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Find the element that appears once

  • Difficulty Level : Hard
  • Last Updated : 26 Jul, 2021
 

Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. The expected time complexity is O(n) and O(1) extra space. 

Examples:

Input: arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3} 
Output:
In the given array all element appear three times except 2 which appears once.

Input: arr[] = {10, 20, 10, 30, 10, 30, 30} 
Output: 20 
In the given array all element appear three times except 20 which appears once. 

We can use sorting to do it in O(nLogn) time. We can also use hashing, it has the worst-case time complexity of O(n), but requires extra space.
The idea is to use bitwise operators for a solution that is O(n) time and uses O(1) extra space. The solution is not easy like other XOR-based solutions, because all elements appear an odd number of times here. The idea is taken from here.
Run a loop for all elements in the array. At the end of every iteration, maintain the following two values.
ones: The bits that have appeared 1st time or 4th time or 7th time .. etc.
twos: The bits that have appeared 2nd time or 5th time or 8th time .. etc.
Finally, we return the value of ‘ones’
How to maintain the values of ‘ones’ and ‘twos’? 
‘ones’ and ‘twos’ are initialized as 0. For every new element in the array, find out the common set bits in the new element and the previous value of ‘ones’. These common set bits are actually the bits that should be added to ‘twos’. So do bitwise XOR of the common set bits with ‘twos’. ‘twos’ also gets some extra bits that appear the third time. These extra bits are removed later. 
Update ‘ones’ by doing XOR of new element with the previous value of ‘ones’. There may be some bits that appear 3rd time. These extra bits are also removed later.
Both ‘ones’ and ‘twos’ contain those extra bits which appear 3rd time. Remove these extra bits by finding out common set bits in ‘ones’ and ‘twos’. 
 



 

Below is the implementation of the above approach: 

C++




// C++ program to find the element
// that occur only once
#include <bits/stdc++.h>
using namespace std;
 
int getSingle(int arr[], int n)
{
    int ones = 0, twos = 0;
 
    int common_bit_mask;
 
    // Let us take the example of
    // {3, 3, 2, 3} to understand
    // this
    for (int i = 0; i < n; i++) {
       
        /* The expression "one & arr[i]" gives the bits that
        are there in both 'ones' and new element from arr[].
        We add these bits to 'twos' using bitwise OR
 
        Value of 'twos' will be set as 0, 3, 3 and 1 after
        1st, 2nd, 3rd and 4th iterations respectively */
        twos = twos | (ones & arr[i]);
 
        /* XOR the new bits with previous 'ones' to get all
        bits appearing odd number of times
 
        Value of 'ones' will be set as 3, 0, 2 and 3 after
        1st, 2nd, 3rd and 4th iterations respectively */
        ones = ones ^ arr[i];
 
        /* The common bits are those bits which appear third
        time So these bits should not be there in both
        'ones' and 'twos'. common_bit_mask contains all
        these bits as 0, so that the bits can be removed
        from 'ones' and 'twos'
 
        Value of 'common_bit_mask' will be set as 00, 00, 01
        and 10 after 1st, 2nd, 3rd and 4th iterations
        respectively */
        common_bit_mask = ~(ones & twos);
 
        /* Remove common bits (the bits that appear third
        time) from 'ones'
 
        Value of 'ones' will be set as 3, 0, 0 and 2 after
        1st, 2nd, 3rd and 4th iterations respectively */
        ones &= common_bit_mask;
 
        /* Remove common bits (the bits that appear third
        time) from 'twos'
 
        Value of 'twos' will be set as 0, 3, 1 and 0 after
        1st, 2nd, 3rd and 4th itearations respectively */
        twos &= common_bit_mask;
 
        // uncomment this code to see intermediate values
        // printf (" %d %d n", ones, twos);
    }
 
    return ones;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 3, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The element with single occurrence is  "
         << getSingle(arr, n);
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// C program to find the element
// that occur only once
#include <stdio.h>
 
int getSingle(int arr[], int n)
{
    int ones = 0, twos = 0;
 
    int common_bit_mask;
 
    // Let us take the example of {3, 3, 2, 3} to understand this
    for (int i = 0; i < n; i++) {
        /* The expression "one & arr[i]" gives the bits that are
           there in both 'ones' and new element from arr[].  We
           add these bits to 'twos' using bitwise OR
 
