Find the repeating and the missing | Added 3 new methods

Difficulty Level : Medium
Last Updated : 06 Oct, 2021

Given an unsorted array of size n. Array elements are in the range from 1 to n. One number from set {1, 2, …n} is missing and one number occurs twice in the array. Find these two numbers.

Examples:

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 Input: arr[] = {3, 1, 3}
Output: Missing = 2, Repeating = 3
Explanation: In the array, 
2 is missing and 3 occurs twice 

Input: arr[] = {4, 3, 6, 2, 1, 1}
Output: Missing = 5, Repeating = 1

Below are various methods to solve the problems:

Method 1 (Use Sorting)
Approach:

Sort the input array.
Traverse the array and check for missing and repeating.

Time Complexity: O(nLogn)

Thanks to LoneShadow for suggesting this method.

Method 2 (Use count array)
Approach:

Create a temp array temp[] of size n with all initial values as 0.
Traverse the input array arr[], and do following for each arr[i]
- if(temp[arr[i]] == 0) temp[arr[i]] = 1;
- if(temp[arr[i]] == 1) output “arr[i]” //repeating
Traverse temp[] and output the array element having value as 0 (This is the missing element)

Time Complexity: O(n)

Auxiliary Space: O(n)

Method 3 (Use elements as Index and mark the visited places)
Approach:
Traverse the array. While traversing, use the absolute value of every element as an index and make the value at this index as negative to mark it visited. If something is already marked negative then this is the repeating element. To find missing, traverse the array again and look for a positive value.

C++

// C++ program to Find the repeating
// and missing elements
 
#include <bits/stdc++.h>
using namespace std;
 
void printTwoElements(int arr[], int size)
{
    int i;
    cout << " The repeating element is ";
 
    for (i = 0; i < size; i++) {
        if (arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
        else
            cout << abs(arr[i]) << "\n";
    }
 
    cout << "and the missing element is ";
    for (i = 0; i < size; i++) {
        if (arr[i] > 0)
            cout << (i + 1);
    }
}
 
/* Driver code */
int main()
{
    int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printTwoElements(arr, n);
}
 
// This code is contributed by Shivi_Aggarwal

C

// C program to Find the repeating
// and missing elements
 
#include <stdio.h>
#include <stdlib.h>
 
void printTwoElements(int arr[], int size)
{
    int i;
    printf("\n The repeating element is");
 
    for (i = 0; i < size; i++) {
        if (arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
        else
            printf(" %d ", abs(arr[i]));
    }
 
    printf("\nand the missing element is ");
    for (i = 0; i < size; i++) {
        if (arr[i] > 0)
            printf("%d", i + 1);
    }
}
 
// Driver code
int main()
{
    int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printTwoElements(arr, n);
    return 0;
}

Java

// Java program to Find the repeating
// and missing elements
 
import java.io.*;
 
class GFG {
 
    static void printTwoElements(int arr[], int size)
    {
        int i;
        System.out.print("The repeating element is ");
 
        for (i = 0; i < size; i++) {
            int abs_val = Math.abs(arr[i]);
            if (arr[abs_val - 1] > 0)
                arr[abs_val - 1] = -arr[abs_val - 1];
            else
                System.out.println(abs_val);
        }
 
        System.out.print("And the missing element is ");
        for (i = 0; i < size; i++) {
            if (arr[i] > 0)
                System.out.println(i + 1);
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
        int n = arr.length;
        printTwoElements(arr, n);
    }
}
 
// This code is contributed by Gitanjali

Python3

# Python3 code to Find the repeating 
# and the missing elements
 
def printTwoElements( arr, size):
    for i in range(size):
        if arr[abs(arr[i])-1] > 0:
            arr[abs(arr[i])-1] = -arr[abs(arr[i])-1]
        else:
            print("The repeating element is", abs(arr[i]))
             
    for i in range(size):
        if arr[i]>0:
            print("and the missing element is", i + 1)
 
