Search an element in a sorted and rotated array with duplicates

27 May, 2022
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. 
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.
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Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the index of the// key in arr[l..h] if the key is present// otherwise return -1int search(int arr[], int l, int h, int key){    if (l > h)        return -1;     int mid = l + (h - l) / 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;        return search(arr, l, h, key);    }     // 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 codeint 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 implementation of the approachclass 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 - l) / 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--;              return search(arr,l,h,key);        }             // If arr[l...mid] is sorted        else 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            else              return search(arr, mid + 1, h, key);        }             // If arr[l..mid] first subarray is not sorted        // then arr[mid... h] must be sorted subarray           else  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 implementation of the approach # Function to return the index of the# key in arr[l..h] if the key is present# otherwise return -1def 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;        return search(arr, l, h, key)    # 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 codeif __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# implementation of the approachusing 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 - l) / 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;            return search(arr, l, h, key)        }             // 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
<script>    // Javascript implementation of the approach         // Function to return the index of the    // key in arr[l..h] if the key is present    // otherwise return -1    function search(arr, l, h, key)    {        if (l > h)            return -1;              let mid = parseInt((l + h) / 2, 10);        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;            return search(arr, l, h, key)        }              // 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);    }         let arr = [ 3, 3, 1, 2, 3, 3 ];    let n = arr.length;    let key = 3;     document.write(search(arr, 0, n - 1, key));         </script>
4
Time Complexity: O(logn), where n represents the size of the given array.Auxiliary Space: O(logn) due to recursive stack space.
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