           Value of 'twos' will be set as 0, 3, 3 and 1 after 1st,
           2nd, 3rd and 4th iterations respectively */
        twos = twos | (ones & arr[i]);
 
        /* XOR the new bits with previous 'ones' to get all bits
           appearing odd number of times
 
           Value of 'ones' will be set as 3, 0, 2 and 3 after 1st,
           2nd, 3rd and 4th iterations respectively */
        ones = ones ^ arr[i];
 
        /* The common bits are those bits which appear third time
           So these bits should not be there in both 'ones' and 'twos'.
           common_bit_mask contains all these bits as 0, so that the bits can
           be removed from 'ones' and 'twos'  
 
           Value of 'common_bit_mask' will be set as 00, 00, 01 and 10
           after 1st, 2nd, 3rd and 4th iterations respectively */
        common_bit_mask = ~(ones & twos);
 
        /* Remove common bits (the bits that appear third time) from 'ones'
             
           Value of 'ones' will be set as 3, 0, 0 and 2 after 1st,
           2nd, 3rd and 4th iterations respectively */
        ones &= common_bit_mask;
 
        /* Remove common bits (the bits that appear third time) from 'twos'
 
           Value of 'twos' will be set as 0, 3, 1 and 0 after 1st,
           2nd, 3rd and 4th itearations respectively */
        twos &= common_bit_mask;
 
        // uncomment this code to see intermediate values
        // printf (" %d %d n", ones, twos);
    }
 
    return ones;
}
 
int main()
{
    int arr[] = { 3, 3, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("The element with single occurrence is %d ",
           getSingle(arr, n));
    return 0;
}

Java




// Java code to find the element
// that occur only once
 
class GFG {
    // Method to find the element that occur only once
    static int getSingle(int arr[], int n)
    {
        int ones = 0, twos = 0;
        int common_bit_mask;
 
        for (int i = 0; i < n; i++) {
            /*"one & arr[i]" gives the bits that are there in
            both 'ones' and new element from arr[]. We
            add these bits to 'twos' using bitwise OR*/
            twos = twos | (ones & arr[i]);
 
            /*"one & arr[i]" gives the bits that are
            there in both 'ones' and new element from arr[].
            We add these bits to 'twos' using bitwise OR*/
            ones = ones ^ arr[i];
 
            /* The common bits are those bits which appear third time
            So these bits should not be there in both 'ones' and 'twos'.
            common_bit_mask contains all these bits as 0, so that the bits can
            be removed from 'ones' and 'twos'*/
            common_bit_mask = ~(ones & twos);
 
            /*Remove common bits (the bits that appear third time) from 'ones'*/
            ones &= common_bit_mask;
 
            /*Remove common bits (the bits that appear third time) from 'twos'*/
            twos &= common_bit_mask;
        }
        return ones;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int arr[] = { 3, 3, 2, 3 };
        int n = arr.length;
        System.out.println("The element with single occurrence is " + getSingle(arr, n));
    }
}
// Code contributed by Rishab Jain

Python3




# Python3 code to find the element that
# appears once
 
def getSingle(arr, n):
    ones = 0
    twos = 0
     
    for i in range(n):
        # one & arr[i]" gives the bits that
        # are there in both 'ones' and new
        # element from arr[]. We add these
        # bits to 'twos' using bitwise XOR
        twos = twos ^ (ones & arr[i])
         
        # one & arr[i]" gives the bits that
        # are there in both 'ones' and new
        # element from arr[]. We add these
        # bits to 'twos' using bitwise XOR
        ones = ones ^ arr[i]
         
        # The common bits are those bits
        # which appear third time. So these
        # bits should not be there in both
        # 'ones' and 'twos'. common_bit_mask
        # contains all these bits as 0, so
        # that the bits can be removed from
        # 'ones' and 'twos'
        common_bit_mask = ~(ones & twos)
         
        # Remove common bits (the bits that
        # appear third time) from 'ones'
        ones &= common_bit_mask
         
        # Remove common bits (the bits that
        # appear third time) from 'twos'
        twos &= common_bit_mask
    return ones
     
