# Subarrays whose sum is a perfect square

Given an array, arr[] of size N, the task is to print the start and end indices of all subarrays whose sum is a perfect square.

Examples:

Input: arr[] = {65, 79, 81}
Output: (0, 1) (0, 2) (2, 2)
Explanation:
Subarray sum whose start and end index is (0, 1) = 65 + 79 = 144 = 122
Subarray sum whose start and end index is (0, 2} = 65 + 79 + 81 = 225 = 152
Subarray sum whose start and end index is {2, 2} = 81 = 92

Input: arr[] = {1, 2, 3, 4, 5}
Output: (0, 0) (1, 3) (3, 3) (3, 4)

Approach: The problem can be solved using the Prefix Sum Array technique. The idea is to find the sum of all subarrays using the Prefix Sum Array. For each subarray, check if the sum of the subarray is a perfect square or not. If found to be true for any subarray, then print the start and end indices of that subarray. Follow the steps below to solve the problem.

1. Initialize a variable, say currSubSum to store the current subarray sum.
2. Iterate over the array to generate all possible subarrays of the given array.
3. Calculate the sum of each subarray and for each subarray sum, check if it is a perfect square or not.
4. If found to be true for any subarray, then print the start and end indices of the subarray.

Below is the implementation of the above approach:

## C++14

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to print the start and end` `// indices of all subarrays whose sum` `// is a perfect square` `void` `PrintIndexes(``int` `arr[], ``int` `N)` `{`   `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Stores the current` `        ``// subarray sum` `        ``int` `currSubSum = 0;`   `        ``for` `(``int` `j = i; j < N; j++) {`   `            ``// Update current subarray sum` `            ``currSubSum += arr[j];`   `            ``// Stores the square root` `            ``// of currSubSum` `            ``int` `sq = ``sqrt``(currSubSum);`   `            ``// Check if currSubSum is` `            ``// a perfect square or not` `            ``if` `(sq * sq == currSubSum) {` `                ``cout << ``"("` `<< i << ``", "` `                     ``<< j << ``") "``;` `            ``}` `        ``}` `    ``}` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `arr[] = { 65, 79, 81 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``PrintIndexes(arr, N);` `}`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.io.*;`   `class` `GFG{` `    `  `// Function to print the start and end ` `// indices of all subarrays whose sum ` `// is a perfect square ` `static` `void` `PrintIndexes(``int` `arr[], ``int` `N) ` `{ ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        `  `        ``// Stores the current ` `        ``// subarray sum ` `        ``int` `currSubSum = ``0``; ` `  `  `        ``for``(``int` `j = i; j < N; j++) ` `        ``{ ` `            `  `            ``// Update current subarray sum ` `            ``currSubSum += arr[j]; ` `  `  `            ``// Stores the square root ` `            ``// of currSubSum ` `            ``int` `sq = (``int``)Math.sqrt(currSubSum); ` `  `  `            ``// Check if currSubSum is ` `            ``// a perfect square or not ` `            ``if` `(sq * sq == currSubSum)` `            ``{ ` `                ``System.out.print(``"("` `+ i + ``","` `+ ` `                                 ``j + ``")"` `+ ``" "``);` `            ``} ` `        ``} ` `    ``} ` `} `   `// Driver code` `public` `static` `void` `main (String[] args)` `throws` `java.lang.Exception` `{` `    ``int` `arr[] = { ``65``, ``79``, ``81` `}; ` `    `  `    ``PrintIndexes(arr, arr.length);` `}` `}`   `// This code is contributed by bikram2001jha`

## Python3

 `# Python3 program to implement` `# the above approach` `import` `math`   `# Function to prthe start and end` `# indices of all subarrays whose sum` `# is a perfect square` `def` `PrintIndexes(arr, N):` ` `  `    ``for` `i ``in` `range``(N):` ` `  `        ``# Stores the current` `        ``# subarray sum` `        ``currSubSum ``=` `0` ` `  `        ``for` `j ``in` `range``(i, N, ``1``):` `            `  `            ``# Update current subarray sum` `            ``currSubSum ``+``=` `arr[j]` ` `  `            ``# Stores the square root` `            ``# of currSubSum` `            ``sq ``=` `int``(math.sqrt(currSubSum))` ` `  `            ``# Check if currSubSum is` `            ``# a perfect square or not` `            ``if` `(sq ``*` `sq ``=``=` `currSubSum):` `                ``print``(``"("``, i, ``","``, ` `                           ``j, ``")"``, end ``=` `" "``)` `                           `  `# Driver Code` `arr ``=` `[ ``65``, ``79``, ``81` `]` `N ``=` `len``(arr) `   `PrintIndexes(arr, N)`   `# This code is contributed by sanjoy_62`

## C#

 `// C# program to implement ` `// the above approach ` `using` `System;` `class` `GFG{` `    `  `// Function to print the start ` `// and end indices of all ` `// subarrays whose sum ` `// is a perfect square ` `static` `void` `PrintIndexes(``int` `[]arr, ` `                         ``int` `N) ` `{ ` `  ``for``(``int` `i = 0; i < N; i++) ` `  ``{ ` `    ``// Stores the current ` `    ``// subarray sum ` `    ``int` `currSubSum = 0; `   `    ``for``(``int` `j = i; j < N; j++) ` `    ``{ ` `      ``// Update current subarray sum ` `      ``currSubSum += arr[j]; `   `      ``// Stores the square root ` `      ``// of currSubSum ` `      ``int` `sq = (``int``)Math.Sqrt(currSubSum); `   `      ``// Check if currSubSum is ` `      ``// a perfect square or not ` `      ``if` `(sq * sq == currSubSum)` `      ``{ ` `        ``Console.Write(``"("` `+ i + ``","` `+ ` `                      ``j + ``")"` `+ ``" "``);` `      ``} ` `    ``} ` `  ``} ` `} `   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `  ``int` `[]arr = {65, 79, 81};` `  ``PrintIndexes(arr, arr.Length);` `}` `}`   `// This code is contributed by shikhasingrajput`

Output:

```(0, 1) (0, 2) (2, 2)

```

Time Complexity: O(N2)
Auxiliary Space: O(1)

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