Search an element in a sorted and rotated array with duplicates

Last Updated : 27 Feb, 2023

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++

// 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 - 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
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 - 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

# 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; 
        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 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 - 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 

Javascript

<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>

Output

Time Complexity: O(log N), where n represents the size of the given array. If all the elements are same, we may end up doing a linear search.
Auxiliary Space: O(logN) due to recursive stack space.

This approach is better than the recursive one as no auxiliary space is required in this case

C++

// C++ implementation of the above approach using iterative
// version of binary search
#include <bits/stdc++.h>
using namespace std;
 
int search(int arr[], int low, int high, int key)
{
 
    while (low <= high) {
        int mid = low + (high - low) / 2;
 
        if (arr[mid] == key) {
            // if we have found our target element
            // return the index of target element
            return mid;
        }
 
        if (arr[mid] == arr[low] && arr[mid] == arr[high]) {
            // It may happen in case of duplicates
 
            ++low;
            --high;
            continue;
        }
 
        if (arr[low] <= arr[mid]) {
            // This means array is sorted from index low to
            // mid We will check that if target element lies
            // in left half or not
 
            if (key >= arr[low] && key < arr[mid])
                high = mid - 1;
 
            else
                // This means that our target lies in other
                // half of array So we shift low to mid+1 to
                // search in right half
                low = mid + 1;
        }
 
        else {
            // This means array is sorted between mid and
            // high index
 
            // This will check our target element is
            // in right half or not
            if (key <= arr[high] && key > arr[mid])
                low = mid + 1;
 
            else
                // Means our target is in left half
                high = mid - 1;
        }
    }
 
    // If target element is not present
 
    return -1;
}
// Driver Code
int main()
{
    int arr[] = { 3, 3, 1, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int key = 3;
    cout << search(arr, 0, n - 1, key) << endl;
 
    return 0;
}

Java

import java.util.Scanner;
 
public class Main {
    public static int search(int[] arr, int low, int high, int key) {
        while (low <= high) {
            int mid = low + (high - low) / 2;
 
            if (arr[mid] == key) {
                // if we have found our target element
                // return the index of target element
                return mid;
            }
 
            if (arr[mid] == arr[low] && arr[mid] == arr[high]) {
                // It may happen in case of duplicates
 
                ++low;
                --high;
                continue;
            }
 
            if (arr[low] <= arr[mid]) {
                // This means array is sorted from index low to
                // mid We will check that if target element lies
                // in left half or not
 
                if (key >= arr[low] && key < arr[mid])
                    high = mid - 1;
 
                else
                    // This means that our target lies in other
                    // half of array So we shift low to mid+1 to
                    // search in right half
                    low = mid + 1;
            }
 
            else {
                // This means array is sorted between mid and
                // high index
 
                // This will check our target element is
                // in right half or not
                if (key <= arr[high] && key > arr[mid])
                    low = mid + 1;
 
                else
                    // Means our target is in left half
                    high = mid - 1;
            }
        }
 
        // If target element is not present
 
        return -1;
    }
 
    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));
    }
}

Python3

def search(arr, low, high, key):
 
    while low <= high:
        mid = low + (high - low) // 2
 
        if arr[mid] == key:
            # if we have found our target element
            # return the index of target element
            return mid
 
        if arr[mid] == arr[low] and arr[mid] == arr[high]:
            # It may happen in case of duplicates
            low += 1
            high -= 1
            continue
 
        if arr[low] <= arr[mid]:
            # This means array is sorted from index low to
            # mid We will check that if target element lies
            # in left half or not
 
            if key >= arr[low] and key < arr[mid]:
                high = mid - 1
            else:
                # This means that our target lies in other
                # half of array So we shift low to mid+1 to
                # search in right half
                low = mid + 1
 
        else:
            # This means array is sorted between mid and
            # high index
 
            # This will check our target element is
            # in right half or not
            if key <= arr[high] and key > arr[mid]:
                low = mid + 1
            else:
                # Means our target is in left half
                high = mid - 1
 
    # If target element is not present
    return -1
 
# Driver Code
arr = [3, 3, 1, 2, 3, 3]
n = len(arr)
 
key = 3
print(search(arr, 0, n - 1, key))
 
# This code is contributed by Prince Kumar

C#

using System;
 
class MainClass {
  public static int Search(int[] arr, int low, int high, int key) {
    while (low <= high) {
      int mid = low + (high - low) / 2;
 
      if (arr[mid] == key) {
        // if we have found our target element
        // return the index of target element
        return mid;
      }
 
      if (arr[mid] == arr[low] && arr[mid] == arr[high]) {
        // It may happen in case of duplicates
        ++low;
        --high;
        continue;
      }
 
      if (arr[low] <= arr[mid]) {
        // This means array is sorted from index low to
        // mid We will check that if target element lies
        // in left half or not
        if (key >= arr[low] && key < arr[mid])
          high = mid - 1;
        else
          // This means that our target lies in other
          // half of array So we shift low to mid+1 to
          // search in right half
          low = mid + 1;
      }
      else {
        // This means array is sorted between mid and
        // high index
        // This will check our target element is
        // in right half or not
        if (key <= arr[high] && key > arr[mid])
          low = mid + 1;
        else
          // Means our target is in left half
          high = mid - 1;
      }
    }
 
    // If target element is not present
    return -1;
  }
 
  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));
  }
}

Javascript

// JavaScript implementation of the above approach
 
function search(arr, low, high, key) {
  while (low <= high) {
    let mid = Math.floor(low + (high - low) / 2);
 
    if (arr[mid] === key) {
      // if we have found our target element
      // return the index of target element
      return mid;
    }
 
    if (arr[mid] === arr[low] && arr[mid] === arr[high]) {
      // It may happen in case of duplicates
      low++;
      high--;
      continue;
    }
 
    if (arr[low] <= arr[mid]) {
      // This means array is sorted from index low to
      // mid We will check that if target element lies
      // in left half or not
 
      if (key >= arr[low] && key < arr[mid]) {
        high = mid - 1;
      } else {
        // This means that our target lies in other
        // half of array So we shift low to mid+1 to
        // search in right half
        low = mid + 1;
      }
    } else {
      // This means array is sorted between mid and
      // high index
 
      // This will check our target element is
      // in right half or not
      if (key <= arr[high] && key > arr[mid]) {
        low = mid + 1;
      } else {
        // Means our target is in left half
        high = mid - 1;
      }
    }
  }
 
  // If target element is not present
 
  return -1;
}
 
// Driver code
let arr = [3, 3, 1, 2, 3, 3];
let n = arr.length;
 
let key = 3;
console.log(search(arr, 0, n - 1, key));

Output

Time Complexity: O(N), where n represents the size of the given array. If all the elements are same, we may end up doing a linear search.
Auxiliary Space: O(1)

Suggest improvement

Find the Peak Element in a 2D Array/Matrix

Search for an element in a Mountain Array

Share your thoughts in the comments

Variants of Binary Search

Implementation of Binary Search in different languages

Comparison with other Searching

Some problems on Binary Search

Search an element in a sorted and rotated array with duplicates

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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