How to Find Missing Number on Integer Array of 1 to 100

Let's understand the problem statement, we have numbers from 1 to 100 that are put into an integer array, what's the best way to find out which number is missing? If Interviewer especially mentions 1 to 100 then you can apply the above trick about the sum of the series as shown below as well. If it has more than one missing element that you can use BitSet class, of course only if your interviewer allows it.

1) Sum of the series: Formula: n (n+1)/2( but only work for one missing number)

2) Use BitSet, if an array has more than one missing elements.

import java.util.Arrays;
import java.util.BitSet;
 
/**
 * Java program to find missing elements in a Integer array containing 
 * numbers from 1 to 100.
 *
 * @author Javin Paul
 */
public class MissingNumberInArray {
 
    public static void main(String args[]) {

        // one missing number
        printMissingNumber(new int[]{1, 2, 3, 4, 6}, 6);
 
        // two missing number
        printMissingNumber(new int[]{1, 2, 3, 4, 6, 7, 9, 8, 10}, 10);
 
        // three missing number
        printMissingNumber(new int[]{1, 2, 3, 4, 6, 9, 8}, 10);
 
        // four missing number
        printMissingNumber(new int[]{1, 2, 3, 4, 9, 8}, 10);
 
        // Only one missing number in array
        int[] iArray = new int[]{1, 2, 3, 5};
        int missing = getMissingNumber(iArray, 5);
        System.out.printf("Missing number in array %s is %d %n", 
                             Arrays.toString(iArray), missing);
    }
    /**
    * A general method to find missing values from an integer array in Java.
    * This method will work even if array has more than one missing element.
    */
    private static void printMissingNumber(int[] numbers, int count) {
        int missingCount = count - numbers.length;
        BitSet bitSet = new BitSet(count);
 
        for (int number : numbers) {
            bitSet.set(number - 1);
        }
 
        System.out.printf("Missing numbers in integer array %s, with total number %d is %n",
        Arrays.toString(numbers), count);
        int lastMissingIndex = 0;

        for (int i = 0; i < missingCount; i++) {
            lastMissingIndex = bitSet.nextClearBit(lastMissingIndex);
            System.out.println(++lastMissingIndex);
        }
 
    }
    /**
    * Java method to find missing number in array of size n containing
    * numbers from 1 to n only.
    * can be used to find missing elements on integer array of 
    * numbers from 1 to 100 or 1 - 1000
    */
    private static int getMissingNumber(int[] numbers, int totalCount) {
        int expectedSum = totalCount * ((totalCount + 1) / 2);
        int actualSum = 0;
        for (int i : numbers) {
            actualSum += i;
        }
 
        return expectedSum - actualSum;
    }
 
}
 

Output

Missing numbers in integer array [1, 2, 3, 4, 6], with total number 6 is
5
Missing numbers in integer array [1, 2, 3, 4, 6, 7, 9, 8, 10], with total number 10 is
5
Missing numbers in integer array [1, 2, 3, 4, 6, 9, 8], with total number 10 is
5
7
10
Missing numbers in integer array [1, 2, 3, 4, 9, 8], with total number 10 is
5
6
7
10
Missing number in array [1, 2, 3, 5] is 4

Source: Java67

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