Missing and Repeating in an Array
Last Updated :
23 Jul, 2025
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Given an unsorted array arr[] of size n, containing elements from the range 1 to n, it is known that one number in this range is missing, and another number occurs twice in the array, find both the duplicate number and the missing number.
Examples:
Input: arr[] = [3, 1, 3]
Output: [3, 2]
Explanation: 3 is occurs twice and 2 is missing.Input: arr[] = [4, 3, 6, 2, 1, 1]
Output: [1, 5]
Explanation: 1 is occurs twice and 5 is missing.
Table of Content
[Approach 1] Using Visited Array - O(n) Time and O(n) Space
The idea is to use a frequency array to keep track of how many times each number appears in the original array. Since we know the numbers should range from 1 to n with each appearing exactly once, any number appearing twice is our repeating number, and any number with zero frequency is our missing number.
#include <iostream>
#include <vector>
using namespace std;
vector<int> findTwoElement(vector<int>& arr) {
int n = arr.size();
// frequency array to count occurrences
vector<int> freq(n + 1, 0);
int repeating = -1, missing = -1;
// count frequency of each element
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// identify repeating and missing elements
for (int i = 1; i <= n; i++) {
if (freq[i] == 0) missing = i;
else if (freq[i] == 2) repeating = i;
}
return {repeating, missing};
}
int main() {
vector<int> arr = {3, 1, 3};
vector<int> ans = findTwoElement(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findTwoElement(int[] arr) {
int n = arr.length;
// frequency array to count occurrences
int[] freq = new int[n + 1];
int repeating = -1;
int missing = -1;
// count frequency of each element
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// identify missing and repeating numbers
for (int i = 1; i <= n; i++) {
if (freq[i] == 0) missing = i;
else if (freq[i] == 2) repeating = i;
}
ArrayList<Integer> result = new ArrayList<>();
result.add(repeating);
result.add(missing);
return result;
}
public static void main(String[] args) {
int[] arr = {3, 1, 3};
ArrayList<Integer> ans = findTwoElement(arr);
System.out.println(ans.get(0) + " " + ans.get(1));
}
}
def findTwoElement(arr):
n = len(arr)
# frequency array to count occurrences
freq = [0] * (n + 1)
repeating = -1
missing = -1
# count frequency of each element
for num in arr:
freq[num] += 1
# identify missing and repeating numbers
for i in range(1, n + 1):
if freq[i] == 0:
missing = i
elif freq[i] == 2:
repeating = i
return [repeating, missing]
if __name__ == "__main__":
arr = [3, 1, 3]
ans = findTwoElement(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
class GfG {
static List<int> findTwoElement(int[] arr) {
int n = arr.Length;
// frequency array to count occurrences
int[] freq = new int[n + 1];
int repeating = -1, missing = -1;
// count frequency of each element
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// identify missing and repeating numbers
for (int i = 1; i <= n; i++) {
if (freq[i] == 0) missing = i;
else if (freq[i] == 2) repeating = i;
}
return new List<int> { repeating, missing };
}
static void Main() {
int[] arr = {3, 1, 3};
List<int> ans = findTwoElement(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function findTwoElement(arr) {
let n = arr.length;
// frequency array to count occurrences
let freq = new Array(n + 1).fill(0);
let repeating = -1, missing = -1;
// count frequency of each element
for (let i = 0; i < n; i++) {
freq[arr[i]]++;
}
// identify missing and repeating numbers
for (let i = 1; i <= n; i++) {
if (freq[i] === 0) missing = i;
else if (freq[i] === 2) repeating = i;
}
return [repeating, missing];
}
// Driver Code
let arr = [3, 1, 3];
let ans = findTwoElement(arr);
console.log(ans[0], ans[1]);
Output
3 2
[Approach 2] Using Array Marking - O(n) Time and O(1) Space
The main idea is to use the input array itself for tracking: it negates the value at the index corresponding to each element to mark it as visited. If it encounters a value that has already been negated, it identifies that number as the repeating one. In a second pass, it finds the index that remains positive, which corresponds to the missing number.
