# Find the maximum elements in the first and the second halves of the Array

Given an array arr[] of N integers. The task is to find the largest elements in the first half and the second half of the array. Note that if the size of the array is odd then the middle element will be included in both the halves.

Examples:

Input: arr[] = {1, 12, 14, 5}
Output: 12, 14
First half is {1, 12} and the second half is {14, 5}.

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Calculate the middle index of the array as mid = N / 2. Now the first halve elements will be present in the subarray arr[0…mid-1] and arr[mid…N-1] if N is even.
If N is odd then the halves are arr[0…mid] and arr[mid…N-1]

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print largest element in ` `// first half and second half of an array ` `void` `findMax(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the maximum element ` `    ``// in the first half ` `    ``int` `maxFirst = INT_MIN; ` ` `  `    ``// Middle index of the array ` `    ``int` `mid = n / 2; ` ` `  `    ``// Calculate the maximum element ` `    ``// in the first half ` `    ``for` `(``int` `i = 0; i < mid; i++) ` `        ``maxFirst = max(maxFirst, arr[i]); ` ` `  `    ``// If the size of array is odd then ` `    ``// the middle element will be included ` `    ``// in both the halves ` `    ``if` `(n % 2 == 1) ` `        ``maxFirst = max(maxFirst, arr[mid]); ` ` `  `    ``// To store the maximum element ` `    ``// in the second half ` `    ``int` `maxSecond = INT_MIN; ` ` `  `    ``// Calculate the maximum element ` `    ``// int the second half ` `    ``for` `(``int` `i = mid; i < n; i++) ` `        ``maxSecond = max(maxSecond, arr[i]); ` ` `  `    ``// Print the found maximums ` `    ``cout << maxFirst << ``", "` `<< maxSecond; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 12, 14, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``findMax(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `findMax(``int` `[]arr, ``int` `n)  ` `    ``{  ` `     `  `        ``// To store the maximum element  ` `        ``// in the first half  ` `        ``int` `maxFirst = Integer.MIN_VALUE;  ` `     `  `        ``// Middle index of the array  ` `        ``int` `mid = n / ``2``;  ` `     `  `        ``// Calculate the maximum element  ` `        ``// in the first half  ` `        ``for` `(``int` `i = ``0``; i < mid; i++)  ` `        ``{ ` `            ``maxFirst = Math.max(maxFirst, arr[i]);  ` `        ``} ` `     `  `        ``// If the size of array is odd then  ` `        ``// the middle element will be included  ` `        ``// in both the halves  ` `        ``if` `(n % ``2` `== ``1``)  ` `        ``{ ` `            ``maxFirst = Math.max(maxFirst, arr[mid]);  ` `        ``} ` `         `  `        ``// To store the maximum element  ` `        ``// in the second half  ` `        ``int` `maxSecond = Integer.MIN_VALUE;  ` `     `  `        ``// Calculate the maximum element  ` `        ``// int the second half  ` `        ``for` `(``int` `i = mid; i < n; i++)  ` `        ``{ ` `            ``maxSecond = Math.max(maxSecond, arr[i]);  ` `        ``} ` `         `  `        ``// Print the found maximums  ` `        ``System.out.print(maxFirst + ``", "` `+ maxSecond); ` `        ``// cout << maxFirst << ", " << maxSecond;  ` `    ``}  ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `[]arr = { ``1``, ``12``, ``14``, ``5` `};  ` `        ``int` `n = arr.length;  ` `     `  `        ``findMax(arr, n);  ` `    ``}  ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach ` `import` `sys ` ` `  `# Function to print largest element in ` `# first half and second half of an array ` `def` `findMax(arr, n) : ` ` `  `    ``# To store the maximum element ` `    ``# in the first half ` `    ``maxFirst ``=` `-``sys.maxsize ``-` `1` ` `  `    ``# Middle index of the array ` `    ``mid ``=` `n ``/``/` `2``; ` ` `  `    ``# Calculate the maximum element ` `    ``# in the first half ` `    ``for` `i ``in` `range``(``0``, mid): ` `        ``maxFirst ``=` `max``(maxFirst, arr[i]) ` ` `  `    ``# If the size of array is odd then ` `    ``# the middle element will be included ` `    ``# in both the halves ` `    ``if` `(n ``%` `2` `=``=` `1``): ` `        ``maxFirst ``=` `max``(maxFirst, arr[mid]) ` ` `  `    ``# To store the maximum element ` `    ``# in the second half ` `    ``maxSecond ``=` `-``sys.maxsize ``-` `1` ` `  `    ``# Calculate the maximum element ` `    ``# int the second half ` `    ``for` `i ``in` `range``(mid, n): ` `        ``maxSecond ``=` `max``(maxSecond, arr[i]) ` ` `  `    ``# Print the found maximums ` `    ``print``(maxFirst, ``","``, maxSecond) ` ` `  `# Driver code ` `arr ``=` `[``1``, ``12``, ``14``, ``5` `] ` `n ``=` `len``(arr) ` ` `  `findMax(arr, n) ` ` `  `# This code is contributed by ihritik `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `findMax(``int` `[]arr, ``int` `n)  ` `    ``{  ` `     `  `        ``// To store the maximum element  ` `        ``// in the first half  ` `        ``int` `maxFirst = ``int``.MinValue;  ` `     `  `        ``// Middle index of the array  ` `        ``int` `mid = n / 2;  ` `     `  `        ``// Calculate the maximum element  ` `        ``// in the first half  ` `        ``for` `(``int` `i = 0; i < mid; i++)  ` `        ``{ ` `            ``maxFirst = Math.Max(maxFirst, arr[i]);  ` `        ``} ` `     `  `        ``// If the size of array is odd then  ` `        ``// the middle element will be included  ` `        ``// in both the halves  ` `        ``if` `(n % 2 == 1)  ` `        ``{ ` `            ``maxFirst = Math.Max(maxFirst, arr[mid]);  ` `        ``} ` `         `  `        ``// To store the maximum element  ` `        ``// in the second half  ` `        ``int` `maxSecond = ``int``.MinValue;  ` `     `  `        ``// Calculate the maximum element  ` `        ``// int the second half  ` `        ``for` `(``int` `i = mid; i < n; i++)  ` `        ``{ ` `            ``maxSecond = Math.Max(maxSecond, arr[i]);  ` `        ``} ` `         `  `        ``// Print the found maximums  ` `        ``Console.WriteLine(maxFirst + ``", "` `+ maxSecond); ` `        ``// cout << maxFirst << ", " << maxSecond;  ` `    ``}  ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = { 1, 12, 14, 5 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``findMax(arr, n);  ` `    ``}  ` `} ` ` `  `// This code is contributed by nidhiva `

Output:

```12, 14
```

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Improved By : nidhiva, vt_m, ihritik