Find the missing and repeating number

import java.io.*;

class Main {

    static void printTwoElements(int[] arr, int n)
    {
        int[] temp
            = new int[n]; // Creating temp array of size n
                          // with initial values as 0.
        int repeatingNumber = -1;
        int missingNumber = -1;

        for (int i = 0; i < n; i++) {
            temp[arr[i] - 1]++;
            if (temp[arr[i] - 1] > 1) {
                repeatingNumber = arr[i];
            }
        }
        for (int i = 0; i < n; i++) {
            if (temp[i] == 0) {
                missingNumber = i + 1;
                break;
            }
        }

        System.out.println("The repeating number is "
                           + repeatingNumber + ".");
        System.out.println("The missing number is "
                           + missingNumber + ".");
    }

    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 vivek1208

// 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

// 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, 5, 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.

// 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);
            }
        }
    }
}

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