Find the element that appears once in a sorted array
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.
Below is the implementation of the above approach.
C++
// C++ program to find the element that // appears only once #include <bits/stdc++.h> using namespace std; // A Linear Search based function to find // the element that appears only once void search( int arr[], int n) { int ans = -1; for ( int i = 0; i < n; i += 2) { if (arr[i] != arr[i + 1]) { ans = arr[i]; break ; } } if (arr[n - 2] != arr[n - 1]) ans = arr[n-1]; // ans = -1 if no such element is present. cout << "The required element is " << ans << "\n" ; } // Driver code int main() { int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = sizeof (arr) / sizeof (arr[0]); search(arr, len); return 0; } // This code is contributed by yashbeersingh42 |
Java
// Java program to find the element that // appears only once import java.io.*; class GFG { // A Linear Search based function to find // the element that appears only once static void search( int arr[], int n) { int ans = - 1 ; for ( int i = 0 ; i < n- 1 ; i += 2 ) { if (arr[i] != arr[i + 1 ]) { ans = arr[i]; break ; } } if (arr[n - 2 ] != arr[n - 1 ]) ans = arr[n- 1 ]; // ans = -1 if no such element is present. System.out.println( "The required element is " + ans); } public static void main(String[] args) { int arr[] = { 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 }; int len = arr.length; search(arr, len); } } |
Python3
# Python3 program to find the element that # appears only once # A Linear Search based function to find # the element that appears only once def search(arr, n): ans = - 1 for i in range ( 0 , n, 2 ): if (arr[i] ! = arr[i + 1 ]): ans = arr[i] break if (arr[n - 2 ] ! = arr[n - 1 ]): ans = arr[n - 1 ] # ans = -1 if no such element is present. print ( "The required element is" , ans) # Driver code arr = [ 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 ] Len = len (arr) search(arr, Len ) # This code is contributed by divyesh072019 |
C#
// C# program to find the element that // appears only once using System; class GFG { // A Linear Search based function to find // the element that appears only once static void search( int [] arr, int n) { int ans = -1; for ( int i = 0; i < n; i += 2) { if (arr[i] != arr[i + 1]) { ans = arr[i]; break ; } } if (arr[n - 2] != arr[n - 1]) ans = arr[n-1]; // ans = -1 if no such element is present. Console.Write( "The required element is " + ans); } public static void Main(String[] args) { int [] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = arr.Length; search(arr, len); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript program to find the element that // appears only once // A Linear Search based function to find // the element that appears only once function search(arr, n) { let ans = -1; for (let i = 0; i < n; i += 2) { if (arr[i] != arr[i + 1]) { ans = arr[i]; break ; } } if (arr[n - 2] != arr[n - 1]) ans = arr[n-1]; // ans = -1 if no such element is present. document.write( "The required element is " + ans + "<br>" ); } // Driver code let arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ]; let len = arr.length; search(arr, len); // This code is contributed by Surbhi Tyagi </script> |
The required element is 2
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
Another Simple Solution is to use the properties of XOR (a ^ a = 0 & a ^ 0 = a). The idea is to find the XOR of the complete array. The XOR of the array is the required answer.
Below is the implementation of the above approach.
C++
// C++ program to find the element that // appears only once #include <bits/stdc++.h> using namespace std; // A XOR based function to find // the element that appears only once void search( int arr[], int n) { int XOR = 0; for ( int i = 0; i < n; i++) { XOR = XOR ^ arr[i]; } cout << "The required element is " << XOR << "\n" ; } // Driver code int main() { int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = sizeof (arr) / sizeof (arr[0]); search(arr, len); return 0; } // This code is contributed by yashbeersingh42 |
Java
// Java program to find the element that // appears only once import java.io.*; class GFG { // A XOR based function to find // the element that appears only once static void search( int arr[], int n) { int XOR = 0 ; for ( int i = 0 ; i < n; i++) { XOR = XOR ^ arr[i]; } System.out.println( "The required element is " + XOR); } // Driver Code public static void main(String[] args) { int arr[] = { 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 }; int len = arr.length; search(arr, len); } } // This code is contributed by yashbeersingh42 |
Python3
# Python3 program to find the element that # appears only once # A XOR based function to find # the element that appears only once def search(arr, n) : XOR = 0 for i in range (n) : XOR = XOR ^ arr[i] print ( "The required element is" , XOR) # Driver code arr = [ 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 ] Len = len (arr) search(arr, Len ) # This code is contributed by divyesh072019 |
C#
// C# program to find the element that // appears only once using System; class GFG{ // A XOR based function to find // the element that appears only once static void search( int []arr, int n) { int XOR = 0; for ( int i = 0; i < n; i++) { XOR = XOR ^ arr[i]; } Console.Write( "The required element is " + XOR); } // Driver Code public static void Main(String[] args) { int []arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = arr.Length; search(arr, len); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript program to find the element that // appears only once // A XOR based function to find // the element that appears only once function search(arr, n) { let XOR = 0; for (let i = 0; i < n; i++) { XOR = XOR ^ arr[i]; } document.write( "The required element is " + XOR + "<br>" ); } // Driver code let arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ]; let len = arr.length; search(arr, len); // This code is contributed by Surbhi Tyagi. </script> |
The required element is 2
Time Complexity: O(n)
Auxiliary Space: O(1)
An Efficient Solution can find the required element in O(Log n) time. The idea is to use Binary Search. Below is an observation on the input array.