# driver code
arr = [3, 3, 2, 3]
n = len(arr)
print("The element with single occurrence is ",
        getSingle(arr, n))
 
# This code is contributed by "Abhishek Sharma 44"

C#




// C# code to find the element
// that occur only once
using System;
class GFG {
    // Method to find the element
    // that occur only once
    static int getSingle(int[] arr, int n)
    {
        int ones = 0, twos = 0;
        int common_bit_mask;
 
        for (int i = 0; i < n; i++) {
            // "one & arr[i]" gives the bits
            // that are there in both 'ones'
            // and new element from arr[].
            // We add these bits to 'twos'
            // using bitwise OR
            twos = twos | (ones & arr[i]);
 
            // "one & arr[i]" gives the bits
            // that are there in both 'ones'
            // and new element from arr[].
            // We add these bits to 'twos'
            // using bitwise OR
            ones = ones ^ arr[i];
 
            // The common bits are those bits
            // which appear third time So these
            // bits should not be there in both
            // 'ones' and 'twos'. common_bit_mask
            // contains all these bits as 0,
            // so that the bits can be removed
            // from 'ones' and 'twos'
            common_bit_mask = ~(ones & twos);
 
            // Remove common bits (the bits that
            // appear third time) from 'ones'
            ones &= common_bit_mask;
 
            // Remove common bits (the bits that
            // appear third time) from 'twos'
            twos &= common_bit_mask;
        }
        return ones;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 3, 3, 2, 3 };
        int n = arr.Length;
        Console.WriteLine("The element with single"
                          + "occurrence is " + getSingle(arr, n));
    }
}
 
// This Code is contributed by vt_m.

PHP




<?php
// PHP program to find the element
// that occur only once
 
function getSingle($arr, $n)
{
    $ones = 0; $twos = 0 ;
 
    $common_bit_mask;
 
    // Let us take the example of
    // {3, 3, 2, 3} to understand this
    for($i = 0; $i < $n; $i++ )
    {
        /* The expression "one & arr[i]"
        gives the bits that are there in
        both 'ones' and new element from
        arr[]. We add these bits to 'twos'
        using bitwise OR
        Value of 'twos' will be set as 0,
        3, 3 and 1 after 1st, 2nd, 3rd
        and 4th iterations respectively */
        $twos = $twos | ($ones & $arr[$i]);
 
 
        /* XOR the new bits with previous
        'ones' to get all bits appearing
        odd number of times
 
        Value of 'ones' will be set as 3,
        0, 2 and 3 after 1st, 2nd, 3rd and
        4th iterations respectively */
        $ones = $ones ^ $arr[$i];
 
        /* The common bits are those bits
        which appear third time. So these
        bits should not be there in both
        'ones' and 'twos'. common_bit_mask
        contains all these bits as 0, so
        that the bits can be removed from
        'ones' and 'twos'
 
        Value of 'common_bit_mask' will be
        set as 00, 00, 01 and 10 after 1st,
        2nd, 3rd and 4th iterations respectively */
        $common_bit_mask = ~($ones & $twos);
 
 
        /* Remove common bits (the bits
        that appear third time) from 'ones'
             
        Value of 'ones' will be set as 3,
        0, 0 and 2 after 1st, 2nd, 3rd
        and 4th iterations respectively */
        $ones &= $common_bit_mask;
 
 
        /* Remove common bits (the bits
        that appear third time) from 'twos'
 
        Value of 'twos' will be set as 0, 3,
        1 and 0 after 1st, 2nd, 3rd and 4th
        itearations respectively */
        $twos &= $common_bit_mask;
 
        // uncomment this code to see
        // intermediate values
        // printf (" %d %d n", ones, twos);
    }
 
    return $ones;
}
 
// Driver Code
$arr = array(3, 3, 2, 3);
$n = sizeof($arr);
echo "The element with single " .
                "occurrence is ",
             getSingle($arr, $n);
 
// This code is contributed by m_kit
?>

Javascript




<script>
 
// Javascript program for the above approach
 
    // Method to find the element that occur only once
    function getSingle(arr, n)
    {
        let ones = 0, twos = 0;
        let common_bit_mask;
 
        for (let i = 0; i < n; i++) {
            /*"one & arr[i]" gives the bits that are there in
            both 'ones' and new element from arr[]. We
            add these bits to 'twos' using bitwise OR*/
            twos = twos | (ones & arr[i]);
 
