JavaScript Program to Find Element that Appears once in Array where Other Elements Appears Twice

Last Updated : 04 Jun, 2024

Given an Array of Integers consisting of n elements where each element appears twice except one element. The task is to find the element which appears only once in the input array in JavaScript.

Example:

Input : arr = [ 5, 9, 2, 7, 4, 8, 6, 1, 5, 1, 4, 6, 3, 7, 8, 2, 9 ] 
Output : 3
Explanation: The input array consists of 17 elements 
where all elements except 3 appears twice i.e. 
(1, 2, 4, 5, 6, 7, 8, 9) are repeating twice. Hence 
the answer will be 3.
Input : arr = [ 4, 7, 1, 4, 1 ]
Output : 7

Approaches to Find the Element that Appears Once in Array where Every Other Element Appears Twice:

Table of Content

Using array filter() and indexOf() Methods
Using forEach() Loop with XOR Operator
Using Map Data Structure

Using array filter() and indexOf() Methods

The first approach to solve the problem is to use the filter() method of arrays. It will return a new array say uniqueArr. Now perform the filter() method on the input array and check if the index of the current element in the array is equal to the last index of the element in the array. Because if they are equal then the element is unique. Print the first element of uniqueArr array.

Example: This example implements array filter() and indexOf() methods to get the element.

JavaScript

const arr = [
    5, 9, 2, 7, 4, 6, 8, 1, 5, 
    1, 4, 6, 3, 7, 8, 2, 9
];

const uniqueArr = arr.filter((num) => {
    return arr.indexOf(num) === arr.lastIndexOf(num);
});

console.log(uniqueArr);

Output

[ 3 ]

Using forEach() Loop with XOR Operator

Another approach to solving the problem is to use the XOR operator (^).

Initialize ans as 0.
Now apply forEach loop in the array and for every element perform its XOR operation with ans and return ans.
Since XOR is commutative and associative it will cancel out any duplicates in the array and only the unique element will be contained in ans.
Print ans.

Example: This example uses Array filter and XOR operator to get required result

JavaScript

const arr = [
    5, 9, 2, 7, 4, 6, 8, 1, 5, 
    1, 4, 6, 3, 7, 8, 2, 9
];

let ans = 0;

arr.forEach((num) => {
    ans ^= num;
});

console.log(ans);

Output

Using Map Data Structure

In this approach, we can utilize a JavaScript Map data structure to keep track of the count of each element in the array. Then, we iterate through the array, updating the count in the map. Finally, we iterate through the map and return the key whose value is 1, indicating that it appeared only once in the array.

Example:

JavaScript

function findSingleElement(arr) {
    const countMap = new Map();
    
    // Count the occurrences of each element in the array
    arr.forEach(num => {
        if (countMap.has(num)) {
            countMap.set(num, countMap.get(num) + 1);
        } else {
            countMap.set(num, 1);
        }
    });
    
    // Iterate through the map and find the element with count 1
    for (const [key, value] of countMap.entries()) {
        if (value === 1) {
            return key;
        }
    }
    
    // If no element with count 1 is found, return null or handle it as needed
    return null;
}

const arr = [5, 9, 2, 7, 4, 6, 8, 1, 5, 1, 4, 6, 3, 7, 8, 2, 9];
console.log(findSingleElement(arr)); // Output: 3