whose square root is present` `// in the same array` `#include` `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);` `    ``cout << (getSum(arr, n));` `    ``return` `0;` `}`   `// This code is contributed by Rajput-Ji`

## Java

 `// 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 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#

 `// 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

 `# 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

 ``

## Javascript

 ``

Output

`13`

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(n)

My Personal Notes arrow_drop_up
Related Articles