            /*"one & arr[i]" gives the bits that are
            there in both 'ones' and new element from arr[].
            We add these bits to 'twos' using bitwise OR*/
            ones = ones ^ arr[i];
 
            /* The common bits are those bits which appear third time
            So these bits should not be there in both 'ones' and 'twos'.
            common_bit_mask contains all these bits as 0, so that the bits can
            be removed from 'ones' and 'twos'*/
            common_bit_mask = ~(ones & twos);
 
            /*Remove common bits (the bits that appear third time) from 'ones'*/
            ones &= common_bit_mask;
 
            /*Remove common bits (the bits that appear third time) from 'twos'*/
            twos &= common_bit_mask;
        }
        return ones;
    }
 
// Driver Code
 
    let arr = [ 3, 3, 2, 3 ];
    let n = arr.length;
    document.write("The element with single occurrence is " + getSingle(arr, n));
 
</script>
Output
The element with single occurrence is  2

Time Complexity: O(n) 
Auxiliary Space: O(1)

Following is another O(n) time complexity and O(1) extra space method suggested by aj. We can sum the bits in the same positions for all the numbers and take modulo with 3. The bits for which sum is not multiple of 3, are the bits of number with single occurrence. 
Let us consider the example array {5, 5, 5, 8}. The 101, 101, 101, 1000 
Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0; 
Sum of second bits%3 = (0 + 0 + 0 + 0)%3 = 0; 
Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0; 
Sum of fourth bits%3 = (1)%3 = 1; 
Hence number which appears once is 1000

Note: this approach won’t work for negative numbers

Below is the implementation of the above approach:

C++




// C++ program to find the element
// that occur only once
#include <bits/stdc++.h>
using namespace std;
#define INT_SIZE 32
 
int getSingle(int arr[], int n)
{
    // Initialize result
    int result = 0;
 
    int x, sum;
 
    // Iterate through every bit
    for (int i = 0; i < INT_SIZE; i++) {
 
        // Find sum of set bits at ith position in all
        // array elements
        sum = 0;
        x = (1 << i);
        for (int j = 0; j < n; j++) {
            if (arr[j] & x)
                sum++;
        }
 
        // The bits with sum not multiple of 3, are the
        // bits of element with single occurrence.
        if ((sum % 3) != 0)
            result |= x;
    }
 
    return result;
}
 
// Driver code
int main()
{
    int arr[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The element with single occurrence is " << getSingle(arr, n);
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// C program to find the element
// that occur only once
#include <stdio.h>
#define INT_SIZE 32
 
int getSingle(int arr[], int n)
{
    // Initialize result
    int result = 0;
 
    int x, sum;
 
    // Iterate through every bit
    for (int i = 0; i < INT_SIZE; i++) {
        // Find sum of set bits at ith position in all
        // array elements
        sum = 0;
        x = (1 << i);
        for (int j = 0; j < n; j++) {
            if (arr[j] & x)
                sum++;
        }
 
        // The bits with sum not multiple of 3, are the
        // bits of element with single occurrence.
        if ((sum % 3) != 0)
            result |= x;
    }
 
    return result;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("The element with single occurrence is %d ",
           getSingle(arr, n));
    return 0;
}

Java




// Java code to find the element
// that occur only once
 
class GFG {
    static final int INT_SIZE = 32;
 
    // Method to find the element that occur only once
    static int getSingle(int arr[], int n)
    {
        int result = 0;
        int x, sum;
 
        // Iterate through every bit
        for (int i = 0; i < INT_SIZE; i++) {
            // Find sum of set bits at ith position in all
            // array elements
            sum = 0;
            x = (1 << i);
            for (int j = 0; j < n; j++) {
                if ((arr[j] & x) != 0)
                    sum++;
            }
            // The bits with sum not multiple of 3, are the
            // bits of element with single occurrence.
            if ((sum % 3) != 0)
                result |= x;
        }
        return result;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int arr[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
        int n = arr.length;
        System.out.println("The element with single occurrence is " + getSingle(arr, n));
    }
}
// Code contributed by Rishab Jain