# Driver program to test above function */
arr = [7, 3, 4, 5, 5, 6, 2]
n = len(arr)
printTwoElements(arr, n)
 
# This code is contributed by "Abhishek Sharma 44"

C#

// C# program to Find the repeating
// and missing elements
 
using System;
 
class GFG {
    static void printTwoElements(int[] arr, int size)
    {
        int i;
        Console.Write("The repeating element is ");
 
        for (i = 0; i < size; i++) {
            int abs_val = Math.Abs(arr[i]);
            if (arr[abs_val - 1] > 0)
                arr[abs_val - 1] = -arr[abs_val - 1];
            else
                Console.WriteLine(abs_val);
        }
 
        Console.Write("And the missing element is ");
        for (i = 0; i < size; i++) {
            if (arr[i] > 0)
                Console.WriteLine(i + 1);
        }
    }
 
    // Driver program
    public static void Main()
    {
        int[] arr = { 7, 3, 4, 5, 5, 6, 2 };
        int n = arr.Length;
        printTwoElements(arr, n);
    }
}
// This code is contributed by Sam007

PHP

<?php
// PHP program to Find the repeating
// and missing elements
 
function printTwoElements($arr, $size)
{
    $i;
    echo "The repeating element is", " ";
 
    for($i = 0; $i < $size; $i++)
    {
        if($arr[abs($arr[$i]) - 1] > 0)
            $arr[abs($arr[$i]) - 1] = 
            - $arr[abs($arr[$i]) - 1];
        else
            echo ( abs($arr[$i]));
    }
 
    echo "\nand the missing element is ";
    for($i = 0; $i < $size; $i++)
    {
        if($arr[$i] > 0)
            echo($i + 1);
    }
}
     
    // Driver Code
    $arr = array(7, 3, 4, 5, 5, 6, 2);
    $n = count($arr);
    printTwoElements($arr, $n);
 
// This code is contributed by anuj_67.
?>

Javascript

<script>
// Program to Find the repeating
// and missing elements
  
    function  printTwoElements(arr,size)
   { 
    var i;
        document.write(" The repeating element is ");
            
    for (i = 0; i < size; i++) 
    {
        var abs_value = Math.abs(arr[i]);
        if (arr[abs_value - 1] > 0)
            arr[abs_value - 1] = -arr[abs_value - 1];
        else
            document.write( abs_value);
    }
              
    document.write("<br> and the missing element is ");
    for (i = 0; i < size; i++) 
    {
        if (arr[i] > 0)
            document.write (i + 1);
    }
}
  
/* Driver code */
    arr = new Array ( 7, 3, 4, 5, 5, 6, 2 );
    n = arr.length;
    printTwoElements(arr, n);
     
    // This code is contributed by simranarora5sos
</script>

Output

 The repeating element is 5
and the missing element is 1

Time Complexity: O(n)
Thanks to Manish Mishra for suggesting this method.

Method 4 (Make two equations)
Approach:

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.

Time Complexity: O(n)
Thanks to disappearedng for suggesting this solution.

Note: This method can cause arithmetic overflow as we calculate product and sum of all array elements.

Method 5 (Use XOR)

Approach:

Let x and y be the desired output elements.
Calculate XOR of all the array elements.

xor1 = arr[0]^arr[1]^arr[2]…..arr[n-1]

XOR the result with all numbers from 1 to n

xor1 = xor1^1^2^…..^n

In the result xor1, all elements would nullify each other except x and y. All the bits that are set in xor1 will be set in either x or y. So if we take any set bit (We have chosen the rightmost set bit in code) of xor1 and divide the elements of the array in two sets – one set of elements with same bit set and other set with same bit not set. By doing so, we will get x in one set and y in another set. Now if we do XOR of all the elements in first set, we will get x, and by doing same in other set we will get y.

