Search for A Range

06/26/2016 Array Binary Search

Question
Solution

Question

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

Result: Accepted Time: 0ms

Here should be some explanations.

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

Complexity Analytics

Time Complexity: $O(log(n))$
Space Complexity: $O(1)$