Find the element that appears once
Last Updated :
17 Apr, 2024
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: 2
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 iterations 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 iterations 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.
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>
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
iterations 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
?>
OutputThe 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
Python3
# 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 by vt_m.
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>
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
?>
OutputThe element with single occurrence is 7
Time Complexity: O(n)
Auxiliary Space: O(1)
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
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>
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
?>
OutputThe element with single occurrence is 7
Time Complexity: O(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
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// function which find number
static void singlenumber(int[] a, int N)
{
// umap for finding frequency
Dictionary<int, int> fmap = new Dictionary<int, int>();
// traverse the array for frequency
for (int i = 0; i < N; i++)
{
if (fmap.ContainsKey(a[i]))
fmap[a[i]]++;
else
fmap.Add(a[i], 1);
}
// iterate over the map
foreach (int it in fmap.Keys.ToList())
{
// check frequency whether it is one or not.
if(fmap[it] == 1)
{
// return it as we got the answer
Console.Write("The element with single occurrence is " + it);
}
}
}
// Driver Code
public static void Main (string[] args) {
// given array
int[] arr = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};
// size of the array
int n= arr.Length;
// printing the returned value
singlenumber(arr, n);
}
}
// This code is contributed by splevel62.
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>
OutputThe element with single occurrence is 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Using count() method.
We can find the occurrence of elements using count() method. If count() method returns 1 then the element has a single occurrence.
C++
// C++ code to find the element that
// appears once
#include<bits/stdc++.h>
using namespace std;
int main()
{
vector<int> arr{3, 3, 2, 3};
int x;
for (int i : arr)
{
if (count(arr.begin(), arr.end(), i) == 1)
{
x = i;
break;
}
}
cout << "The element with single occurrence is " << x << endl;
return 0;
}
// This code contributed by Srj_27
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
List<Integer> arr = new ArrayList<>(Arrays.asList(3, 3, 2, 3));
int x = 0;
for (int i : arr) {
if (Collections.frequency(arr, i) == 1) {
x = i;
break;
}
}
System.out.println("The element with single occurrence is " + x);
}
}
Python3
# Python3 code to find the element that
# appears once
arr = [3, 3, 2, 3]
for i in arr:
if(arr.count(i)==1):
x=i
break
print("The element with single occurrence is ",x)
C#
// C# code to find the element that
// appears once
using System;
using System.Linq;
public class GFG
{
public static void Main(string[] args)
{
int[] nums = { 3, 3, 2, 3};
foreach(int i in nums)
{
if (nums.Count(n => n==i)==1)
Console.WriteLine("The element with single occurrence is "+i);
}
}
}
// This code contributed by Akshay
// Tripathi(akshaytripathi19410)
Javascript
// Javascript code to find the element that
// appears once
const arr = [3, 3, 2, 3];
let x;
for (let i of arr) {
if (arr.filter(num => num === i).length === 1) {
x = i;
break;
}
}
console.log(`The element with single occurrence is ${x}`);
// This code is contributed by akashish__
OutputThe element with single occurrence is 2
Time Complexity: O(n^2)
Auxiliary Space: O(1)
This article is compiled by Sumit Jain and reviewed by the GeeksforGeeks team.
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