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].

 

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 Array Binary Search

 

class Solution {
public:
    int find(int a[],int left,int right,int key,bool isLeft){
        if(left>right)
            return -1;
        int mid=(right+left)/2;
        if(a[mid]==key){
            int pos=isLeft ? find(a,left,mid-1,key,isLeft) : find(a,mid+1,right,key,isLeft);
            return pos==-1 ? mid : pos; 
        }else if(a[mid]>key)
            return find(a,left,mid-1,key,isLeft);
        else
            return find(a,mid+1,right,key,isLeft);
    }

    vector<int> searchRange(int A[], int n, int target) {
        int pos1=find(A,0,n-1,target,true);
        int pos2=find(A,0,n-1,target,false);
        vector<int> range;
        range.push_back(pos1);
        range.push_back(pos2);
        return range;
    }
};