Below is the implementation of the above approach:

C++

// C++ program to Find the repeating
// and missing elements
 
#include <bits/stdc++.h>
using namespace std;
 
/* The output of this function is stored at
*x and *y */
void getTwoElements(int arr[], int n,
                    int* x, int* y)
{
    /* Will hold xor of all elements 
    and numbers from 1 to n */
    int xor1;
 
    /* Will have only single set bit of xor1 */
    int set_bit_no;
 
    int i;
    *x = 0;
    *y = 0;
 
    xor1 = arr[0];
 
    /* Get the xor of all array elements */
    for (i = 1; i < n; i++)
        xor1 = xor1 ^ arr[i];
 
    /* XOR the previous result with numbers 
    from 1 to n*/
    for (i = 1; i <= n; i++)
        xor1 = xor1 ^ i;
 
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = xor1 & ~(xor1 - 1);
 
    /* Now divide elements into two 
    sets by comparing a rightmost set
    bit of xor1 with the bit at the same 
    position in each element. Also, 
    get XORs of two sets. The two
    XORs are the output elements. 
    The following two for loops 
    serve the purpose */
    for (i = 0; i < n; i++) {
        if (arr[i] & set_bit_no)
            /* arr[i] belongs to first set */
            *x = *x ^ arr[i];
 
        else
            /* arr[i] belongs to second set*/
            *y = *y ^ arr[i];
    }
    for (i = 1; i <= n; i++) {
        if (i & set_bit_no)
            /* i belongs to first set */
            *x = *x ^ i;
 
        else
            /* i belongs to second set*/
            *y = *y ^ i;
    }
 
    /* *x and *y hold the desired
        output elements */
}
 
/* Driver code */
int main()
{
    int arr[] = { 1, 3, 4, 5, 5, 6, 2 };
    int* x = (int*)malloc(sizeof(int));
    int* y = (int*)malloc(sizeof(int));
    int n = sizeof(arr) / sizeof(arr[0]);
 
    getTwoElements(arr, n, x, y);
    cout << " The missing element is " << *x << " and the repeating"
         << " number is " << *y;
    getchar();
}
 
// This code is contributed by Code_Mech

C

// C program to Find the repeating
// and missing elements
 
#include <stdio.h>
#include <stdlib.h>
 
/* The output of this function is stored at
   *x and *y */
void getTwoElements(int arr[], int n, int* x, int* y)
{
    /* Will hold xor of all elements and numbers 
       from 1 to n */
    int xor1;
 
    /* Will have only single set bit of xor1 */
    int set_bit_no;
 
    int i;
    *x = 0;
    *y = 0;
 
    xor1 = arr[0];
 
    /* Get the xor of all array elements */
    for (i = 1; i < n; i++)
        xor1 = xor1 ^ arr[i];
 
    /* XOR the previous result with numbers 
       from 1 to n*/
    for (i = 1; i <= n; i++)
        xor1 = xor1 ^ i;
 
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = xor1 & ~(xor1 - 1);
 
    /* Now divide elements in two sets by comparing 
    rightmost set bit of xor1 with bit at same 
    position in each element. Also, get XORs of two 
    sets. The two XORs are the output elements. The
    following two for loops serve the purpose */
    for (i = 0; i < n; i++) {
        if (arr[i] & set_bit_no)
            /* arr[i] belongs to first set */
            *x = *x ^ arr[i];
 
        else
            /* arr[i] belongs to second set*/
            *y = *y ^ arr[i];
    }
    for (i = 1; i <= n; i++) {
        if (i & set_bit_no)
            /* i belongs to first set */
            *x = *x ^ i;
 
        else
            /* i belongs to second set*/
            *y = *y ^ i;
    }
 
    /* *x and *y hold the desired output elements */
}
 
/* Driver program to test above function */
int main()
{
    int arr[] = { 1, 3, 4, 5, 5, 6, 2 };
    int* x = (int*)malloc(sizeof(int));
    int* y = (int*)malloc(sizeof(int));
    int n = sizeof(arr) / sizeof(arr[0]);
 
    getTwoElements(arr, n, x, y);
    printf(" The missing element is %d"
           " and the repeating number"
           " is %d",
           *x, *y);
    getchar();
}

Java

// Java program to Find the repeating
// and missing elements
 
import java.io.*;
 
class GFG {
    static int x, y;
 
    static void getTwoElements(int arr[], int n)
    {
        /* Will hold xor of all elements
       and numbers from 1 to n  */
        int xor1;
 
        /* Will have only single set bit of xor1 */
        int set_bit_no;
 
        int i;
        x = 0;
        y = 0;
 
        xor1 = arr[0];
 
        /* Get the xor of all array elements  */
        for (i = 1; i < n; i++)
            xor1 = xor1 ^ arr[i];
 