#include <iostream>
#include <vector>
using namespace std;
vector<int> findTwoElement(vector<int>& arr) {
int n = arr.size();
int repeating = -1;
// mark visited indices by negating the value at
// that index
for (int i = 0; i < n; i++) {
int val = abs(arr[i]);
if (arr[val - 1] > 0) {
arr[val - 1] = -arr[val - 1];
} else {
// found the repeating element
repeating = val;
}
}
int missing = -1;
// the index with a positive value is the
// missing number
for (int i = 0; i < n; i++) {
if (arr[i] > 0) {
missing = i + 1;
break;
}
}
return {repeating, missing};
}
int main() {
vector<int> arr = {3, 1, 3};
vector<int> ans = findTwoElement(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findTwoElement(int[] arr) {
int n = arr.length;
int repeating = -1;
// traverse the array and mark visited indices
// by negating the value at index arr[i] - 1
for (int i = 0; i < n; i++) {
int val = Math.abs(arr[i]);
// if the value at index val - 1 is already negative
// it means we've seen this value before
if (arr[val - 1] > 0) {
arr[val - 1] = -arr[val - 1];
} else {
// if it's already negative, this value is
// the repeating one
repeating = val;
}
}
int missing = -1;
// after marking, the index with a positive value
// corresponds to the missing number
for (int i = 0; i < n; i++) {
if (arr[i] > 0) {
missing = i + 1;
break;
}
}
// return result: first repeating, then missing
ArrayList<Integer> result = new ArrayList<>();
result.add(repeating);
result.add(missing);
return result;
}
public static void main(String[] args) {
int[] arr = {3, 1, 3};
ArrayList<Integer> ans = findTwoElement(arr);
System.out.println(ans.get(0) + " " + ans.get(1));
}
}
def findTwoElement(arr):
n = len(arr)
repeating = -1
# mark visited indices by negating the value
# at that index
for i in range(n):
val = abs(arr[i])
# if value at index val - 1 is already negative,
# val is repeating
if arr[val - 1] > 0:
arr[val - 1] = -arr[val - 1]
else:
# Already visited → repeating element
repeating = val
missing = -1
# the index with a positive value corresponds
# to the missing number
for i in range(n):
if arr[i] > 0:
missing = i + 1
break
return [repeating, missing]
if __name__ == "__main__":
arr = [3, 1, 3]
ans = findTwoElement(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
class GfG {
static List<int> findTwoElement(int[] arr) {
int n = arr.Length;
int repeating = -1;
// Mark visited indices by negating the
// value at that index
for (int i = 0; i < n; i++) {
int val = Math.Abs(arr[i]);
// If already negative, this value is repeating
if (arr[val - 1] > 0) {
arr[val - 1] = -arr[val - 1];
}
else {
// Repeated element found
repeating = val;
}
}
int missing = -1;
// the index with a positive value corresponds
// to the missing number
for (int i = 0; i < n; i++) {
if (arr[i] > 0) {
missing = i + 1;
break;
}
}
return new List<int> { repeating, missing };
}
static void Main() {
int[] arr = {3, 1, 3};
List<int> ans = findTwoElement(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function findTwoElement(arr) {
let n = arr.length;
let repeating = -1;
// Mark visited indices by negating the value
// at the target index
for (let i = 0; i < n; i++) {
let val = Math.abs(arr[i]);
// If already negative, the number is repeating
if (arr[val - 1] > 0) {
arr[val - 1] = -arr[val - 1];
} else {
repeating = val;
}
}
let missing = -1;
// The index with a positive value corresponds
// to the missing number
for (let i = 0; i < n; i++) {
if (arr[i] > 0) {
missing = i + 1;
break;
}
}
return [repeating, missing];
}
// Driver Code
let arr = [3, 1, 3];
let ans = findTwoElement(arr);
console.log(ans[0], ans[1]);
Output
3 2
[Approach 3] Making Two Math Equations - O(n) Time and O(1) Space
The idea is to use mathematical equations based on the sum and sum of squares of numbers from 1 to n. The difference between expected and actual sums will give us one equation, and the difference between expected and actual sum of squares will give us another equation. Solving these equations yields our missing and repeating numbers.
