# Search an element in a sorted and rotated array with duplicates

Given an array arr[] which is sorted and rotated, the task is to find an element in the rotated array (with duplicates) in O(log n) time.
Note: Print the index where the key exists. In case of multiple answer print any of them

Examples:

Input: arr[] = {3, 3, 3, 1, 2, 3}, key = 3
Output: 0
arr[0] = 3

Input: arr[] = {3, 3, 3, 1, 2, 3}, key = 11
Output: -1
11 is not present in the given array.

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

Approach: The idea is the same as the previous one without duplicates. The only difference is that due to the existence of duplicates, arr[low] == arr[mid] could be possible, the first half could be out of order (i.e. not in the ascending order, e.g. {3, 1, 2, 3, 3, 3, 3}) and we have to deal this case separately.
In that case, it is guaranteed that arr[high] also equal to arr[mid], so the condition arr[mid] == arr[low] == arr[high] can be checked before the original logic, and if so then move left and right both towards the middle by 1 and repeat.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the index of the ` `// key in arr[l..h] if the key is present ` `// otherwise return -1 ` `int` `search(``int` `arr[], ``int` `l, ``int` `h, ``int` `key) ` `{ ` `    ``if` `(l > h) ` `        ``return` `-1; ` ` `  `    ``int` `mid = (l + h) / 2; ` `    ``if` `(arr[mid] == key) ` `        ``return` `mid; ` ` `  `    ``// The tricky case, just update left and right ` `    ``if` `((arr[l] == arr[mid]) ` `        ``&& (arr[h] == arr[mid])) { ` `        ``++l; ` `        ``--h; ` `    ``} ` ` `  `    ``// If arr[l...mid] is sorted ` `    ``if` `(arr[l] <= arr[mid]) { ` ` `  `        ``// As this subarray is sorted, we can quickly ` `        ``// check if key lies in any of the halves ` `        ``if` `(key >= arr[l] && key <= arr[mid]) ` `            ``return` `search(arr, l, mid - 1, key); ` ` `  `        ``// If key does not lie in the first half ` `        ``// subarray then divide the other half ` `        ``// into two subarrays such that we can ` `        ``// quickly check if key lies in the other half ` `        ``return` `search(arr, mid + 1, h, key); ` `    ``} ` ` `  `    ``// If arr[l..mid] first subarray is not sorted ` `    ``// then arr[mid... h] must be sorted subarray ` `    ``if` `(key >= arr[mid] && key <= arr[h]) ` `        ``return` `search(arr, mid + 1, h, key); ` ` `  `    ``return` `search(arr, l, mid - 1, key); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 3, 1, 2, 3, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``int` `key = 3; ` ` `  `    ``cout << search(arr, 0, n - 1, key); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to return the index of the  ` `    ``// key in arr[l..h] if the key is present  ` `    ``// otherwise return -1 ` `    ``static` `int` `search(``int` `arr[], ``int` `l, ``int` `h, ``int` `key)  ` `    ``{  ` `        ``if` `(l > h)  ` `            ``return` `-``1``;  ` `     `  `        ``int` `mid = (l + h) / ``2``;  ` `        ``if` `(arr[mid] == key)  ` `            ``return` `mid;  ` `     `  `        ``// The tricky case, just update left and right  ` `        ``if` `((arr[l] == arr[mid])  ` `            ``&& (arr[h] == arr[mid])) ` `        ``{  ` `            ``++l;  ` `            ``--h;  ` `        ``}  ` `     `  `        ``// If arr[l...mid] is sorted ` `        ``if` `(arr[l] <= arr[mid]) ` `        ``{  ` `     `  `            ``// As this subarray is sorted, we can quickly  ` `            ``// check if key lies in any of the halves  ` `            ``if` `(key >= arr[l] && key <= arr[mid])  ` `                ``return` `search(arr, l, mid - ``1``, key);  ` `     `  `            ``// If key does not lie in the first half  ` `            ``// subarray then divide the other half  ` `            ``// into two subarrays such that we can  ` `            ``// quickly check if key lies in the other half  ` `            ``return` `search(arr, mid + ``1``, h, key);  ` `        ``}  ` `     `  `        ``// If arr[l..mid] first subarray is not sorted ` `        ``// then arr[mid... h] must be sorted subarray  ` `        ``if` `(key >= arr[mid] && key <= arr[h])  ` `            ``return` `search(arr, mid + ``1``, h, key);  ` `     `  `        ``return` `search(arr, l, mid - ``1``, key);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `arr[] = { ``3``, ``3``, ``1``, ``2``, ``3``, ``3` `};  ` `        ``int` `n = arr.length;  ` `        ``int` `key = ``3``;  ` `     `  `        ``System.out.println(search(arr, ``0``, n - ``1``, key));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the index of the  ` `# key in arr[l..h] if the key is present  ` `# otherwise return -1  ` `def` `search(arr, l, h, key) :  ` ` `  `    ``if` `(l > h) : ` `        ``return` `-``1``; ` `         `  `    ``mid ``=` `(l ``+` `h) ``/``/` `2``;  ` `    ``if` `(arr[mid] ``=``=` `key) : ` `        ``return` `mid;  ` ` `  `    ``# The tricky case, just update left and right  ` `    ``if` `((arr[l] ``=``=` `arr[mid]) ``and` `(arr[h] ``=``=` `arr[mid])) : ` `        ``l ``+``=` `1``;  ` `        ``h ``-``=` `1``;  ` `     `  `    ``# If arr[l...mid] is sorted  ` `    ``if` `(arr[l] <``=` `arr[mid]) : ` ` `  `        ``# As this subarray is sorted, we can quickly  ` `        ``# check if key lies in any of the halves  ` `        ``if` `(key >``=` `arr[l] ``and` `key <``=` `arr[mid]) : ` `            ``return` `search(arr, l, mid ``-` `1``, key);  ` ` `  `        ``# If key does not lie in the first half  ` `        ``# subarray then divide the other half  ` `        ``# into two subarrays such that we can  ` `        ``# quickly check if key lies in the other half  ` `        ``return` `search(arr, mid ``+` `1``, h, key);  ` ` `  `    ``# If arr[l..mid] first subarray is not sorted  ` `    ``# then arr[mid... h] must be sorted subarray  ` `    ``if` `(key >``=` `arr[mid] ``and` `key <``=` `arr[h]) : ` `        ``return` `search(arr, mid ``+` `1``, h, key);  ` ` `  `    ``return` `search(arr, l, mid ``-` `1``, key);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``3``, ``3``, ``1``, ``2``, ``3``, ``3` `];  ` `    ``n ``=` `len``(arr);  ` `    ``key ``=` `3``;  ` ` `  `    ``print``(search(arr, ``0``, n ``-` `1``, key));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to return the index of the  ` `    ``// key in arr[l..h] if the key is present  ` `    ``// otherwise return -1  ` `    ``static` `int` `search(``int` `[]arr, ``int` `l, ``int` `h, ``int` `key)  ` `    ``{  ` `        ``if` `(l > h)  ` `            ``return` `-1;  ` `     `  `        ``int` `mid = (l + h) / 2;  ` `        ``if` `(arr[mid] == key)  ` `            ``return` `mid;  ` `     `  `        ``// The tricky case, just update left and right  ` `        ``if` `((arr[l] == arr[mid])  ` `            ``&& (arr[h] == arr[mid]))  ` `        ``{  ` `            ``++l;  ` `            ``--h;  ` `        ``}  ` `     `  `        ``// If arr[l...mid] is sorted  ` `        ``if` `(arr[l] <= arr[mid])  ` `        ``{  ` `     `  `            ``// As this subarray is sorted, we can quickly  ` `            ``// check if key lies in any of the halves  ` `            ``if` `(key >= arr[l] && key <= arr[mid])  ` `                ``return` `search(arr, l, mid - 1, key);  ` `     `  `            ``// If key does not lie in the first half  ` `            ``// subarray then divide the other half  ` `            ``// into two subarrays such that we can  ` `            ``// quickly check if key lies in the other half  ` `            ``return` `search(arr, mid + 1, h, key);  ` `        ``}  ` `     `  `        ``// If arr[l..mid] first subarray is not sorted  ` `        ``// then arr[mid... h] must be sorted subarray  ` `        ``if` `(key >= arr[mid] && key <= arr[h])  ` `            ``return` `search(arr, mid + 1, h, key);  ` `     `  `        ``return` `search(arr, l, mid - 1, key);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main ()  ` `    ``{  ` `        ``int` `[]arr = { 3, 3, 1, 2, 3, 3 };  ` `        ``int` `n = arr.Length;  ` `        ``int` `key = 3;  ` `     `  `        ``Console.WriteLine(search(arr, 0, n - 1, key));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output:

```4
```

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