All elements before the required have the first occurrence at even index (0, 2, ..) and the next occurrence at odd index (1, 3, …). And all elements after the required elements have the first occurrence at an odd index and the next occurrence at an even index.
- Find the middle index, say ‘mid’.
- If ‘mid’ is even, then compare arr[mid] and arr[mid + 1]. If both are the same, then the required element after ‘mid’ and else before mid.
- If ‘mid’ is odd, then compare arr[mid] and arr[mid – 1]. If both are the same, then the required element after ‘mid’ and else before mid.
Below is the implementation based on the 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 |
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 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; } |
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 ); } } // 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 Tanisha Mittal |
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 ) # Driver Code # 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" |
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. |
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 ?> |
Javascript
<script> // Javascript implementation // 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) { document.write( "The required element is " + arr[low]); return ; } // Find the middle point var mid = Math.floor((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 var arr = [1, 1, 2, 4, 4, 5, 5, 6, 6]; var len = arr.length; search(arr, 0, len - 1) // This is code is contributed // by shubhamsingh10 </script> |
The required element is 2
Time Complexity: O(Log n)
Auxiliary Space: O(1)
Note: Other Solutions to the question are slight variations of the approaches discussed in this post.
Another Approach: An Efficient Solution can find the required element in O(Log n) time. The idea is to use Binary Search without recursion. All elements before the required have the first occurrence at even index (0, 2, ..and so on) and the next occurrence at odd index (1, 3, ..and so on).
The approach will be as follows:
Find the middle index assuming mid using start pointer and end pointer. And check the mid element in the following cases
- Case 1) If mid element is not equal to mid+1 element and mid-1 element. This case returns the answer.
- Case 2) When mid element is even and equal to mid+1 element this means number is not present in the left side of the array. In this case start pointer will change to mid+1.
- Case 3) When mid element is odd and equal to mid-1 element this means number is not present in the left side of the array. In this case start pointer will change to mid+1.
- Case 4) When mid element is odd and equal to mid+1 element this means number is not present in the right side of the array. In this case end pointer will change to mid-1.
- Case 5) When mid element is even and equal to mid-1 element this means number is not present in the right side of the array. In this case end pointer will change to mid-1.
Check for all case for possible values of mid till start<=end..
If all checks fail there is no such element.
This solution requires extra checks for edge cases.
- Edge Case 1) If only one element is present in the array. Therefore return the only element of the array.
- Edge Case 2) If last element of the array is the required element. Therefore return the last element of the array.
- Edge Case 3) If first element of the array is the required element. Therefore return the first element of the array.
Below is the implementation based on the above idea:
C++
#include <iostream> using namespace std; int search( int nums[], int n) { // A Binary Search based method to find the element // that appears only once int start = 0, end = n - 1, mid; // For Edge Cases if (n == 1) // If only one element is in the array return nums[0]; if (nums[start] != nums[start + 1]) // If the first element // is the element that // appears only once return nums[start]; if (nums[end] != nums[end - 1]) // If Last element is the element // that appears only once return nums[end]; // Binary Search while (start <= end) { mid = start + (end - start) / 2; // CASE 1 if (nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1]) return nums[mid]; // CASE 2 and CASE 3 else if ((nums[mid] == nums[mid + 1] && mid % 2 == 0) || (nums[mid] == nums[mid - 1] && mid % 2 != 0)) start = mid + 1; // CASE 4 and CASE 5 else end = mid - 1; } // If no such element found return -1; } // Driver code int main() { int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int n = sizeof (arr) / sizeof (arr[0]); int element = search(arr, n); if (element != -1) cout << "The required element is " << element; else cout << "There is no such element" ; } // This code is contributed by umadevi9616 |
Java
