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.

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++

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// 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 -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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output:

4

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