Given an integer array of size N - 1, containing all the numbers from 1 to N except one, find the missing number.
Assumptions
The given array is not null, and N >= 1
Examples
A = {2, 1, 4}, the missing number is 3
A = {1, 2, 3}, the missing number is 4
A = {}, the missing number is 1
1 //time o(n) space o(n)
2 public int missing_method1(int[] array) {
3 // Write your solution here
4 //corner case
5 if (array == null){
6 return -1 ;
7 }
8 // since we know for sure there is one number missing from array
9 int size = array.length ;
10 //1: put for 1 to size into hashset
11 Set<Integer> dic = new HashSet<>(size) ;
12 for (int i = 0 ; i <size ; i++){
13 dic.add(array[i]);
14 }
15 //2: and then loop through the array and cross check with the dic
16 for (int i = 1; i <= size +1 ; i++) {
17 if (!dic.contains(i)){
18 return i;
19 }
20 }
21 return -1 ;
22 }
23
24 // time o(n) space o(1)
25 //use sum
26 public int missing_method2(int[] array) {
27 // Write your solution here
28 //corner case
29 if (array == null){
30 return -1 ;
31 }
32 int n = array.length +1 ;
33 long targetSum = (n +0L) * (n+1)/2 ;
34 long actualSum = 0L ;
35 for(int num : array){
36 actualSum += num ;
37 }
38 return (int)(targetSum - actualSum);
39 }
40
41 //xor bit operation - time o(n) sapce: o(1)
42 public int missing_method3(int[] array) {
43 // Write your solution here
44 //corner case
45 if (array == null){
46 return -1 ;
47 }
48 int n = array.length + 1 ;
49 int xor = 0;
50 //xor from 1 to n
51 for (int i = 1; i <= n; i++) {
52 xor ^= i ;
53 }
54 //then iterate all the items in array
55 /*
56 after this operation, all the numbers from 1 to n
57 are pair xor'ed except for the missing number.
58 since x ^ x = 0, the remaining number is the result
59 */
60 for(int num : array){
61 xor ^= num ;
62 }
63 return xor ;
64 }
