Given an array **arr[]** of **N** integers representing the heights of the sticks. The task is to find the area of the largest square that can be formed using these sticks and the count of such squares. **Note** that a single side of the square can only use a single stick.

**Examples:**

Input:arr[] = {5, 3, 2, 3, 6, 3, 3}

Output:

Area = 9

Count = 1

Side of the square will be 3 and

only one such square is possible.

Input:arr[] = {2, 2, 2, 9, 2, 2, 2, 2, 2}

Output:

Area = 4

Count = 2

**Approach:** Count the frequencies of all the elements of the array. Now, starting from the maximum (in order to maximize the area) find the first frequency which is at least 4 so that a square can be formed then the area can be calculated as freq[i] * freq[i] and the count of such squares will be freq[i] / 4.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the area of the largest ` `// square that can be formed ` `// and the count of such squares ` `void` `findMaxSquare(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// Maximum value from the array ` ` ` `int` `maxVal = *max_element(arr, arr + n); ` ` ` ` ` `// Update the frequencies of ` ` ` `// the array elements ` ` ` `int` `freq[maxVal + 1] = { 0 }; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `freq[arr[i]]++; ` ` ` ` ` `// Starting from the maximum length sticks ` ` ` `// in order to maximize the area ` ` ` `for` `(` `int` `i = maxVal; i > 0; i--) { ` ` ` ` ` `// The count of sticks with the current ` ` ` `// length has to be at least 4 ` ` ` `// in order to form a square ` ` ` `if` `(freq[i] >= 4) { ` ` ` `cout << ` `"Area = "` `<< (` `pow` `(i, 2)); ` ` ` `cout << ` `"\nCount = "` `<< (freq[i] / 4); ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Impossible to form a square ` ` ` `cout << ` `"-1"` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 2, 2, 2, 9, 2, 2, 2, 2, 2 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `findMaxSquare(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to find the area of the largest ` `// square that can be formed ` `// and the count of such squares ` `static` `void` `findMaxSquare(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// Maximum value from the array ` ` ` `int` `maxVal = Arrays.stream(arr).max().getAsInt(); ` ` ` ` ` `// Update the frequencies of ` ` ` `// the array elements ` ` ` `int` `[]freq = ` `new` `int` `[maxVal + ` `1` `]; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `freq[arr[i]]++; ` ` ` ` ` `// Starting from the maximum length sticks ` ` ` `// in order to maximize the area ` ` ` `for` `(` `int` `i = maxVal; i > ` `0` `; i--) ` ` ` `{ ` ` ` ` ` `// The count of sticks with the current ` ` ` `// length has to be at least 4 ` ` ` `// in order to form a square ` ` ` `if` `(freq[i] >= ` `4` `) ` ` ` `{ ` ` ` `System.out.print(` `"Area = "` `+ ` ` ` `(Math.pow(i, ` `2` `))); ` ` ` `System.out.print(` `"\nCount = "` `+ ` ` ` `(freq[i] / ` `4` `)); ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Impossible to form a square ` ` ` `System.out.print(` `"-1"` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `arr[] = { ` `2` `, ` `2` `, ` `2` `, ` `9` `, ` `2` `, ` `2` `, ` `2` `, ` `2` `, ` `2` `}; ` ` ` `int` `n = arr.length; ` ` ` ` ` `findMaxSquare(arr, n); ` `} ` `} ` ` ` `// This code is contributed by Princi Singh ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to find the area of the largest ` `# square that can be formed ` `# and the count of such squares ` `def` `findMaxSquare(arr, n) : ` ` ` ` ` `# Maximum value from the array ` ` ` `maxVal ` `=` `max` `(arr); ` ` ` ` ` `# Update the frequencies of ` ` ` `# the array elements ` ` ` `freq ` `=` `[` `0` `] ` `*` `(maxVal ` `+` `1` `) ; ` ` ` `for` `i ` `in` `range` `(n) : ` ` ` `freq[arr[i]] ` `+` `=` `1` `; ` ` ` ` ` `# Starting from the maximum length sticks ` ` ` `# in order to maximize the area ` ` ` `for` `i ` `in` `range` `(maxVal, ` `0` `, ` `-` `1` `) : ` ` ` ` ` `# The count of sticks with the current ` ` ` `# length has to be at least 4 ` ` ` `# in order to form a square ` ` ` `if` `(freq[i] >` `=` `4` `) : ` ` ` `print` `(` `"Area = "` `, ` `pow` `(i, ` `2` `)); ` ` ` `print` `(` `"Count ="` `, freq[i] ` `/` `/` `4` `); ` ` ` `return` `; ` ` ` ` ` `# Impossible to form a square ` ` ` `print` `(` `"-1"` `); ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `arr ` `=` `[ ` `2` `, ` `2` `, ` `2` `, ` `9` `, ` `2` `, ` `2` `, ` `2` `, ` `2` `, ` `2` `]; ` ` ` `n ` `=` `len` `(arr); ` ` ` ` ` `findMaxSquare(arr, n); ` ` ` `# This code is contributed by AnkitRai01 ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to find the area of the largest ` `// square that can be formed ` `// and the count of such squares ` `static` `void` `findMaxSquare(` `int` `[]arr, ` `int` `n) ` `{ ` ` ` ` ` `// Maximum value from the array ` ` ` `int` `maxVal = arr.Max(); ` ` ` ` ` `// Update the frequencies of ` ` ` `// the array elements ` ` ` `int` `[]freq = ` `new` `int` `[maxVal + 1]; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `freq[arr[i]]++; ` ` ` ` ` `// Starting from the maximum length sticks ` ` ` `// in order to maximize the area ` ` ` `for` `(` `int` `i = maxVal; i > 0; i--) ` ` ` `{ ` ` ` ` ` `// The count of sticks with the current ` ` ` `// length has to be at least 4 ` ` ` `// in order to form a square ` ` ` `if` `(freq[i] >= 4) ` ` ` `{ ` ` ` `Console.Write(` `"Area = "` `+ ` ` ` `(Math.Pow(i, 2))); ` ` ` `Console.Write(` `"\nCount = "` `+ ` ` ` `(freq[i] / 4)); ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Impossible to form a square ` ` ` `Console.Write(` `"-1"` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `[]arr = { 2, 2, 2, 9, 2, 2, 2, 2, 2 }; ` ` ` `int` `n = arr.Length; ` ` ` ` ` `findMaxSquare(arr, n); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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**Output:**

Area = 4 Count = 2

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