ObjecÂtive: – Given an Integer array. Array contains duplicates of all the numbers in array except one number . Find that number.
Example :
int [] A = { 2,1,3,5,5,3,2,1,6,7,7,8,8};
Output : Missing duplicate is 6
Appraoch:
- Naive solution is use Hash Table ..space complexity – O(n)
- Better solution – XOR
- A^A = 0 and A^B^A = B, so if we XOR all the elements, answer will be the missing no
- If we have only one element, the missing no will be that no
Code:
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| public class MissingDuplicate { | |
| // naive solution is use Hash Table ..space complexity – O(n) | |
| // better solutiion – XOR | |
| // A^A = 0 and A^B^A = B, so if we XOR all the elements, answer will be the | |
| // missing no | |
| public int find(int[] A) { | |
| int miss = A[0]; // if we have only one element, the missing no will be | |
| // that no | |
| for (int i = 1; i < A.length; i++) { | |
| miss = miss ^ A[i]; | |
| } | |
| return miss; | |
| } | |
| public static void main(String[] args) { | |
| int[] A = { 2, 1, 3, 5, 5, 3, 2, 1, 6, 7, 7, 8, 8 }; | |
| MissingDuplicate i = new MissingDuplicate(); | |
| System.out.println("Missing duplicate is " + i.find(A)); | |
| } | |
| } |
Output:
Missing duplicate is 6
Nice One