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++ 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 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 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# 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 |
<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++ 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 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 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# 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 |
<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++ 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 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 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 |
# 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# 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. |
<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> |
<?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 ?> |
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:
#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 |
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 |
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 |
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 |
<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)
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++ 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 |
// Java program to find the element that // appears only once import java.util.*;
class GFG {
// function to find element using hashmap
static void search( int arr[], int n)
{
// taking hashmap to store frequency
HashMap<Integer,Integer> mm= new HashMap<Integer,Integer>();
// iterating over array
for ( int i = 0 ; i < arr.length; i++) {
if (mm.containsKey(arr[i])) {
int count = mm.get(arr[i]) + 1 ;
mm.put(arr[i], count);
} else {
mm.put(arr[i], 1 );
}
}
// iterating over map
for (Map.Entry<Integer,Integer> x : mm.entrySet()){
int c= x.getValue();
// if element found
if (c == 1 ){
// printing element
System.out.println(x.getKey());
break ;
}
}
}
// Driver Code
public static void main(String[] args) {
int arr[] = { 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 };
int len = arr.length;
// function calling
search(arr, len);
return ;
}
} // This code is contributed by bhardwajji |
# Python3 program to find the element that # appears only once # function to find element using dictionary def search(arr):
# taking dictionary to store frequency
freq = {}
# iterating over array
for i in arr:
# storing frequency
if i in freq:
freq[i] + = 1
else :
freq[i] = 1
# iterating over dictionary
for key, value in freq.items():
# if element found
if value = = 1 :
# printing element
print (key)
break
# Driver code if __name__ = = '__main__' :
arr = [ 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 ]
search(arr)
|
using System;
using System.Collections.Generic;
class GFG {
// function to find element using Dictionary
static void Search( int [] arr, int n) {
// taking Dictionary to store frequency
Dictionary< int , int > mm = new Dictionary< int , int >();
// iterating over array
for ( int i = 0; i < arr.Length; i++) {
if (mm.ContainsKey(arr[i])) {
int count = mm[arr[i]] + 1;
mm[arr[i]] = count;
} else {
mm.Add(arr[i], 1);
}
}
// iterating over Dictionary
foreach (KeyValuePair< int , int > x in mm) {
int c = x.Value;
// if element found
if (c == 1){
// printing element
Console.WriteLine(x.Key);
break ;
}
}
}
// Driver Code
public static void Main( string [] args) {
int [] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.Length;
// function calling
Search(arr, len);
return ;
}
} |
// Javascript equivalent of above code // function to find element using dictionary function search(arr){
// taking dictionary to store frequency
let freq = {};
// iterating over array
for (let i of arr){
// storing frequency
if (i in freq){
freq[i] += 1;
}
else {
freq[i] = 1;
}
}
// iterating over dictionary
for (let [key, value] of Object.entries(freq)){
// if element found
if (value == 1){
// printing element
console.log(key);
break ;
}
}
} // Driver code arr = [1, 1, 2, 4, 4, 5, 5, 6, 6];
search(arr);
|
2
Time Complexity:- O(N)
Auxiliary Space:- O(N)
Another Approach Using Stacks:
We know that the given array is sorted in non-decreasing (ascending) order, and that every element occurs twice except for 1 element. The approach involves using a stack while iterating through the entire array. While pushing in elements, we may push this singular element on to the stack. By the end of the iteration we will have that singular element on the top of the stack which is returned.
While iterating through the array, if we have no element in the stack, or the consective element is not equal to the top most element in the stack, we can push that element to the stack. Else we can pop the top element from the stack. Finally, we can return the top most element, which is the singular element in the array.