        /* XOR the previous result with numbers from 
       1 to n*/
        for (i = 1; i <= n; i++)
            xor1 = xor1 ^ i;
 
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = xor1 & ~(xor1 - 1);
 
        /* Now divide elements into two sets by comparing
    rightmost set bit of xor1 with the bit at the same 
    position in each element. Also, get XORs of two
    sets. The two XORs are the output elements. The 
    following two for loops serve the purpose */
        for (i = 0; i < n; i++) {
            if ((arr[i] & set_bit_no) != 0)
                /* arr[i] belongs to first set */
                x = x ^ arr[i];
 
            else
                /* arr[i] belongs to second set*/
                y = y ^ arr[i];
        }
        for (i = 1; i <= n; i++) {
            if ((i & set_bit_no) != 0)
                /* i belongs to first set */
                x = x ^ i;
 
            else
                /* i belongs to second set*/
                y = y ^ i;
        }
 
        /* *x and *y hold the desired output elements */
    }
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 4, 5, 1, 6, 2 };
 
        int n = arr.length;
        getTwoElements(arr, n);
        System.out.println(" The missing element is  "
                           + x + "and the "
                           + "repeating number is "
                           + y);
    }
}
 
// This code is contributed by Gitanjali.

Python3

# Python3 program to find the repeating 
# and missing elements 
 
# The output of this function is stored 
# at x and y 
def getTwoElements(arr, n):
     
    global x, y
    x = 0
    y = 0
     
    # Will hold xor of all elements 
    # and numbers from 1 to n 
    xor1 = arr[0]
     
    # Get the xor of all array elements
    for i in range(1, n):
        xor1 = xor1 ^ arr[i]
         
    # XOR the previous result with numbers 
    # from 1 to n
    for i in range(1, n + 1):
        xor1 = xor1 ^ i
     
    # Will have only single set bit of xor1
    set_bit_no = xor1 & ~(xor1 - 1)
     
    # Now divide elements into two 
    # sets by comparing a rightmost set 
    # bit of xor1 with the bit at the same 
    # position in each element. Also, 
    # get XORs of two sets. The two 
    # XORs are the output elements. 
    # The following two for loops 
    # serve the purpose
    for i in range(n):
        if (arr[i] & set_bit_no) != 0:
             
            # arr[i] belongs to first set
            x = x ^ arr[i]
        else:
             
            # arr[i] belongs to second set
            y = y ^ arr[i]
             
    for i in range(1, n + 1):
        if (i & set_bit_no) != 0:
             
            # i belongs to first set
            x = x ^ i
        else:
             
            # i belongs to second set
            y = y ^ i 
         
    # x and y hold the desired 
    # output elements 
     
# Driver code
arr = [ 1, 3, 4, 5, 5, 6, 2 ]
n = len(arr)
     
getTwoElements(arr, n)
 
print("The missing element is", x,
      "and the repeating number is", y)
     
# This code is contributed by stutipathak31jan

C#

// C# program to Find the repeating
// and missing elements
 
using System;
 
class GFG {
    static int x, y;
 
    static void getTwoElements(int[] arr, int n)
    {
        /* Will hold xor of all elements
        and numbers from 1 to n */
        int xor1;
 
        /* Will have only single set bit of xor1 */
        int set_bit_no;
 
        int i;
        x = 0;
        y = 0;
 
        xor1 = arr[0];
 
        /* Get the xor of all array elements */
        for (i = 1; i < n; i++)
            xor1 = xor1 ^ arr[i];
 
        /* XOR the previous result with numbers from 
        1 to n*/
        for (i = 1; i <= n; i++)
            xor1 = xor1 ^ i;
 
        /* Get the rightmost set bit in set_bit_no */
        set_bit_no = xor1 & ~(xor1 - 1);
 
        /* Now divide elements in two sets by comparing
        rightmost set bit of xor1 with bit at same 
        position in each element. Also, get XORs of two
        sets. The two XORs are the output elements.The 
        following two for loops serve the purpose */
        for (i = 0; i < n; i++) {
            if ((arr[i] & set_bit_no) != 0)
 
