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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 Searchclass 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; }};
Search for a Range
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