// C++ program to find the element that // appears only once #include <bits/stdc++.h> using namespace std;
// function to find element using stack int singleNonDuplicate(vector< int >& nums) {
//declaring the stack for stroing elements
stack< int > stk;
//traversing linearly over the entire array
for ( int i = 0; i < nums.size(); i++) {
/*checking if element is presnt in the stack
or the current array element equal to stack's top element*/
if (stk.empty() or nums[i] != stk.top()) {
//pushing the element in the stack if conditions satisfied
stk.push(nums[i]);
} else {
//If condition doesn't match pop elements from the stack
stk.pop();
}
}
//return the topmost element from the stack
return stk.top();
}
// Driver code int main()
{ //declaring a vector
vector< int > arr{ 1, 1, 2, 4, 4, 5, 5, 6, 6 };
//calling function
cout<<singleNonDuplicate(arr)<<endl;
return 0;
} // This code is contributed by shubhamrajput6156 |
// Java program to find the element that // appears only once import java.util.Stack;
public class GFG {
// function to find element using stack
public static int singleNonDuplicate( int [] nums) {
// declaring the stack for storing elements
Stack<Integer> stk = new Stack<>();
// traversing linearly over the entire array
for ( int i = 0 ; i < nums.length; i++) {
/* checking if the stack is empty or the current array element is not equal to
stack's top element */
if (stk.isEmpty() || nums[i] != stk.peek()) {
// pushing the element into the stack if the conditions are satisfied
stk.push(nums[i]);
} else {
// If condition doesn't match, pop elements from the stack
stk.pop();
}
}
// return the topmost element from the stack which is the single non-duplicate element
return stk.peek();
}
// Driver code
public static void main(String[] args) {
// declaring an array
int [] arr = { 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 };
// calling function
System.out.println(singleNonDuplicate(arr));
}
} |
# Python3 program to find the element that # appears only once # function to find element using stack def singleNonDuplicate(nums):
# Declaring an empty list to act as a stack for storing elements
stack = []
# Traversing linearly over the entire array
for num in nums:
# Checking if the stack is empty or if the current array element is not equal to the stack's top element
if not stack or num ! = stack[ - 1 ]:
# Pushing the element into the stack if the conditions are satisfied
stack.append(num)
else :
# If the condition doesn't match, it means the current element is a duplicate of the top element in the stack
# So, popping the top element from the stack effectively removes the pair from the stack
stack.pop()
# The remaining element in the stack is the element that appears only once
return stack[ - 1 ]
# Driver code if __name__ = = "__main__" :
# Declaring a list
arr = [ 1 , 1 , 2 , 4 , 4 , 5 , 5 , 6 , 6 ]
# Calling the function
print (singleNonDuplicate(arr))
|
using System;
using System.Collections.Generic;
namespace SingleNonDuplicate
{ class Program
{
static int SingleNonDuplicate(List< int > nums)
{
// declaring the stack for storing elements
Stack< int > stk = new Stack< int >();
// traversing linearly over the entire array
foreach ( int num in nums)
{
// checking if element is present in the stack
// or the current array element equal to stack's top element
if (stk.Count == 0 || num != stk.Peek())
{
// pushing the element in the stack if conditions satisfied
stk.Push(num);
}
else
{
// if condition doesn't match, pop elements from the stack
stk.Pop();
}
}
// return the topmost element from the stack
return stk.Peek();
}
static void Main( string [] args)
{
// declaring a list
List< int > arr = new List< int > { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
// calling function
Console.WriteLine(SingleNonDuplicate(arr));
}
}
} // This code is contributed by rambabuguphka |
function singleNonDuplicate(nums) {
// Initialize an empty stack
const stack = [];
// Traverse the array linearly
for (let i = 0; i < nums.length; i++) {
// Check if the stack is empty or the current
// array element is not equal to the stack's top element
if (stack.length === 0 || nums[i] !== stack[stack.length - 1]) {
// Push the element onto the stack if the conditions are satisfied
stack.push(nums[i]);
} else {
// If the condition doesn't match, pop elements from the stack
stack.pop();
}
}
// Return the topmost element from the stack
return stack[0];
} // Driver code const arr = [1, 1, 2, 4, 4, 5, 5, 6, 6]; console.log(singleNonDuplicate(arr)); // Output: 2
|
Output:
2
Time Complexity:- O(N) for traversing through entire array linearly.
Auxiliary Space:- O(N) for stroing the elements in stack.