Illustration:
Note: This method can cause arithmetic overflow as we calculate the sum of squares (or product) and sum of all array elements.
#include <iostream>
#include <vector>
using namespace std;
vector<int> findTwoElement(vector<int>& arr) {
int n = arr.size();
// Sum of first n natural numbers
long long s = (n * (n + 1)) / 2;
// Sum of squares of first n natural numbers
int ssq = (n * (n + 1) * (2 * n + 1)) / 6;
// Subtract actual values from expected sums
for (int i = 0; i < n; i++) {
s -= arr[i];
ssq -= arr[i] * arr[i];
}
// Using the equations: missing - repeating = s
// missing^2 - repeating^2 = ssq
int missing = (s + ssq / s) / 2;
int repeating = missing - s;
return {repeating, missing};
}
int main() {
vector<int> arr = {3, 1, 3};
vector<int> ans = findTwoElement(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findTwoElement(int[] arr) {
int n = arr.length;
// Expected sum and sum of squares for numbers from 1 to n
int s = (n * (n + 1)) / 2;
int ssq = (n * (n + 1) * (2 * n + 1)) / 6;
int missing = 0, repeating = 0;
// Subtract actual values from expected sums
for (int i = 0; i < n; i++) {
s -= arr[i];
ssq -= arr[i] * arr[i];
}
// Let x = missing, y = repeating
// s = x - y
// ssq = x^2 - y^2 = (x - y)(x + y) = s * (x + y)
// => x + y = ssq / s
// => x = (s + ssq / s) / 2
// => y = x - s
missing = (s + ssq / s) / 2;
repeating = missing - s;
ArrayList<Integer> res = new ArrayList<>();
res.add(repeating);
res.add(missing);
return res;
}
public static void main(String[] args) {
int[] arr = {3, 1, 3};
ArrayList<Integer> ans = findTwoElement(arr);
System.out.println(ans.get(0) + " " + ans.get(1));
}
}
def findTwoElement(arr):
n = len(arr)
# Expected sum and sum of squares for numbers from 1 to n
s = (n * (n + 1)) // 2
ssq = (n * (n + 1) * (2 * n + 1)) // 6
missing = 0
repeating = 0
# Subtract actual sum and sum of squares from expected values
for num in arr:
s -= num
ssq -= num * num
# Let s = x - y and ssq = x^2 - y^2 = (x - y)(x + y)
# => x = (s + ssq // s) // 2, y = x - s
missing = (s + ssq // s) // 2
repeating = missing - s
return [repeating, missing]
if __name__ == "__main__":
arr = [3, 1, 3]
ans = findTwoElement(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
class GfG {
static List<int> findTwoElement(int[] arr) {
int n = arr.Length;
// Expected sum and sum of squares from 1 to n
int s = (n * (n + 1)) / 2;
int ssq = (n * (n + 1) * (2 * n + 1)) / 6;
int missing = 0, repeating = 0;
// Subtract actual values from expected ones
for (int i = 0; i < n; i++) {
s -= arr[i];
ssq -= arr[i] * arr[i];
}
// Let s = x - y, ssq = x^2 - y^2 = (x - y)(x + y)
// => x = (s + ssq / s) / 2, y = x - s
missing = (s + ssq / s) / 2;
repeating = missing - s;
return new List<int> { repeating, missing };
}
static void Main() {
int[] arr = { 3, 1, 3 };
List<int> ans = findTwoElement(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function findTwoElement(arr) {
let n = arr.length;
// Expected sum and sum of squares from 1 to n
let s = (n * (n + 1)) / 2;
let ssq = (n * (n + 1) * (2 * n + 1)) / 6;
let missing = 0, repeating = 0;
// Subtract actual values from the expected ones
for (let i = 0; i < n; i++) {
s -= arr[i];
ssq -= arr[i] * arr[i];
}
// Let s = x - y, ssq = x^2 - y^2 = (x - y)(x + y)
// Solve: x = (s + ssq / s) / 2, y = x - s
missing = (s + ssq / s) / 2;
repeating = missing - s;
return [repeating, missing];
}
// Driver Code
let arr = [3, 1, 3];
let ans = findTwoElement(arr);
console.log(ans[0], ans[1]);
Output
3 2
An Alternate way to make two equations:
- Let x be the missing and y be the repeating element.
