/*
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].
*/
class Solution {
public:
int binarySearch(int A[], int n, int target)
{
int i = 0, j = n - 1;
while (i <= j)
{
int m = i + (j - i) / 2;
if (A[m] == target) return m;
if (A[m] > target) j = m - 1;
if (A[m] < target) i = m + 1;
}
return -1;
}
vector<int> searchRange(int A[], int n, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int pos_target = -1;
int left = 0, right = n - 1;
while (left <= right)
{
int m = left + (right - left) / 2;
if (A[m] == target) {pos_target = m; break;}
if (A[m] > target) right = m - 1;
if (A[m] < target) left = m + 1;
}
if (pos_target == -1) return vector<int>{-1, -1};
int start = pos_target, i = left, j = pos_target - 1;
while (i <= j)
{
int m = i + (j - i) / 2;
if (A[m] == target)
{
j = m;
if (i == j)
{
start = i;
break;
}
}
else
{
i = m + 1;
}
}
int end = pos_target; i = pos_target + 1, j = right;
while (i <= j)
{
int m = i + (j - i) / 2;
if (A[m] == target)
{
i = m;
if (i == j)
{
end = i;
break;
}
}
else
{
j = m - 1;
}
}
return vector<int>{start, end};
}
};
/*
A better approach
*/
class Solution {
public:
vector<int> searchRange(int A[], int n, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int lower = -1, upper = -1, i = 0, j = n - 1;
//find lowerbound
while (i <= j)
{
int m = i + (j - i) / 2;
if (A[m] == target)
{
lower = m;
j = m - 1;
}
else if (A[m] > target)
{
j = m - 1;
}
else
{
i = m + 1;
}
}
if (lower == -1) return vector<int>{-1, -1};
i = lower, j = n - 1;
while (i <= j)
{
int m = i + (j - i) / 2;
if (A[m] == target)
{
i = m + 1;
upper = m;
}
else
{
j = m - 1;
}
}
return vector<int>{lower, upper};
}
};