Matrix Search

• Time : $O(logn + logm)$
• Space : $O(1)$

int Solution::searchMatrix(vector<vector<int> > &A, int B) {
    int l = 0, r = A.size() - 1;
    while(l <= r) {
        int m = l + (r - l) / 2;
        if(A[m].front() <= B and B <= A[m].back()) return *lower_bound(begin(A[m]), end(A[m]), B) == B;
        else if(A[m].back() < B) l = m + 1;
        else r = m - 1;
    }
    return 0;
}