Problem description

Given an array of integers nums sorted in ascending order, 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].

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Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

solution

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class Solution {
public:
int binary_search_left(vector<int>& nums, int target){
int low = 0;
int high = nums.size();
int mid;
while(low < high) {
mid = low + (high - low) / 2;
if(nums[mid] < target)
low = mid + 1;
else high = mid;
}
if(nums[low] == target && low != nums.size())
return low;
else return -1;
}

int binary_search_right(vector<int>& nums, int target){
int low = 0;
int high = nums.size();
int mid;
while(low < high) {
mid = low + (high - low) / 2;
if(nums[mid] > target)
high = mid;
else low = mid + 1;
}
if(nums[low -1] == target && (low-1) != nums.size())
return low - 1;
else return -1;
}

vector<int> searchRange(vector<int>& nums, int target) {
if(nums.size() == 0) return vector<int>{-1,-1};

int left = binary_search_left(nums, target);
int right = binary_search_right(nums, target);
return vector<int>{left, right};
}
};