Given a sorted array in which all elements appear twice (one after one) and one element appears only once. Find that element in O(log n) complexity.
Example:
Input: arr[] = {1, 1, 3, 3, 4, 5, 5, 7, 7, 8, 8}
Output: 4
Input: arr[] = {1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8}
Output: 8
A Simple Solution is to traverse the array from left to right. Since the array is sorted, we can easily figure out the required element.
An Efficient Solution can find the required element in O(Log n) time. The idea is to use Binary Search. Below is an observation in the input array.
All elements before the required have the first occurrence at even index (0, 2, ..) and next occurrence at odd index (1, 3, …). And all elements after the required elements have the first occurrence at odd index and next occurrence at even index.
1) Find the middle index, say ‘mid’.
2) If ‘mid’ is even, then compare arr[mid] and arr[mid + 1]. If both are the same, then the required element after ‘mid’ else before mid.
3) If ‘mid’ is odd, then compare arr[mid] and arr[mid – 1]. If both are the same, then the required element after ‘mid’ else before mid.
Below is the implementation based on above idea.
C++
// C++ program to find the element that // appears only once #include <iostream> using namespace std; // A Binary Search based function to find // the element that appears only once void search(int arr[], int low, int high) { // Base cases if (low > high) return; if (low == high) { cout << "The required element is " << arr[low]; return; } // Find the middle point int mid = (low + high) / 2; // If mid is even and element next to mid is // same as mid, then output element lies on // right side, else on left side if (mid % 2 == 0) { if (arr[mid] == arr[mid + 1]) search(arr, mid + 2, high); else search(arr, low, mid); } // If mid is odd else { if (arr[mid] == arr[mid - 1]) search(arr, mid + 1, high); else search(arr, low, mid - 1); } } // Driver code int main() { int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = sizeof(arr)/sizeof(arr[0]); search(arr, 0, len - 1); return 0; } // This code is contributed by ShubhamCoder |
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C
// C program to find the element that appears only once #include<stdio.h> // A Binary Search based function to find the element // that appears only once void search(int *arr, int low, int high) { // Base cases if (low > high) return; if (low==high) { printf("The required element is %d ", arr[low]); return; } // Find the middle point int mid = (low + high) / 2; // If mid is even and element next to mid is // same as mid, then output element lies on // right side, else on left side if (mid%2 == 0) { if (arr[mid] == arr[mid+1]) search(arr, mid+2, high); else search(arr, low, mid); } else // If mid is odd { if (arr[mid] == arr[mid-1]) search(arr, mid+1, high); else search(arr, low, mid-1); } } // Driver program int main() { int arr[] = {1, 1, 2, 4, 4, 5, 5, 6, 6}; int len = sizeof(arr)/sizeof(arr[0]); search(arr, 0, len-1); return 0; } |
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Java
// Java program to find the element that appears only once public class Main { // A Binary Search based method to find the element // that appears only once public static void search(int[] arr, int low, int high) { if(low > high) return; if(low == high) { System.out.println("The required element is "+arr[low]); return; } // Find the middle point int mid = (low + high)/2; // If mid is even and element next to mid is // same as mid, then output element lies on // right side, else on left side if(mid % 2 == 0) { if(arr[mid] == arr[mid+1]) search(arr, mid+2, high); else search(arr, low, mid); } // If mid is odd else if(mid % 2 == 1) { if(arr[mid] == arr[mid-1]) search(arr, mid+1, high); else search(arr, low, mid-1); } } public static void main(String[] args) { int[] arr = {1, 1, 2, 4, 4, 5, 5, 6, 6}; search(arr, 0, arr.length-1); } } // This code is contributed by Tanisha Mittal |
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Python
# A Binary search based function to find # the element that appears only once def search(arr, low, high): # Base cases if low > high: return None if low == high: return arr[low] # Find the middle point mid = low + (high - low)/2 # If mid is even and element next to mid is # same as mid, then output element lies on # right side, else on left side if mid%2 == 0: if arr[mid] == arr[mid+1]: return search(arr, mid+2, high) else: return search(arr, low, mid) else: # if mid is odd if arr[mid] == arr[mid-1]: return search(arr, mid+1, high) else: return search(arr, low, mid-1) # Test Array arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ] # Function call result = search(arr, 0, len(arr)-1) if result is not None: print "The required element is %d" % result else: print "Invalid Array" |
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C#
// C# program to find the element // that appears only once using System; class GFG { // A Binary Search based // method to find the element // that appears only once public static void search(int[] arr, int low, int high) { if(low > high) return; if(low == high) { Console.WriteLine("The required element is " +arr[low]); return; } // Find the middle point int mid = (low + high)/2; // If mid is even and element // next to mid is same as mid // then output element lies on // right side, else on left side if(mid % 2 == 0) { if(arr[mid] == arr[mid + 1]) search(arr, mid + 2, high); else search(arr, low, mid); } // If mid is odd else if(mid % 2 == 1) { if(arr[mid] == arr[mid - 1]) search(arr, mid + 1, high); else search(arr, low, mid - 1); } } // Driver Code public static void Main(String[] args) { int[] arr = {1, 1, 2, 4, 4, 5, 5, 6, 6}; search(arr, 0, arr.Length - 1); } } // This code is contributed by Nitin Mittal. |
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PHP
<?php // PHP program to find the element // that appears only once // A Binary Search based function // to find the element that // appears only once function search($arr, $low, $high) { // Base cases if ($low > $high) return; if ($low==$high) { echo("The required element is " ); echo $arr[$low] ; return; } // Find the middle point $mid = ($low + $high) / 2; // If mid is even and element // next to mid is same as mid, // then output element lies on // right side, else on left side if ($mid % 2 == 0) { if ($arr[$mid] == $arr[$mid + 1]) search($arr, $mid + 2, $high); else search($arr, $low, $mid); } // If mid is odd else { if ($arr[$mid] == $arr[$mid - 1]) search($arr, $mid + 1, $high); else search($arr, $low, $mid - 1); } } // Driver Code $arr = array(1, 1, 2, 4, 4, 5, 5, 6, 6); $len = sizeof($arr); search($arr, 0, $len - 1); // This code is contributed by nitin mittal ?> |
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Output:
The required element is 2
Time Complexity: O(Log n)
This article is contributed by Mehboob Elahi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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