Find the Missing Number

You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in list. One of the integers is missing in the list. Write an efficient code to find the missing integer.


Solution 1: using the formula n(n+1) / 2

Code:

package microsoftInterviewPrep;

public class MissingNumber {
   
    static void findMissingNumber(int[] a){

        //we add one to the length since we are missing one element in the array
        // we should include that also to get our results correct
        int n = a.length + 1;
        int total = n* (n+1) /2;
       
        int arraytotal = 0;
        for(int i=0; i<a.length; i++){
            arraytotal += a[i];
        }
       
        int res = total - arraytotal;
       
        System.out.println("Missing number is "+ res);
    }
   
    public static void main(String[] args){
       
        int[] a= {1, 2, 4, 6, 3, 7, 8};
        findMissingNumber(a);
       
    }

}

Output:
Missing number is 5


Solution 2: Using XOR

Code:

package microsoftInterviewPrep;

public class MissingNumber2 {
   
    static void findMissingNumber(int[] a){

        int x1 = a[0];
        int x2 = 1;
       
        for(int i=1; i<a.length; i++){
           
            x1 = x1 ^ a[i];
           
        }
       
        // 1 is added to length since the number of elements will be one more when we
        // include the count of missing element
       
        for(int i=2; i<= a.length+1; i++){
           
            x2 = x2 ^ i;
        }
       
        int res = x1 ^ x2;
       
        System.out.println(res);
       
    }
   
    public static void main(String[] args){
       
        int[] a= {1, 2, 4, 6, 3, 7, 8};
        findMissingNumber(a);
       
    }

}


Output:

5



 

Comments

Post a Comment

Popular posts from this blog

Rearrange Array in Maximum-Minimum form

Given a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on. Naive Solution: Here we will use an auxiliary array to store the elements in correct order. For even indices of the array  Large numbers from the given array are added. For odd indices small numbers are added. Code: import java.util.*; public class Test{        public static void main(String[] args) {          Scanner in = new Scanner(System.in);     int[] a = {1,2,3,4,5,6,7,8,9};     int n = a.length;     int[] temp = new int[n];     int small = 0;     int large = n-1;     for(int i=0; i<n; i++){         if(i%2 == 0){             temp[i] = a[large--];         }  ...
Read more

Find zeroes to be flipped so that number of consecutive 1's is maximized

Image
Input: arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1} m = 2 Output: 5 7 We are allowed to flip maximum 2 zeroes. If we flip arr[5] and arr[7], we get 8 consecutive 1's which is maximum possible under given constraints Input: arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1} m = 1 Output: 7 We are allowed to flip maximum 1 zero. If we flip arr[7], we get 5 consecutive 1's which is maximum possible under given constraints. Input: arr[] = {0, 0, 0, 1} m = 4 Output: 0 1 2 Since m is more than number of zeroes, we can flip all zeroes. Steps: It is based on sliding window method. The window at any time should consists of only m zeroes. If the zerocount is below m wr is incremented. If the zerocount is more wl is incremented. Code: import java.util.*; public class Solution {       public static void main(String[] args) {         int[] a = {1,1,0,1,1,0,0,1,1,1};         int m = 2;   ...
Read more

Count substrings with same first and last character

Given a string we have check contiguous substrings with same first and last characer. Input: abcab Output: 7 (a,abca,b,bcab, c, a, b) Naive Solution: T: O(n^2) S: O(1)      We initialize the count to length of the string as each and every element contribute to its length.       In the outer loop we will pick a element. In the inner loop we will compare it other elements and if they are equal we will increment the counter.  Code: import java.util.*; public class Solution {          public static void main(String[] args) {         String str = "aba";         char[] s = str.toCharArray();         int len = s.length;         int count = len;         for(int i=0; i<len-1; i++){             for(int j=i+1; j<len; j++){               ...
Read more