- Get the sum of all numbers using formula S = n(n+1)/2 - x + y
- Get product of all numbers using formula P = 1*2*3*...*n * y / x
- The above two steps give us two equations, we can solve the equations and get the values of x and y.
[Approach 4] Using XOR - O(n) Time and O(1) Space
The idea is to use XOR operations to isolate the missing and repeating numbers. By XORing all array elements with numbers 1 to n, we get the XOR of our missing and repeating numbers. Then, using a set bit in this XOR result, we can divide all numbers into two groups, which helps us separate the missing and repeating numbers.
Refer to Find the two numbers with odd occurrences in an unsorted array to understand how groups will be created.
Step by step approach:
- XOR all array elements and numbers from 1 to n to get XOR of missing and repeating numbers.
- Find the rightmost set bit in this XOR result using
xorVal & ~(xorVal-1). - Use this set bit to divide array elements and numbers 1 to n into two groups.
- XOR elements of first group to get x and second group to get y.
- Count occurrences of x in original array to determine which is missing and which is repeating.
- If x appears in array, x is repeating and y is missing; otherwise vice versa.
- Return both the repeating and missing numbers.
#include <iostream>
#include <vector>
using namespace std;
vector<int> findTwoElement(vector<int>& arr) {
int n = arr.size();
int xorVal = 0;
// get the xor of all array elements
// and numbers from 1 to n
for (int i = 0; i < n; i++) {
xorVal ^= arr[i];
xorVal ^= (i + 1); // 1 to n numbers
}
// get the rightmost set bit in xorVal
int setBitIndex = xorVal & ~(xorVal - 1);
int x = 0, y = 0;
// now divide elements into two sets
// by comparing rightmost set bit
for (int i = 0; i < n; i++) {
// decide whether arr[i] is in first set
// or second
if (arr[i] & setBitIndex) {
x ^= arr[i];
}
else {
y ^= arr[i];
}
// decide whether (i+1) is in first set
// or second
if ((i+1) & setBitIndex) {
x ^= (i + 1);
}
else {
y ^= (i + 1);
}
}
// x and y are the repeating and missing values.
// to know which one is what, traverse the array
int missing, repeating;
int xCnt = 0;
for (int i=0; i<n; i++) {
if (arr[i] == x) {
xCnt++;
}
}
if (xCnt == 0) {
missing = x;
repeating = y;
}
else {
missing = y;
repeating = x;
}
return {repeating, missing};
}
int main() {
vector<int> arr = {3, 1, 3};
vector<int> ans = findTwoElement(arr);
cout << ans[0] << " " << ans[1] << endl;
return 0;
}
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findTwoElement(int[] arr) {
int n = arr.length;
int xorVal = 0;
// get the xor of all array elements
// And numbers from 1 to n
for (int i = 0; i < n; i++) {
xorVal ^= arr[i];
xorVal ^= (i + 1); // 1 to n numbers
}
// get the rightmost set bit in xorVal
int setBitIndex = xorVal & ~(xorVal - 1);
int x = 0, y = 0;
// now divide elements into two sets
// by comparing rightmost set bit
for (int i = 0; i < n; i++) {
// decide whether arr[i] is in first set
// or second
if ((arr[i] & setBitIndex) != 0) {
x ^= arr[i];
}
else {
y ^= arr[i];
}
// decide whether (i+1) is in first set
// or second
if (((i + 1) & setBitIndex) != 0) {
x ^= (i + 1);
}
else {
y ^= (i + 1);
}
}
// x and y are the repeating and missing values.