Python 3




# Python3 code to find the element
# that occur only once
INT_SIZE = 32
 
def getSingle(arr, n) :
     
    # Initialize result
    result = 0
     
    # Iterate through every bit
    for i in range(0, INT_SIZE) :
         
        # Find sum of set bits
        # at ith position in all
        # array elements
        sm = 0
        x = (1 << i)
        for j in range(0, n) :
            if (arr[j] & x) :
                sm = sm + 1
                 
        # The bits with sum not
        # multiple of 3, are the
        # bits of element with
        # single occurrence.
        if ((sm % 3)!= 0) :
            result = result | x
     
    return result
     
# Driver program
arr = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7]
n = len(arr)
print("The element with single occurrence is ", getSingle(arr, n))
 
 
# This code is contributed
# by Nikita Tiwari.

C#




// C# code to find the element
// that occur only once
using System;
 
class GFG {
    static int INT_SIZE = 32;
 
    // Method to find the element
    // that occur only once
    static int getSingle(int[] arr, int n)
    {
        int result = 0;
        int x, sum;
 
        // Iterate through every bit
        for (int i = 0; i < INT_SIZE; i++) {
            // Find sum of set bits at ith
            // position in all array elements
            sum = 0;
            x = (1 << i);
            for (int j = 0; j < n; j++) {
                if ((arr[j] & x) != 0)
                    sum++;
            }
 
            // The bits with sum not multiple
            // of 3, are the bits of element
            // with single occurrence.
            if ((sum % 3) != 0)
                result |= x;
        }
        return result;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 12, 1, 12, 3, 12, 1,
                      1, 2, 3, 2, 2, 3, 7 };
        int n = arr.Length;
        Console.WriteLine("The element with single "
                          + "occurrence is " + getSingle(arr, n));
    }
}
 
// This code is contributed ny vt_m.

PHP




<?php
// PHP code to find the element
// that occur only once
$INT_SIZE= 32;
 
function getSingle($arr, $n)
{
    global $INT_SIZE;
     
    // Initialize result
    $result = 0;
 
    $x; $sum;
 
    // Iterate through every bit
    for ($i = 0; $i < $INT_SIZE; $i++)
    {
    // Find sum of set bits at ith
    // position in all array elements
    $sum = 0;
    $x = (1 << $i);
    for ($j = 0; $j < $n; $j++ )
    {
        if ($arr[$j] & $x)
            $sum++;
    }
 
    // The bits with sum not multiple
    // of 3, are the bits of element
    // with single occurrence.
    if (($sum % 3)  !=0 )
        $result |= $x;
    }
 
    return $result;
}
 
// Driver Code
$arr = array (12, 1, 12, 3, 12, 1,
              1, 2, 3, 2, 2, 3, 7);
$n = sizeof($arr);
echo "The element with single occurrence is ",
                          getSingle($arr, $n);
         
// This code is contributed by ajit
?>

Javascript




<script>
    // Javascript program to find the element
    // that occur only once
    let INT_SIZE = 32;
     
    function getSingle(arr, n)
    {
     
        // Initialize result
        let result = 0;
        let x, sum;
 
        // Iterate through every bit
        for (let i = 0; i < INT_SIZE; i++)
        {
 
            // Find sum of set bits at ith position in all
            // array elements
            sum = 0;
            x = (1 << i);
            for (let j = 0; j < n; j++)
            {
                if (arr[j] & x)
                    sum++;
            }
 
            // The bits with sum not multiple of 3, are the
            // bits of element with single occurrence.
            if ((sum % 3) != 0)
                result |= x;
        }
        return result;
    }
 
// Driver code
    let arr = [ 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 ];
    let n = arr.length;
    document.write("The element with single occurrence is " + getSingle(arr, n));
 
// This code is contributed by mukesh07.
</script>
Output
The element with single occurrence is 7

