algorithms/cpp/search2DMatrix/search2DMatrix.cpp

// Source : https://oj.leetcode.com/problems/search-a-2d-matrix/
// Author : Hao Chen
// Date   : 2014-06-23

/********************************************************************************** 
* 
* Write an efficient algorithm that searches for a value in an m x n matrix. 
* This matrix has the following properties:
* 
* Integers in each row are sorted from left to right.
* The first integer of each row is greater than the last integer of the previous row.
* 
* For example,
* 
* Consider the following matrix:
* 
* [
*   [1,   3,  5,  7],
*   [10, 11, 16, 20],
*   [23, 30, 34, 50]
* ]
* 
* Given target = 3, return true.
*               
**********************************************************************************/

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        
        int idx = vertical_binary_search(matrix, target);
        if (idx<0){
            return false;
        }
        
        idx = binary_search(matrix[idx], target);
        
        return (idx < 0 ? false : true);
        
    }
    int vertical_binary_search(vector< vector<int> > v, int key){
        int low = 0;
        int high = v.size()-1;
        while(low <= high){
            int mid = low + (high-low)/2;
            if (v[mid][0] == key){
                return mid;
            }
            if (key < v[mid][0]){
                high = mid - 1;
                continue;
            }
            if (key > v[mid][0]){
                low = mid + 1;
                continue;
            }
        }
        return low-1;        
    }
    
    int binary_search(vector<int> v, int key) {
        int low = 0;
        int high = v.size()-1;
        while(low <= high){
            int mid = low + (high-low)/2;
            if (v[mid] == key){
                return mid;
            }
            if (key < v[mid]){
                high = mid - 1;
                continue;
            }
            if (key > v[mid]){
                low = mid + 1;
                continue;
            }
        }
        return -1;
    }
};