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?
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 element.
import java.util.Arrays; import java.util.BitSet; public class temp{ private static void Missing_Number(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("\nMissing 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.printf(" %d \n",++lastMissingIndex); } } public static void main(String args[]){ Missing_Number(new int[]{1, 2, 3, 4, 6}, 6); Missing_Number(new int[]{1, 2, 3, 4, 6, 7, 9, 8, 10}, 10); Missing_Number(new int[]{1, 2, 3, 4, 6, 9, 8}, 10); Missing_Number(new int[]{1, 2, 3, 4, 9, 8}, 10); } }
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
ScrShot
![](https://www.fcukthecode.com/wp-content/uploads/2024/03/Screenshot-from-2024-03-21-20-00-47.png)