74. Search a 2D Matrix

Problem Statement

You are given an m x n integer matrix matrix with the following two properties:

Each row is sorted in non-decreasing order.
The first integer of each row is greater than the last integer of the previous row.

Given an integer target, return true if target is in matrix or false otherwise.

You must write a solution in O(log(m * n)) time complexity.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-104 <= matrix[i][j], target <= 104

Intuition

Approach:

Treating the entire 2D array as one
Now, using mid/n, mid%n 
Trick acces the elemets

Links

https://leetcode.com/problems/search-a-2d-matrix/description/

Video Links

https://www.youtube.com/watch?v=JXU4Akft7yk&ab_channel=takeUforward

Approach 1:

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int low=0, high=matrix.size()*(matrix[0].size())-1;
        int n=matrix[0].size();

        while(low<=high){
            int mid=low+(high-low)/2;

            if(matrix[mid/n][mid%n]==target)
                return true;
            else if(matrix[mid/n][mid%n]<target)
                low=mid+1;
            else
                high=mid-1;
        }

        return false;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++