LF.24.Total Occurance
Given a target integer T and an integer array A sorted in ascending order, Find the total number of occurrences of T in A.
Examples
A = {1, 2, 3, 4, 5}, T = 3, return 1
A = {1, 2, 2, 2, 3}, T = 2, return 3
A = {1, 2, 2, 2, 3}, T = 4, return 0
Corner Cases
What if A is null? We should return 0 in this case.
1 public class Solution {
2 public int totalOccurrence(int[] array, int target) {
3 // Write your solution here
4 if (array == null || array.length == 0 ) {
5 return 0 ;
6 }
7 //find the first occurance and the last occurance then +1 is the # of items in between
8 int leftIndex = firstOccurence(array, target);
9 int rightIndex = lastOccurence(array, target);
10 //if either side not found, means there is no match
11 if (leftIndex == -1 || rightIndex == -1) {
12 return 0 ;
13 }
14 // 1 2 3 : 1st = 1 last= 3 totally should be 3 -1 +1 = 3 occurance
15 if(leftIndex <= rightIndex){
16 return (rightIndex - leftIndex) +1 ;
17 }
18 return 0;
19 }
20
21 private int firstOccurence(int[] array, int target){
22 int left = 0, right = array.length - 1 ;
23 while(left + 1 < right){
24 int mid = left + (right - left)/2 ;
25 if (array[mid] == target) {
26 right = mid ;
27 } else if(array[mid] < target){
28 left = mid ;
29 } else {
30 right = mid ;
31 }
32 }
33 if (array[left] == target) {
34 return left;
35 }
36 if (array[right] == target) {
37 return right ;
38 }
39 return -1 ;
40 }
41
42 // -1 means there is no target
43 private int lastOccurence(int[] array, int target){
44 int left = 0, right = array.length - 1 ;
45 while(left + 1 < right){
46 int mid = left + (right - left)/2 ;
47 if (array[mid] == target) {
48 left = mid ;
49 } else if(array[mid] < target){
50 left = mid ;
51 } else {
52 right = mid ;
53 }
54 }
55 if (array[right] == target) {
56 return right ;
57 }
58 if (array[left] == target) {
59 return left;
60 }
61 return -1 ;
62 }
63 }
