Given an array arr[] containing 2*n + 2 positive numbers, out of which 2*n numbers exist in pairs whereas the other two number occur exactly once and are distinct. The task is to find the other two numbers.
Note: Return the numbers in increasing order.
Example:
Input: arr[] = [1, 2, 3, 2, 1, 4]
Output: 3 4
Explanation: 3 and 4 occur exactly once.
Input: arr[] = [2, 1, 3, 2]
Output: 1 3
Explanation: 1 and 3 occur exactly once.
[Naive Approach] Nested Loop Frequency Counting – O(n^2) Time and O(1) Space
This approach iterates through the array and counts the frequency of each element using a nested loop. For each element, the inner loop counts how many times it appears in the array. If an element appears exactly once, it is returned as the result. This method ensures that the correct element is identified but is inefficient due to the nested loop, finally return them in sorted order.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> singleNum(vector<int>& arr) {
int n = arr.size();
vector<int> ans;
// Iterate over every element
for (int i = 0; i < n; i++) {
// Initialize count to 0
int count = 0;
for (int j = 0; j < n; j++) {
// Count the frequency of the element
if (arr[i] == arr[j]) {
count++;
}
}
// If the frequency of the element is one
if (count == 1) {
ans.push_back(arr[i]);
}
}
sort(ans.begin(), ans.end());
// If no element exists at most once
return ans;
}
int main() {
vector<int> arr = { 1, 2, 3, 2, 1, 4 };
vector<int> ans = singleNum(arr);
for(auto it:ans)
cout<<it<<" ";
return 0;
}
import java.util.*;
class GfG {
static int[] singleNum(int[] arr) {
// Collect elements that appear exactly once
List<Integer> result = new ArrayList<>();
for(int i=0;i<arr.length;i++){
int count = 0;
for(int j=0;j<arr.length;j++){
// count the frequency of element arr[i]
if(arr[i]==arr[j])
count++;
}
if(count == 1)result.add(arr[i]);
}
// Sort the result list
Collections.sort(result);
// Convert List<Integer> to int[]
int[] resArray = new int[result.size()];
for (int i = 0; i < result.size(); i++) {
resArray[i] = result.get(i);
}
return resArray;
}
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 2, 1, 4 };
int[] result = singleNum(arr);
// Print the result array
for(int i=0;i<result.length;i++){
System.out.print(result[i] + " ");
}
}
}
def singleNum(arr):
n = len(arr)
result = []
# Iterate over every element
for i in range(n):
count = 0
for j in range(n):
if arr[i] == arr[j]:
count += 1
# If the frequency of the element is one and not already in result
if count == 1:
result.append(arr[i])
# Return sorted result
return sorted(result)
if __name__ == "__main__":
arr = [1, 2, 3, 2, 1, 4]
print(*singleNum(arr))
using System;
using System.Collections.Generic;
class GfG {
static List<int> singleNum(int[] arr) {
int n = arr.Length;
List<int> result = new List<int>();
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
// If the frequency is one and not already in the result list
if (count == 1 && !result.Contains(arr[i])) {
result.Add(arr[i]);
}
}
// Sort the result before returning
result.Sort();
return result;
}
static void Main(string[] args) {
int[] arr = { 1, 2, 3, 2, 1, 4 };
List<int> singles = singleNum(arr);
foreach (int num in singles) {
Console.Write(num + " ");
}
}
}
function singleNum(arr) {
let n = arr.length;
let result = [];
for (let i = 0; i < n; i++) {
let count = 0;
for (let j = 0; j < n; j++) {
if (arr[i] === arr[j]) {
count++;
}
}