                /* arr[i] belongs to first set */
                x = x ^ arr[i];
 
            else
 
                /* arr[i] belongs to second set*/
                y = y ^ arr[i];
        }
        for (i = 1; i <= n; i++) {
            if ((i & set_bit_no) != 0)
 
                /* i belongs to first set */
                x = x ^ i;
 
            else
 
                /* i belongs to second set*/
                y = y ^ i;
        }
 
        /* *x and *y hold the desired output elements */
    }
 
    // Driver program
    public static void Main()
    {
        int[] arr = { 1, 3, 4, 5, 1, 6, 2 };
 
        int n = arr.Length;
        getTwoElements(arr, n);
        Console.Write(" The missing element is "
                      + x + "and the "
                      + "repeating number is "
                      + y);
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to Find the repeating
// and missing elements
 
function getTwoElements(&$arr, $n)
{
 
    /* Will hold xor of all elements
    and numbers from 1 to n */
    $xor1;
 
    /* Will have only single set bit of xor1 */
    $set_bit_no;
 
    $i;
    $x = 0;
    $y = 0;
 
    $xor1 = $arr[0];
 
    /* Get the xor of all array elements */
    for ($i = 1; $i < $n; $i++)
        $xor1 = $xor1 ^ $arr[$i];
 
    /* XOR the previous result with numbers 
    from 1 to n*/
    for ($i = 1; $i <= $n; $i++)
        $xor1 = $xor1 ^ $i;
 
    /* Get the rightmost set bit in set_bit_no */
    $set_bit_no = $xor1 & ~($xor1 - 1);
 
    /* Now divide elements in two sets by comparing
    rightmost set bit of xor1 with bit at same 
    position in each element. Also, get XORs of two
    sets. The two XORs are the output elements.The 
    following two for loops serve the purpose */
    for ($i = 0; $i < $n; $i++) 
    {
        if (($arr[$i] & $set_bit_no) != 0)
         
            /* arr[i] belongs to first set */
            $x = $x ^ $arr[$i];
 
        else
         
            /* arr[i] belongs to second set*/
            $y = $y ^ $arr[$i];
    }
     
    for ($i = 1; $i <= $n; $i++)
    {
        if (($i & $set_bit_no) != 0)
         
            /* i belongs to first set */
            $x = $x ^ $i;
 
        else
         
            /* i belongs to second set*/
            $y = $y ^ $i;
    }
 
    /* *x and *y hold the desired output elements */
    echo("The missing element is " . $x . 
         " and the repeating number is " . $y);
}
 
// Driver Code
$arr = array( 1, 3, 4, 5, 1, 6, 2 );
$n = sizeof($arr);
getTwoElements($arr, $n);
 
// This code is contributed by Code_Mech

Output

 The missing element is 7 and the repeating number is 5

Time Complexity: O(n)
This method doesn’t cause overflow, but it doesn’t tell which one occurs twice and which one is missing. We can add one more step that checks which one is missing and which one is repeating. This can be easily done in O(n) time.

Method 6 (Use a Map)
Approach:
This method involves creating a Hashtable with the help of Map. In this, the elements are mapped to their natural index. In this process, if an element is mapped twice, then it is the repeating element. And if an element’s mapping is not there, then it is the missing element.

Below is the implementation of the above approach:

C++

// C++ program to find the repeating
// and missing elements using Maps 
#include <iostream>
#include <unordered_map>
using namespace std;
 
int main()
{
    int arr[] = { 4, 3, 6, 2, 1, 1 };
    int n = 6;
     
    unordered_map<int, bool> numberMap;
     
    for(int i : arr) 
    {
        if (numberMap.find(i) == 
            numberMap.end())
        {
            numberMap[i] = true;
        }
        else
        {
            cout << "Repeating = " << i;
        }
    }
    cout << endl;
     
    for(int i = 1; i <= n; i++)
    {
        if (numberMap.find(i) == 
            numberMap.end()) 
        {
            cout << "Missing = " << i;
        }
    }
    return 0;
}
 
// This code is contributed by RohitOberoi

Java

// Java program to find the
// repeating and missing elements
// using Maps
 
import java.util.*;
 
public class Test1 {
 
    public static void main(String[] args)
    {
 
        int[] arr = { 4, 3, 6, 2, 1, 1 };
 