Another approach suggested by Abhishek Sharma 44. Add each number once and multiply the sum by 3, we will get thrice the sum of each element of the array. Store it as thrice_sum. Subtract the sum of the whole array from the thrice_sum and divide the result by 2. The number we get is the required number (which appears once in the array).
Array [] : [a, a, a, b, b, b, c, c, c, d] 
Mathematical Equation = ( 3*(a+b+c+d) – (a + a + a + b + b + b + c + c + c + d) ) / 2
In more simple words: ( 3*(sum_of_array_without_duplicates) – (sum_of_array) ) / 2
 



let arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}
Required no = ( 3*(sum_of_array_without_duplicates) - (sum_of_array) ) / 2
            = ( 3*(12 + 1 + 3 + 2) - (12 + 1 + 12 + 3 + 12 + 1 + 1 + 2 + 3 + 3))/2 
            = ( 3*     18          -      50) / 2
            = (54 - 50) / 2
            = 2 (required answer)

As we know that set does not contain any duplicate elements, 
But, std::set is commonly implemented as a red-black binary search tree. Insertion on this data structure has a worst-case of O(log(n)) complexity, as the tree is kept balanced. we will be using set here.

Below is the implementation of above approach: 

C++




// C++ program to find the element
// that occur only once
 
#include <bits/stdc++.h>
using namespace std;
 
// function which find number
int singleNumber(int a[], int n)
{
    unordered_set<int> s(a, a + n);
 
    int arr_sum = accumulate(a, a + n, 0); // sum of array
 
    int set_sum = accumulate(s.begin(), s.end(), 0); // sum of set
 
    // applying the formula.
    return (3 * set_sum - arr_sum) / 2;
}
 
// driver function
int main()
{
    int a[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << "The element with single occurrence is " << singleNumber(a, n);
}
 
// This code is contributed by Mohit Kumar 29 (IIIT gwalior)

Java




// Java program to find the element
// that occur only once
import java.util.*;
 
class GFG {
 
    // function which find number
    static int singleNumber(int a[], int n)
    {
        HashSet<Integer> s = new HashSet<Integer>();
        for (int i : a) {
            s.add(i);
        }
 
        int arr_sum = 0; // sum of array
        for (int i : a) {
            arr_sum += i;
        }
 
        int set_sum = 0; // sum of set
        for (int i : s) {
            set_sum += i;
        }
 
        // applying the formula.
        return (3 * set_sum - arr_sum) / 2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
        int n = a.length;
        System.out.println("The element with single "
                           + "occurrence is " + singleNumber(a, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to find the element
# that occur only once
 
# function which find number
def singleNumber(nums):
     
    # applying the formula.
    return (3 * sum(set(nums)) - sum(nums)) / 2
 
# driver function.
a = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7]
print ("The element with single occurrence is ",
                          int(singleNumber(a)))

C#




// C# program to find the element
// that occur only once
using System;
using System.Collections.Generic;
 
class GFG {
 
    // function which find number
    static int singleNumber(int[] a, int n)
    {
        HashSet<int> s = new HashSet<int>();
        foreach(int i in a)
        {
            s.Add(i);
        }
 
        int arr_sum = 0; // sum of array
        foreach(int i in a)
        {
            arr_sum += i;
        }
 
        int set_sum = 0; // sum of set
        foreach(int i in s)
        {
            set_sum += i;
        }
 
        // applying the formula.
        return (3 * set_sum - arr_sum) / 2;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] a = { 12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 };
        int n = a.Length;
        Console.WriteLine("The element with single "
                          + "occurrence is " + singleNumber(a, n));
    }
}
 
// This code is contributed by PrinciRaj1992

PHP




<?php
// PHP program to find the element
// that occur only once
 
//function which find number
function singleNumber($a, $n)
{
    $s = array();
    for ($i = 0; $i < count($a); $i++)
        array_push($s, $a[$i]);
    $s = array_values(array_unique($s));
         
    $arr_sum = 0; // sum of array
    for ($i = 0; $i < count($a); $i++)
    {
        $arr_sum += $a[$i];
    }
     
    $set_sum = 0; // sum of set
    for ($i = 0; $i < count($s); $i++)
    {
        $set_sum += $s[$i];
    }
 
    // applying the formula.
    return (int)(((3 * $set_sum) -
                       $arr_sum) / 2);
}
 
// Driver code
$a = array(12, 1, 12, 3, 12, 1,
           1, 2, 3, 2, 2, 3, 7);
$n = count($a);
print("The element with single occurrence is " .
                          singleNumber($a, $n));
 
