Find the Element That Appears Once in an Array

Find the element that occurs only once in a given set of integers while all the other numbers occur thrice.
Example -  
Input : 3 3 3 4
Output: 4

Input : 5 5 4 8 4 5 8 9 4 8
Output: 9

Please try solving this problem before jumping on the solution

Click to learn

Subscribe for more updates

Preparing for interviews? IDeserve team is here to help.


The idea is to count the set bits in all the given numbers. As we know, all the elements(except one) occur 3 times so their respective set bits will also occur 3 times. Thus when we take modulus 3 of the sum of all set bits we will be left with only the bits of that number which occurs once.

The steps of the algorithm are as following:
1: Create countSetBits[] array of size 32(for representing 32 bit integer) where,
   countSetBits[i] represents count of ith set bit of all elements in the input array.
   Initially all elements of countSetBits[] array are 0.
2. Traverse all the elements of the input array to populate countSetBits, by doing step #3 for each of them.
3. Take the element and check for its set bits. If the ith bit is found to be set, then in the countSetBits[] array increment the count of the element at the index 'i'.

4. After finishing the above operation for all the elements of the input array, the elements of countSetBits[] would represent count of all set bits in the elements of input array. Perform the modulus 3 operation on each element of the countSetBits[] array. Taking mod with 3 will determine the number of times that respective index was set in all the elements of given input array. If we get a remainder 1 at an index 'j', then that means in the number that occurs only once, we have a set bit of index 'j'. We will get a remainder 2 only if the given question is violated with two instances of a number.

5. Finally after modulus of each element in countSetBits[], simply transform the binary representation into decimal.

Hone your coding skills!

AngryNerds Practice platform »

Algorithm Visualization

Code Snippet

package com.ideserve.questions.abhishek;

 * <b>IDeserve <br>
 * <a href=""></a>
 * O(n) solution for finding out the number that occurs once.
 * @author Abhishek
public class BitWiseMod {

	 * Method to find the element that occurs only once in the input array
	 * while all the other numbers occur N times.
	 * @param arr is input array.
	 * @param N is number of times all elements, except one, are occurring in the input array.
	 * @return the number which occurs only once in the input array.
	public int findRequiredNum(int arr[], int N) {
		// For counting set bits in all given numbers.
		int countSetBit[] = new int[32];

		// For each element run the loop.
		for (int i = 0; i < arr.length; i++) {
			// Find the set bits in the current element.
			for (int k = 0; k < 32; k++) {
				int kthSetBit = 1 << k;
				// If the kth bit is set, then increment the count of countSetBit[k].
				if ((arr[i] & kthSetBit) == kthSetBit)
		// Required number.
		int occurredOnce = 0;

		// Iterate over countSetBit.
		for (int i = 0; i < 32; i++) {
			 *  Find the remainder of each element with N so as to see what is
			 *  the representation of the single occurred element.
			countSetBit[i] = countSetBit[i] % N;

			 * After mod operation, each element of countSetBit represent bits
			 * of required element(occurredOnce). Therefore, set ith bit in 
			 * occurredOnce if countSetBit[i] is 1.
			if (countSetBit[i] == 1)
				occurredOnce = occurredOnce | (1 << i);
		return occurredOnce;

	 * Method for testing the algorithm.
	 * @param args
	public static void main(String args[]) {
		int[] arr = { 5, 5, 4, 8, 4, 5, 8, 9, 4, 8 };
		System.out.print("Input Matrix : ");

		for (int i = 0; i < arr.length; i++)
			System.out.print(arr[i] + "  ");

		BitWiseMod solution = new BitWiseMod();

		 *  Since all elements in the input array, except one, occur 3 times,
		 *  pass 3 as argument in call to findRequiredNum.
		System.out.println("\n\nThe number which occured only once is: " + solution.findRequiredNum(arr, 3));

Order of the Algorithm

Time Complexity is O(32.n) = O(n)
Space Complexity is O(1) As constant size array is required.


  • Sincere thanks from IDeserve community to Abhishek Somani for compiling current post.

    Abhishek Somani

    Ninja Programmer