// to know which one is what, traverse the array
int missing, repeating;
int xCnt = 0;
for (int i = 0; i < n; i++) {
if (arr[i] == x) {
xCnt++;
}
}
if (xCnt == 0) {
missing = x;
repeating = y;
}
else {
missing = y;
repeating = x;
}
ArrayList<Integer> result = new ArrayList<>();
result.add(repeating);
result.add(missing);
return result;
}
public static void main(String[] args) {
int[] arr = {3, 1, 3};
ArrayList<Integer> ans = findTwoElement(arr);
System.out.println(ans.get(0) +" "+ ans.get(1));
}
}
def findTwoElement(arr):
n = len(arr)
xorVal = 0
# get the xor of all array elements
# And numbers from 1 to n
for i in range(n):
xorVal ^= arr[i]
xorVal ^= (i + 1) # 1 to n numbers
# get the rightmost set bit in xorVal
setBitIndex = xorVal & ~(xorVal - 1)
x, y = 0, 0
# now divide elements into two sets
# by comparing rightmost set bit
for i in range(n):
# decide whether arr[i] is in first set
# or second
if arr[i] & setBitIndex:
x ^= arr[i]
else:
y ^= arr[i]
# decide whether (i+1) is in first set
# or second
if (i + 1) & setBitIndex:
x ^= (i + 1)
else:
y ^= (i + 1)
# x and y are the repeating and missing values.
# to know which one is what, traverse the array
xCnt = sum(1 for num in arr if num == x)
if xCnt == 0:
missing, repeating = x, y
else:
missing, repeating = y, x
return [repeating, missing]
if __name__ == "__main__":
arr = [3, 1, 3]
ans = findTwoElement(arr)
print(ans[0], ans[1])
using System;
using System.Collections.Generic;
class GfG {
static List<int> findTwoElement(int[] arr) {
int n = arr.Length;
int xorVal = 0;
// get the xor of all array elements
// And numbers from 1 to n
for (int i = 0; i < n; i++) {
xorVal ^= arr[i];
xorVal ^= (i + 1); // 1 to n numbers
}
// get the rightmost set bit in xorVal
int setBitIndex = xorVal & ~(xorVal - 1);
int x = 0, y = 0;
// now divide elements into two sets
// by comparing rightmost set bit
for (int i = 0; i < n; i++) {
// decide whether arr[i] is in first set
// or second
if ((arr[i] & setBitIndex) != 0) {
x ^= arr[i];
}
else {
y ^= arr[i];
}
// decide whether (i+1) is in first set
// or second
if (((i + 1) & setBitIndex) != 0) {
x ^= (i + 1);
}
else {
y ^= (i + 1);
}
}
// x and y are the repeating and missing values.
// to know which one is what, traverse the array
int missing, repeating;
int xCnt = 0;
for (int i = 0; i < n; i++) {
if (arr[i] == x) {
xCnt++;
}
}
if (xCnt == 0) {
missing = x;
repeating = y;
}
else {
missing = y;
repeating = x;
}
return new List<int> { repeating, missing };
}
static void Main() {
int[] arr = {3, 1, 3};
List<int> ans = findTwoElement(arr);
Console.WriteLine(ans[0] + " " + ans[1]);
}
}
function findTwoElement(arr) {
let n = arr.length;
let xorVal = 0;
// Get the xor of all array elements
// And numbers from 1 to n
for (let i = 0; i < n; i++) {
xorVal ^= arr[i];
xorVal ^= (i + 1); // 1 to n numbers
}
// Get the rightmost set bit in xorVal
let setBitIndex = xorVal & ~(xorVal - 1);
let x = 0, y = 0;
// Now divide elements into two sets
// by comparing rightmost set bit
for (let i = 0; i < n; i++) {
// Decide whether arr[i] is in first set
// or second
if (arr[i] & setBitIndex) {
x ^= arr[i];
}
else {
y ^= arr[i];
}
// Decide whether (i+1) is in first set
// or second
if ((i + 1) & setBitIndex) {
x ^= (i + 1);
}
else {
y ^= (i + 1);
}
}
let xCnt = arr.filter(num => num === x).length;
let missing = xCnt === 0 ? x : y;
let repeating = xCnt === 0 ? y : x;
return [repeating, missing];
}
// Driver Code
let arr = [3, 1, 3];
let ans = findTwoElement(arr);
console.log(ans[0], ans[1]);
Output
3 2