// This code is contributed by mits
?>

Javascript




<script>
// Javascript program to find the element
// that occur only once
 
    // function which find number
   function singleNumber(a,n)
   {
       let s = new Set(a);
    let arr_sum = 0; // sum of array
     
    for(let i=0;i<a.length;i++)
    {
        arr_sum += a[i];
    }
     
    let set_sum = 0; // sum of set
    for (let i of s)
    {
         set_sum +=i;
    }
     
    // applying the formula.
    return Math.floor((3 * set_sum - arr_sum) / 2);
   }
    
   // Driver code
   let arr=[12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7 ];
   let n = arr.length;
   document.write("The element with single "
                           + "occurrence is " + singleNumber(arr, n));
     
    // This code is contributed by unknown2108
</script>
Output
The element with single occurrence is 7

Time Complexity: O(Nlog(N)) 
Auxiliary Space: O(N)

Method #4:Using Counter() function

  • Calculate the frequency of array using Counter function
  • Traverse in this Counter dictionary and check if any key has value 1
  • If the value of any key is 1 return the key

Below is the implementation:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function which find number
int singlenumber(int a[],int N)
{
    // umap for finding frequency
    unordered_map<int,int>fmap;
   
    // traverse the array for frequency
    for(int i = 0; i < N;i++)
    {
        fmap[a[i]]++;
    }
   
    // iterate over the map
    for(auto it:fmap)
    {
       
        // check frequency whether it is one or not.
        if(it.second == 1)
        {
           
            // return it as we got the answer
            return it.first;
        }
    }
}
 
// Driver code
int main()
{
   
    // given array
    int a[]={12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};
   
    // size of the array
    int N=sizeof(a)/sizeof(a[0]);
   
    // printing the returned value
    cout << singlenumber(a,N);
    return 0;
}
 
// This Code is contributed by
// Murarishetty Santhosh Charan

Java




// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
     
    // function which find number
    static int singlenumber(int a[],int N)
    {
        // umap for finding frequency
        Map<Integer, Integer> fmap
            = new HashMap<Integer, Integer>();
             
         
        // traverse the array for frequency
        for(int i = 0; i < N;i++)
        {
            if(!fmap.containsKey(a[i]))
                fmap.put(a[i],0);
              
            fmap.put(a[i],fmap.get(a[i])+1);
        }
         
        // iterate over the map
        for(Map.Entry<Integer, Integer> me : fmap.entrySet())
        {
             
            // check frequency whether it is one or not.
            if(me.getValue()==1)
            {
                 
                // return it as we got the answer
                return me.getKey();
            }
        }
        return -1;
         
    }
     
    // Driver code
    public static void main (String[] args) {
         
         
        // given array
    int a[]={12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};
     
    // size of the array
    int N= a.length;
     
     
        // printing the returned value
        System.out.println("The element with single occurrence is "+singlenumber(a,N));
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




from collections import Counter
# Python3 program to find the element
# that occur only once
 
# function which find number
def singleNumber(nums):
   
    # storing the frequencies using Counter
    freq = Counter(nums)
     
    # traversing the Counter dictionary
    for i in freq:
       
        # check if any value is 1
        if(freq[i] == 1):
            return i
 
 
# driver function.
a = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7]
print("The element with single occurrence is ",
      int(singleNumber(a)))
# This code is contributed by vikkycirus

Javascript




<script>
// Javascript program for the above approach
 
// function which find number
function singlenumber(a,N)
{
 
    // umap for finding frequency
    let fmap=new Map();
    
    // traverse the array for frequency
    for(let i = 0; i < N;i++)
    {
        if(!fmap.has(a[i]))
            fmap.set(a[i],0);
         
        fmap.set(a[i],fmap.get(a[i])+1);
    }
    
    // iterate over the map
    for(let [key, value] of fmap.entries())
    {
        
        // check frequency whether it is one or not.
        if(value==1)
        {
            
            // return it as we got the answer
            return key;
        }
    }
}
 
// Driver code
 
// given array
let a = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7];
 
// size of the array
let N = a.length;
 
// printing the returned value
document.write("The element with single occurrence is "+singlenumber(a,N));
 
// This code is contributed by rag2127
</script>
Output
The element with single occurrence is  7

Time Complexity: O(n)

This article is compiled by Sumit Jain and reviewed by the GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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