class GFG { public static int search( int [] nums) { // A Binary Search based method to find the element // that appears only once int start = 0 , end = nums.length - 1 , mid; // For Edge Cases if (nums.length == 1 ) // If only one element is in the array return nums[ 0 ]; if (nums[start] != nums[start + 1 ]) // If the first element // is the element that // appears only once return nums[start]; if (nums[end] != nums[end - 1 ]) // If Last element is the element // that appears only once return nums[end]; // Binary Search while (start <= end) { mid = start + (end - start) / 2 ; // CASE 1 if (nums[mid] != nums[mid - 1 ] && nums[mid] != nums[mid + 1 ]) return nums[mid]; // CASE 2 and CASE 3 else if ((nums[mid] == nums[mid + 1 ] && mid % 2 == 0 ) || (nums[mid] == nums[mid - 1 ] && mid % 2 != 0 )) start = mid + 1 ; // CASE 4 and CASE 5 else end = mid - 1 ; } // If no such element found return - 1 ; } public static void main(String[] args) { int [] arr = { 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 }; int element = search(arr); if (element != - 1 ) System.out.println( "The required element is " + element); else System.out.println( "There is no such element" ); } } // Code Contributed by Arnav Sharma |
Python3
def search(nums): # A Binary Search based method to find the element # that appears only once start = 0 ; end = len (nums) - 1 ; mid = 0 ; # For Edge Cases if ( len (nums) = = 1 ): # If only one element is in the array return nums[ 0 ]; if (nums[start] ! = nums[start + 1 ]): # If the first element # is the element that # appears only once return nums[start]; if (nums[end] ! = nums[end - 1 ]): # If Last element is the element # that appears only once return nums[end]; # Binary Search while (start < = end): mid = start + (end - start) / / 2 ; # CASE 1 if (nums[mid] ! = nums[mid - 1 ] and nums[mid] ! = nums[mid + 1 ]): return nums[mid]; # CASE 2 and CASE 3 elif ((nums[mid] = = nums[mid + 1 ] and mid % 2 = = 0 ) or (nums[mid] = = nums[mid - 1 ] and mid % 2 ! = 0 )): start = mid + 1 ; # CASE 4 and CASE 5 else : end = mid - 1 ; # If no such element found return - 1 ; # Driver code if __name__ = = '__main__' : arr = [ 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 ]; element = search(arr); if (element ! = - 1 ): print ( "The required element is " , element); else : print ( "There is no such element" ); # This code is contributed by umadevi9616 |
C#
using System; public class GFG { public static int search( int [] nums) { // A Binary Search based method to find the element // that appears only once int start = 0, end = nums.Length - 1, mid; // For Edge Cases if (nums.Length == 1) // If only one element is in the array return nums[0]; if (nums[start] != nums[start + 1]) // If the first element // is the element that // appears only once return nums[start]; if (nums[end] != nums[end - 1]) // If Last element is the element // that appears only once return nums[end]; // Binary Search while (start <= end) { mid = start + (end - start) / 2; // CASE 1 if (nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1]) return nums[mid]; // CASE 2 and CASE 3 else if ((nums[mid] == nums[mid + 1] && mid % 2 == 0) || (nums[mid] == nums[mid - 1] && mid % 2 != 0)) start = mid + 1; // CASE 4 and CASE 5 else end = mid - 1; } // If no such element found return -1; } public static void Main(String[] args) { int [] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int element = search(arr); if (element != -1) Console.WriteLine( "The required element is " + element); else Console.WriteLine( "There is no such element" ); } } // This code is contributed by gauravrajput1 |
Javascript
<script> function search(nums) { // A Binary Search based method to find the element // that appears only once var start = 0, end = nums.length - 1, mid; // For Edge Cases if (nums.length == 1) // If only one element is in the array return nums[0]; if (nums[start] != nums[start + 1]) // If the first element // is the element that // appears only once return nums[start]; if (nums[end] != nums[end - 1]) // If Last element is the element // that appears only once return nums[end]; // Binary Search while (start <= end) { mid = start + (end - start) / 2; // CASE 1 if (nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1]) return nums[mid]; // CASE 2 and CASE 3 else if ((nums[mid] == nums[mid + 1] && mid % 2 == 0) || (nums[mid] == nums[mid - 1] && mid % 2 != 0)) start = mid + 1; // CASE 4 and CASE 5 else end = mid - 1; } // If no such element found return -1; } var arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ]; var element = search(arr); if (element = 2) document.write( "The required element is " + element); else document.write( "There is no such element" ); // This code is contributed by shivanisinghss2110 </script> |
The required element is 2
This solution is contributed by Arnav Sharma.
Time Complexity: O(logn)
Auxiliary Space: O(1)
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.
Another Approach:-
We can simply use hashmap to store the frequency of the elements and after that we can just traverse the hashmap to find the element with frequency 1.
Implementation:
C++
// C++ program to find the element that // appears only once #include <bits/stdc++.h> using namespace std; //function to find element using hashmap void search( int arr[], int n) { //taking hashmap to store frequency unordered_map< int , int > mm; //iterating over array for ( int i=0;i<n;i++) { //storing frequency mm[arr[i]]++; } //iterating over map for ( auto x:mm) { //if element found if (x.second==1) { //printing element cout<<x.first<<endl; break ; } } } // Driver code int main() { int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 }; int len = sizeof (arr) / sizeof (arr[0]); search(arr, len); return 0; } // This code is contributed by shubhamrajput6156 |
2
Time Complexity:- O(N)
Auxiliary Space:- O(N)
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