# Maximum element in a sorted and rotated array

Given a sorted array arr[] of distinct elements which is rotated at some unknown point, the task is to find the maximum element in it.

Examples:

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

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

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

Approach: A simple solution is to traverse the complete array and find maximum. This solution requires O(n) time.
We can do it in O(Logn) using Binary Search. If we take a closer look at above examples, we can easily figure out the following pattern:

• The maximum element is the only element whose next is smaller than it. If there is no next smaller element, then there is no rotation (last element is the maximum). We check this condition for middle element by comparing it with elements at mid – 1 and mid + 1.
• If maximum element is not at middle (neither mid nor mid + 1), then maximum element lies in either left half or right half.
1. If middle element is greater than the last element, then the maximum element lies in the left half.
2. Else maximum element lies in the right half.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum element ` `int` `findMax(``int` `arr[], ``int` `low, ``int` `high) ` `{ ` ` `  `    ``// This condition is for the case when ` `    ``// array is not rotated at all ` `    ``if` `(high < low) ` `        ``return` `arr; ` ` `  `    ``// If there is only one element left ` `    ``if` `(high == low) ` `        ``return` `arr[low]; ` ` `  `    ``// Find mid ` `    ``int` `mid = low + (high - low) / 2; ` ` `  `    ``// Check if mid itself is maximum element ` `    ``if` `(mid < high && arr[mid + 1] < arr[mid]) { ` `        ``return` `arr[mid]; ` `    ``} ` ` `  `    ``// Check if element at (mid - 1) is maximum element ` `    ``// Consider the cases like {4, 5, 1, 2, 3} ` `    ``if` `(mid > low && arr[mid] < arr[mid - 1]) { ` `        ``return` `arr[mid - 1]; ` `    ``} ` ` `  `    ``// Decide whether we need to go to ` `    ``// the left half or the right half ` `    ``if` `(arr[low] > arr[mid]) { ` `        ``return` `findMax(arr, low, mid - 1); ` `    ``} ` `    ``else` `{ ` `        ``return` `findMax(arr, mid + 1, high); ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, 6, 1, 2, 3, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << findMax(arr, 0, n - 1); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the maximum element ` `static` `int` `findMax(``int` `arr[], ``int` `low, ``int` `high) ` `{ ` ` `  `    ``// This condition is for the case when ` `    ``// array is not rotated at all ` `    ``if` `(high < low) ` `        ``return` `arr[``0``]; ` ` `  `    ``// If there is only one element left ` `    ``if` `(high == low) ` `        ``return` `arr[low]; ` ` `  `    ``// Find mid ` `    ``int` `mid = low + (high - low) / ``2``; ` ` `  `    ``// Check if mid itself is maximum element ` `    ``if` `(mid < high && arr[mid + ``1``] < arr[mid]) ` `    ``{ ` `        ``return` `arr[mid]; ` `    ``} ` ` `  `    ``// Check if element at (mid - 1) is maximum element ` `    ``// Consider the cases like {4, 5, 1, 2, 3} ` `    ``if` `(mid > low && arr[mid] < arr[mid - ``1``]) ` `    ``{ ` `        ``return` `arr[mid - ``1``]; ` `    ``} ` ` `  `    ``// Decide whether we need to go to ` `    ``// the left half or the right half ` `    ``if` `(arr[low] > arr[mid]) ` `    ``{ ` `        ``return` `findMax(arr, low, mid - ``1``); ` `    ``} ` `    ``else`  `    ``{ ` `        ``return` `findMax(arr, mid + ``1``, high); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``5``, ``6``, ``1``, ``2``, ``3``, ``4` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(findMax(arr, ``0``, n - ``1``)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the maximum element ` `def` `findMax(arr, low, high): ` ` `  `    ``# This condition is for the case when ` `    ``# array is not rotated at all ` `    ``if` `(high < low): ` `        ``return` `arr[``0``] ` ` `  `    ``# If there is only one element left ` `    ``if` `(high ``=``=` `low): ` `        ``return` `arr[low] ` ` `  `    ``# Find mid ` `    ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2` ` `  `    ``# Check if mid itself is maximum element ` `    ``if` `(mid < high ``and` `arr[mid ``+` `1``] < arr[mid]): ` `        ``return` `arr[mid] ` `     `  `    ``# Check if element at (mid - 1) is maximum element ` `    ``# Consider the cases like {4, 5, 1, 2, 3} ` `    ``if` `(mid > low ``and` `arr[mid] < arr[mid ``-` `1``]): ` `        ``return` `arr[mid ``-` `1``] ` ` `  `    ``# Decide whether we need to go to ` `    ``# the left half or the right half ` `    ``if` `(arr[low] > arr[mid]): ` `        ``return` `findMax(arr, low, mid ``-` `1``) ` `    ``else``: ` `        ``return` `findMax(arr, mid ``+` `1``, high) ` ` `  `# Driver code ` `arr ``=` `[``5``, ``6``, ``1``, ``2``, ``3``, ``4``] ` `n ``=` `len``(arr) ` `print``(findMax(arr, ``0``, n ``-` `1``)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the maximum element ` `static` `int` `findMax(``int` `[]arr,  ` `                   ``int` `low, ``int` `high) ` `{ ` ` `  `    ``// This condition is for the case  ` `    ``// when array is not rotated at all ` `    ``if` `(high < low) ` `        ``return` `arr; ` ` `  `    ``// If there is only one element left ` `    ``if` `(high == low) ` `        ``return` `arr[low]; ` ` `  `    ``// Find mid ` `    ``int` `mid = low + (high - low) / 2; ` ` `  `    ``// Check if mid itself is maximum element ` `    ``if` `(mid < high && arr[mid + 1] < arr[mid]) ` `    ``{ ` `        ``return` `arr[mid]; ` `    ``} ` ` `  `    ``// Check if element at (mid - 1)  ` `    ``// is maximum element ` `    ``// Consider the cases like {4, 5, 1, 2, 3} ` `    ``if` `(mid > low && arr[mid] < arr[mid - 1]) ` `    ``{ ` `        ``return` `arr[mid - 1]; ` `    ``} ` ` `  `    ``// Decide whether we need to go to ` `    ``// the left half or the right half ` `    ``if` `(arr[low] > arr[mid]) ` `    ``{ ` `        ``return` `findMax(arr, low, mid - 1); ` `    ``} ` `    ``else` `    ``{ ` `        ``return` `findMax(arr, mid + 1, high); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 5, 6, 1, 2, 3, 4 }; ` `    ``int` `n = arr.Length; ` `     `  `    ``Console.WriteLine(findMax(arr, 0, n - 1)); ` `} ` `} ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` ``\$low` `&& ``\$arr``[``\$mid``] < ``\$arr``[``\$mid` `- 1]) ` `    ``{ ` `        ``return` `\$arr``[``\$mid` `- 1]; ` `    ``} ` ` `  `    ``// Decide whether we need to go to ` `    ``// the left half or the right half ` `    ``if` `(``\$arr``[``\$low``] > ``\$arr``[``\$mid``]) ` `    ``{ ` `        ``return` `findMax(``\$arr``, ``\$low``, ``\$mid` `- 1); ` `    ``} ` `    ``else` `    ``{ ` `        ``return` `findMax(``\$arr``, ``\$mid` `+ 1, ``\$high``); ` `    ``} ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(5, 6, 1, 2, 3, 4); ` `\$n` `= sizeof(``\$arr``); ` `echo` `findMax(``\$arr``, 0, ``\$n` `- 1); ` ` `  `// This code is contributed ` `// by Akanksha Rai `

Output:

```6
```

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