// If the frequency is one and not already added
if (count === 1 && !result.includes(arr[i])) {
result.push(arr[i]);
}
}
// Sort the result in ascending order
result.sort((a, b) => a - b);
return result;
}
// dirver code
let arr = [1, 2, 3, 2, 1, 4];
let ans = singleNum(arr);
console.log(ans.join(" "));
[Better Approach 1] Using Sorting – O(n logn) time and O(1) space
This approach first sorts the array to group identical elements next to each other. Then, it iterates through the sorted array, checking for pairs of identical elements and skipping them. If an element is unique (appears only once), it is added to the result. This method ensures that the two unique elements are found and returned efficiently after sorting.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> singleNum(vector<int> &arr) {
int n = arr.size();
sort(arr.begin(), arr.end());
vector<int> ans;
int i = 0;
while (i < n) {
// If element is present twice, skip
if (i + 1 < n && arr[i + 1] == arr[i]) {
i = i + 2;
}
// If unique, append to result
else {
ans.push_back(arr[i]);
i++;
}
}
return ans;
}
int main() {
vector<int> arr = { 1, 2, 3, 2, 1, 4 };
vector<int> ans = singleNum(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.Arrays;
public class GfG {
public static int[] singleNum(int[] arr) {
Arrays.sort(arr);
int n = arr.length;
// Temporary array to store unique elements
int[] temp = new int[n];
int idx = 0;
int i = 0;
while (i < n) {
// If element is present twice, skip both
if (i + 1 < n && arr[i + 1] == arr[i]) {
i += 2;
} else {
// If unique, store in temp
temp[idx++] = arr[i];
i++;
}
}
// Copy valid elements to a properly-sized array
int[] result = new int[idx];
for (int j = 0; j < idx; j++) {
result[j] = temp[j];
}
return result;
}
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 2, 1, 4 };
int[] ans = singleNum(arr);
// Print the result array
for (int i = 0; i < ans.length; i++) {
System.out.print(ans[i] + " ");
}
}
}
def singleNum(arr):
arr.sort()
ans = []
i = 0
n = len(arr)
while i < n:
# If element is present twice, skip
if i + 1 < n and arr[i + 1] == arr[i]:
i += 2
else:
# If unique, append to result
ans.append(arr[i])
i += 1
return ans
if __name__ == '__main__':
arr = [1, 2, 3, 2, 1, 4]
ans = singleNum(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
using System.Linq;
class GfG {
static List<int> SingleNum(List<int> arr) {
arr.Sort();
List<int> ans = new List<int>();
int i = 0;
while (i < arr.Count) {
// If element is present twice, skip
if (i + 1 < arr.Count && arr[i + 1] == arr[i]) {
i += 2;
}
// If unique, append to result
else {
ans.Add(arr[i]);
i++;
}
}
return ans;
}
static void Main() {
List<int> arr = new List<int> { 1, 2, 3, 2, 1, 4 };
List<int> ans = SingleNum(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function singleNum(arr) {
arr.sort((a, b) => a - b);
let ans = [];
let i = 0;
while (i < arr.length) {
// If element is present twice, skip
if (i + 1 < arr.length && arr[i + 1] === arr[i]) {
i += 2;
}
// If unique, append to result
else {
ans.push(arr[i]);
i++;
}
}
return ans;
}
const arr = [1, 2, 3, 2, 1, 4];
const ans = singleNum(arr);
console.log(ans[0] + ' ' + ans[1]);
[Better Approach 2] Using HashMap – O(n) time and O(n) space
This approach uses an unordered map (freq) to track the frequency of elements in the array. After building the frequency map, it collects elements that occur exactly once and returns them in ascending order.
