Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution: It takes O(lgN) to find both the lower-bound and upper-bound.

 1 class Solution {
 2 public:
 3     int lowerBound(int A[], int n, int target) {
 4         int i = 0, j = n-1;
 5         while(i <= j) {
 6             int mid = i + (j-i)/2;
 7             if(A[mid] < target)
 8                 i = mid+1;
 9             else
10                 j = mid-1;
11         }
12         if(i == -1 || i == n || A[i] != target) return -1;
13         else return i;
14     }
15     
16     int upperBound(int A[], int n, int target) {
17         int i = 0, j = n-1;
18         while(i <= j) {
19             int mid = i + (j-i)/2;
20             if(A[mid] <= target) {
21                 i = mid+1;
22             }
23             else {
24                 j = mid-1;
25             }
26         }
27         if(j == -1 || j == n || A[j] != target) return -1;
28         else return j;
29     }
30     
31     vector<int> searchRange(int A[], int n, int target) {
32         vector<int> res(2,-1);
33         res[0] = lowerBound(A, n, target);
34         res[1] = upperBound(A, n, target);
35         return res;
36     }
37 };