How to find the missing number in a given Array from number 1 to n in Java?

Java 8 Object Oriented Programming Programming

If a single number is missing in an integer array that contains a sequence of numbers values, you can find it basing of the sum of numbers or, basing on the xor of the numbers.

Based on the sum of the numbers −

The sum of n sequential numbers will be [n*(n+1)]/2. Using this get the sum of the numbers the n numbers.
Add all the elements in the array.
Subtract the sum of the numbers in the array from the sum of the n numbers.

Example

import java.util.Scanner;
public class MissingNumber {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      System.out.println("Enter the n value: ");
      int n = sc.nextInt();
      int inpuArray[] = new int[n];
      System.out.println("Enter (n-1) numbers: ");
      for(int i=0; i<=n-2; i++) {
         inpuArray[i] = sc.nextInt();
      }
      //Finding the missing number
      int sumOfAll = (n*(n+1))/2;
      int sumOfArray = 0;
      for(int i=0; i<=n-2; i++) {
         sumOfArray = sumOfArray+inpuArray[i];
      }
      int missingNumber = sumOfAll-sumOfArray;
      System.out.println("Missing number is: "+missingNumber);
   }
}

Output

Enter the n value:
5
Enter (n-1) numbers:
1
2
4
5
Missing number is: 3

Using XOR operation − Another way to find the missing number is using XOR.

Find the XOR of all the numbers up ton.
Find the XOR of all the numbers in the array.
Then find the XOR of both results.

Example

import java.util.Scanner;
public class MissingNumber {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      System.out.println("Enter the n value: ");
      int n = sc.nextInt();
      int inpuArray[] = new int[n];
      System.out.println("Enter (n-1) numbers: ");
      for(int i=0; i<=n-2; i++) {
         inpuArray[i] = sc.nextInt();
      }
      //Finding the missing number
      int xorArray = inpuArray[0];
      //XOR of elements of the array
      for(int i=1; i<=n-1; i++) {
         xorArray = xorArray ^ i;
      }
      int xorAll = inpuArray[0];
      //XOR of elements of the array
      for(int i=1; i<=n+1; i++) {
         xorAll = xorAll ^ i;
      }
      int missingNumber = xorArray ^ xorAll;
      System.out.println("Missing number is: "+missingNumber);
   }
}

Output

Enter the n value:
5
Enter (n-1) numbers:
1
2
4
5
Missing number is: 3

Krishna Kasyap

Published on 02-Aug-2019 13:38:27

Previous Page Print Page