C++
#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> singleNum(const vector<int>& arr) {
unordered_map<int, int> freq;
// Count the frequency of each element
for (int x : arr) {
freq[x]++;
}
vector<int> res;
// Collect numbers that appear exactly once
for (const auto& p : freq) {
if (p.second == 1) {
res.push_back(p.first);
}
}
if (res[0] > res[1]) {
swap(res[0], res[1]);
}
return res;
}
int main() {
vector<int> arr = {1, 2, 3, 2, 1, 4};
vector<int> res = singleNum(arr);
for (int x : res) {
cout << x << " ";
}
cout << endl;
return 0;
}
import java.util.*;
public class GfG {
public static int[] singleNum(int[] arr) {
Map<Integer, Integer> freq = new HashMap<>();
// Count the frequency of each element
for (int x : arr) {
freq.put(x, freq.getOrDefault(x, 0) + 1);
}
List<Integer> temp = new ArrayList<>();
// Collect numbers that appear exactly once
for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
if (entry.getValue() == 1) {
temp.add(entry.getKey());
}
}
// Sort the result
Collections.sort(temp);
// Convert to int[]
int[] result = new int[temp.size()];
for (int i = 0; i < temp.size(); i++) {
result[i] = temp.get(i);
}
return result;
}
public static void main(String[] args) {
int[] arr = {1, 2, 3, 2, 1, 4};
int[] res = singleNum(arr);
for (int x : res) {
System.out.print(x + " ");
}
System.out.println();
}
}
def singleNum(arr):
freq = {}
# Count the frequency of each element
for x in arr:
freq[x] = freq.get(x, 0) + 1
res = []
# Collect numbers that appear exactly once
for key, value in freq.items():
if value == 1:
res.append(key)
if res[0] > res[1]:
res[0], res[1] = res[1], res[0]
return res
if __name__ == '__main__':
arr = [1, 2, 3, 2, 1, 4]
res = singleNum(arr)
for x in res:
print(x, end=' ')
print()
using System;
using System.Collections.Generic;
using System.Linq;
class GfG {
public static List<int> SingleNum(int[] arr) {
Dictionary<int, int> freq = new Dictionary<int, int>();
// Count the frequency of each element
foreach (int x in arr) {
if (freq.ContainsKey(x)) {
freq[x]++;
} else {
freq[x] = 1;
}
}
List<int> res = new List<int>();
// Collect numbers that appear exactly once
foreach (var p in freq) {
if (p.Value == 1) {
res.Add(p.Key);
}
}
if (res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
return res;
}
static void Main() {
int[] arr = {1, 2, 3, 2, 1, 4};
List<int> res = SingleNum(arr);
foreach (int x in res) {
Console.Write(x + " ");
}
Console.WriteLine();
}
}
function singleNum(arr) {
const freq = {};
// Count the frequency of each element
for (let x of arr) {
freq[x] = (freq[x] || 0) + 1;
}
const res = [];
// Collect numbers that appear exactly once
for (const key in freq) {
if (freq[key] === 1) {
res.push(Number(key));
}
}
if (res[0] > res[1]) {
[res[0], res[1]] = [res[1], res[0]];
}
return res;
}
const arr = [1, 2, 3, 2, 1, 4];
const res = singleNum(arr);
console.log(res.join(" "));
[Expected Approach] Using Bit Manipulation – O(n) time and O(1) space
The key idea is that by XOR-ing all elements in the array, we eliminate the effect of all numbers that appear twice (since a ^ a = 0), leaving us with x ^ y, the XOR of the two unique elements. We then find any set bit (typically the rightmost set bit) in this result to determine a position where these two numbers differ. Using this bit, we split the numbers into two groups, XOR each group separately, and get the two unique numbers.
To find the rightmost set bit in xorVal, we use the expression xorVal & (-xorVal). This is a common bit manipulation trick that isolates the lowest (rightmost) bit that is set to 1. It works because in two’s complement, -xorVal is the bitwise inverse of xorVal plus one, which flips all bits after the rightmost 1, thus leaving only that 1 bit in the AND operation. This bit tells us a position where the two numbers (say, x and y) differ , one has a 1 at this position, and the other has a 0.
This trick is particularly efficient and often used in problems involving two unique elements among duplicates. However, if we don’t specifically need the rightmost set bit, we can alternatively find any set bit by scanning bit positions, or using other logical operations
Illustration:
Given the array [2, 4, 7, 9, 2, 4], XOR all elements to get 14. The rightmost set bit of 14 is at position 1. Divide the array into two groups based on this bit:
- Group 1 (bit set at position 1):
[2, 7, 2] - Group 2 (bit not set):
[4, 9, 4]
XOR the elements in each group:
- Group 1:
2 ^ 7 ^ 2 = 7 - Group 2:
4 ^ 9 ^ 4 = 9
The two non-repeating elements are 7 and 9.