        Map<Integer, Boolean> numberMap
            = new HashMap<>();
 
        int max = arr.length;
 
        for (Integer i : arr) {
 
            if (numberMap.get(i) == null) {
                numberMap.put(i, true);
            }
            else {
                System.out.println("Repeating = " + i);
            }
        }
        for (int i = 1; i <= max; i++) {
            if (numberMap.get(i) == null) {
                System.out.println("Missing = " + i);
            }
        }
    }
}

Python3

# Python3 program to find the 
# repeating and missing elements 
# using Maps
def main():
     
    arr = [ 4, 3, 6, 2, 1, 1 ]
     
    numberMap = {}
     
    max = len(arr)
    for i in arr:
        if not i in numberMap:
            numberMap[i] = True
             
        else:
            print("Repeating =", i)
     
    for i in range(1, max + 1):
        if not i in numberMap:
            print("Missing =", i)
main()
 
# This code is contributed by stutipathak31jan

C#

// C# program to find the
// repeating and missing elements
// using Maps
using System;
using System.Collections.Generic;
 
class GFG
{
    public static void Main(String[] args)
    {
        int[] arr = { 4, 3, 6, 2, 1, 1 };
 
        Dictionary<int, Boolean> numberMap =
                   new Dictionary<int, Boolean>();
 
        int max = arr.Length;
 
        foreach (int i in arr) 
        {
            if (!numberMap.ContainsKey(i)) 
            {
                numberMap.Add(i, true);
            }
            else
            {
                Console.WriteLine("Repeating = " + i);
            }
        }
        for (int i = 1; i <= max; i++) 
        {
            if (!numberMap.ContainsKey(i)) 
            {
                Console.WriteLine("Missing = " + i);
            }
        }
    }
}
 
// This code is contributed by PrinciRaj1992

Output

Repeating = 1
Missing = 5

Method 7 (Make two equations using sum and sum of squares)
Approach:

Let x be the missing and y be the repeating element.
Let N is the size of array.
Get the sum of all numbers using formula S = N(N+1)/2
Get the sum of square of all numbers using formula Sum_Sq = N(N+1)(2N+1)/6
Iterate through a loop from i=1….N
S -= A[i]
Sum_Sq -= (A[i]*A[i])
It will give two equations
x-y = S – (1)
x^2 – y^2 = Sum_sq
x+ y = (Sum_sq/S) – (2)

Time Complexity: O(n)

C++

#include <bits/stdc++.h> 
 
using namespace std;
 
vector<int>repeatedNumber(const vector<int> &A) {
    long long int len = A.size();
    long long int Sum_N = (len * (len+1) ) /2, Sum_NSq = (len * (len +1) *(2*len +1) )/6;
    long long int missingNumber=0, repeating=0;
     
    for(int i=0;i<A.size(); i++){
       Sum_N -= (long long int)A[i];
       Sum_NSq -= (long long int)A[i]*(long long int)A[i];
    }
     
    missingNumber = (Sum_N + Sum_NSq/Sum_N)/2;
    repeating= missingNumber - Sum_N;
    vector <int> ans;
    ans.push_back(repeating);
    ans.push_back(missingNumber);
    return ans;
     
}
 
 
int main(void){
        std::vector<int> v = {4, 3, 6, 2, 1, 6,7};
    vector<int> res = repeatedNumber(v);
    for(int x: res){
        cout<< x<<"  ";
    }
    cout<<endl;
}

Java

import java.util.*;
import java.math.BigInteger;
class GFG 
{
    static Vector<Integer> repeatedNumber(int[] a) 
    {
        
        BigInteger n=BigInteger.valueOf(a.length);
   
        BigInteger s=BigInteger.valueOf(0);
        BigInteger ss=BigInteger.valueOf(0);
 
        for(int x : a)
        {
            s=  s.add(BigInteger.valueOf(x));
            ss= ss.add(BigInteger.valueOf(x).multiply(BigInteger.valueOf(x))); 
        }
 
        BigInteger as= n.multiply(n.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2));
        BigInteger ass= as.multiply(BigInteger.valueOf(2).multiply(n).add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(3));
 
