Search a 2D matrix II

/*
Logic:
If we start from top right then we will find that left side elements are smaller
bottom side elements are bigger and we can reach at any element by comparing
current elment with target.
*/
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
       int r=matrix.size(),c=matrix[0].size();
       int i=0,j=c-1;
       while(i<r && j>=0){
           if(matrix[i][j]==target) return true;
           else if(matrix[i][j]>target){
               j--;
           }
           else i++;
       }
       return false;
    }
};

#Python
class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        i=0
        j=len(matrix[0])-1
        while i<len(matrix) and j>=0:
            if matrix[i][j]==target:
                return True
            if matrix[i][j]>target:
                j-=1
            else:
                i+=1
        return False