# My simple and easy-understanding solution with O(logn) time complexity

• ``````class Solution {
public:
int searchStart( int A[], int start, int end, int target ) {
while ( start <= end ) {
int mid = ( start + end ) / 2;
if ( A[mid] == target )
end = mid-1;
else
start = mid+1;
}
return start;
}
int searchEnd( int A[], int start, int end, int target ) {
while ( start <= end ) {
int mid = ( start + end ) / 2;
if ( A[mid] == target )
start = mid+1;
else
end = mid-1;
}
return end;
}
vector<int> searchRange(int A[], int n, int target) {
int start = 0;
int end = n-1;
while ( start <= end ) {
int mid = (start + end ) / 2;
if ( A[mid] < target )
start = mid+1;
else if ( A[mid] > target )
end = mid-1;
else {
int startIndex = searchStart( A, start, mid, target );
int endIndex = searchEnd( A, mid, end, target );
vector<int> v;
v.push_back(startIndex);
v.push_back(endIndex);
return v;
}
}
vector<int> v;
v.push_back(-1);
v.push_back(-1);
return v;
}
};``````

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