        BigInteger sub=as.subtract(s);
        BigInteger add=(ass.subtract(ss)).divide(sub);
        //(ass-ss)/sub;
 
        int b = sub.add(add).divide(BigInteger.valueOf(2)).intValue();
        //(sub+add)/2;
        int A = BigInteger.valueOf(b).subtract(sub).intValue();
        Vector<Integer> ans = new Vector<>();
        ans.add(A);
        ans.add(b);
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args) 
    {
        int[] v = { 4, 3, 6, 2, 1, 6, 7 };
        Vector<Integer> res = repeatedNumber(v);
        for (int x : res) 
        {
            System.out.print(x + " ");
        }
    }
}
 
// This code is contributed by Rajput-Ji

Python3

def repeatedNumber(A):
     
    length = len(A)
    Sum_N = (length * (length + 1)) // 2
    Sum_NSq = ((length * (length + 1) *
                     (2 * length + 1)) // 6)
     
    missingNumber, repeating = 0, 0
     
    for i in range(len(A)):
        Sum_N -= A[i]
        Sum_NSq -= A[i] * A[i]
         
    missingNumber = (Sum_N + Sum_NSq //
                             Sum_N) // 2
    repeating = missingNumber - Sum_N
     
    ans = []
    ans.append(repeating)
    ans.append(missingNumber)
     
    return ans
 
# Driver code
v = [ 4, 3, 6, 2, 1, 6, 7 ]
res = repeatedNumber(v)
 
for i in res:
    print(i, end = " ")
 
# This code is contributed by stutipathak31jan 

C#

using System;
using System.Collections.Generic;
 
class GFG 
{
    static List<int> repeatedNumber(int[] A) 
    {
        int len = A.Length;
        int Sum_N = (len * (len + 1)) / 2;
        int Sum_NSq = (len * (len + 1) * 
                        (2 * len + 1)) / 6;
        int missingNumber = 0, repeating = 0;
 
        for (int i = 0; i < A.Length; i++) 
        {
            Sum_N -= A[i];
            Sum_NSq -= A[i] * A[i];
        }
 
        missingNumber = (Sum_N + Sum_NSq / 
                                 Sum_N) / 2;
        repeating = missingNumber - Sum_N;
        List<int> ans = new List<int>();
        ans.Add(repeating);
        ans.Add(missingNumber);
        return ans;
    }
 
    // Driver Code
    public static void Main(String[] args) 
    {
        int[] v = { 4, 3, 6, 2, 1, 6, 7 };
        List<int> res = repeatedNumber(v);
        foreach (int x in res) 
        {
            Console.Write(x + " ");
        }
    }
}
 
// This code is contributed by PrinciRaj1992

Output

6  5

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

Method 8 (Using OR Operator):

Approach:

Given an input array

Performing OR operation on input array.

At the same time checking if that number has occurred before, by determining if the position is already set or not. We will get the repeating number in this step.

To find missing value we have to check the bit containing 0 using OR again.

C++

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    // Input:
    vector<int> arr = {4, 3, 6, 2, 1, 1};
    int n = arr.size();
     
    // Declaring output variables
    // Note : arr[i]-1 is used instead of arr[i] as we want to use all 64 bits
    int bitOr = (1 << (arr[0]-1));
    int repeating, missing;
     
    // Performing XOR as well as Checking repeating number
    for(int i=1; i<n; i++){
        // If OR operation with 1 gives same output that means, we already have 1 at that position
        if((bitOr | (1 << (arr[i]-1))) == bitOr) {
            repeating = arr[i];
            continue;
        }
        bitOr = (bitOr | (1 << (arr[i]-1)));
    }
     
    // Checking missing number
    for(int i=0; i<n; i++){
        // property: OR with 0 yield 1 hence value of bitOr changes
        if((bitOr | (1 << i)) != bitOr) {
            missing = i+1;
            break;
        }
    }
     
    cout << "Repeating : " << repeating << "\nMissing : " << missing;
    return 0;
}

Output

Repeating : 1
Missing : 5

Time Complexity : O(n)
Auxiliary Complexity : O(1)

Limitations of the approach: it only works on size of array <= 64 if we use long and size of array <= 32

My Personal Notes

Related Articles

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