240. Search a 2D Matrix II

Problem Statement

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

Constraints:

m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-109 <= matrix[i][j] <= 109
All the integers in each row are sorted in ascending order.
All the integers in each column are sorted in ascending order.
-109 <= target <= 109

Intuition

Links

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

Video Links

Approach 1:

MlogN time , 1 Space
Binary Search on all rows

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& arr, int target) {
        
        for(auto &it: arr){
            int low=0, high=arr[0].size()-1;
            while(low<=high){
                int mid = low+(high-low)/2;
                if(it[mid] == target)
                    return true;
                else if(it[mid] > target)
                    high = mid-1;
                else
                    low = mid+1;
            }
        }

        return false;
    }
};

Approach 2:

m+n time 
Start from the edge 0,0 row 
This will clearly state you whether to go bottom or left

Do this until you reach target

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& arr, int target) {
        int i=0, j=arr[0].size()-1;
        while(i<arr.size() and j>=0){
            if(arr[i][j] == target)
                return true;
            else if(arr[i][j] > target)
                j--;
            else
                i++;
        }

        return false;
    }
};

Approach 3:

C++

Approach 4:

C++