C++
#include <iostream>
#include <vector>
using namespace std;
vector<int> singleNum(vector<int> &arr) {
// Get the XOR of the two numbers we need to find
int xorVal = 0;
for (auto i : arr) {
xorVal = i ^ xorVal;
}
// Get its last set bit
xorVal &= -xorVal;
vector<int> res(2, 0);
for (int num : arr) {
// If bit is not set, it belongs to the first set
if ((num & xorVal) == 0) {
res[0] ^= num;
}
// If bit is set, it belongs to the second set
else {
res[1] ^= num;
}
}
// Ensure the order of the returned numbers is consistent
if (res[0] > res[1]) {
swap(res[0], res[1]);
}
return res;
}
int main() {
vector<int> arr = { 1, 2, 3, 2, 1, 4 };
vector<int> ans = singleNum(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.Arrays;
import java.util.List;
public class GfG {
public static int[] singleNum(int[] arr) {
// Get the XOR of the two numbers we need to find
int xorVal = 0;
for (int i : arr) {
xorVal ^= i;
}
// Get its last set bit
xorVal &= -xorVal;
int[] res = new int[2];
for (int num : arr) {
// If bit is not set, it belongs to the first set
if ((num & xorVal) == 0) {
res[0] ^= num;
}
// If bit is set, it belongs to the second set
else {
res[1] ^= num;
}
}
// Ensure the order of the returned numbers is consistent
if (res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
return res;
}
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 2, 1, 4 };
int[] ans = singleNum(arr);
System.out.println(ans[0] + " " + ans[1]);
}
}
def singleNum(arr):
# Get the XOR of the two numbers we need to find
xor_val = 0
for i in arr:
xor_val ^= i
# Get its last set bit
xor_val &= -xor_val
res = [0, 0]
for num in arr:
# If bit is not set, it belongs to the first set
if (num & xor_val) == 0:
res[0] ^= num
# If bit is set, it belongs to the second set
else:
res[1] ^= num
# Ensure the order of the returned numbers is consistent
if res[0] > res[1]:
res[0], res[1] = res[1], res[0]
return res
# Main execution
arr = [1, 2, 3, 2, 1, 4]
ans = singleNum(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
class Solution {
public int[] SingleNum(int[] arr) {
// Get the XOR of the two numbers we need to find
int xorVal = 0;
foreach (int i in arr) {
xorVal ^= i;
}
// Get its last set bit
xorVal &= -xorVal;
int[] res = new int[2];
foreach (int num in arr) {
// If bit is not set, it belongs to the first set
if ((num & xorVal) == 0) {
res[0] ^= num;
}
// If bit is set, it belongs to the second set
else {
res[1] ^= num;
}
}
// Ensure the order of the returned numbers is consistent
if (res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
return res;
}
static void Main() {
int[] arr = { 1, 2, 3, 2, 1, 4 };
Solution sol = new Solution();
int[] ans = sol.SingleNum(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function singleNum(arr) {
// Get the XOR of the two numbers we need to find
let xorVal = 0;
for (let i of arr) {
xorVal ^= i;
}
// Get its last set bit
xorVal &= -xorVal;
let res = [0, 0];
for (let num of arr) {
// If bit is not set, it belongs to the first set
if ((num & xorVal) === 0) {
res[0] ^= num;
}
// If bit is set, it belongs to the second set
else {
res[1] ^= num;
}
}
// Ensure the order of the returned numbers is consistent
if (res[0] > res[1]) {
[res[0], res[1]] = [res[1], res[0]];
}
return res;
}
const arr = [1, 2, 3, 2, 1, 4];
const ans = singleNum(arr);
console.log(ans[0] + ' ' + ans[1]);
Please refer below post for detailed explanation:
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