Search a 2D Matrix II | February Leetcoding challenge 2021 | Day 23
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Manas Sinha
Developer | Designer

By Manas | 23 February 2021

February LeetCoding Challenge 2021

Day 23

Search a 2D Matrix II

PROBLEM  STATEMENT:
Write an efficient algorithm that searches for a target value in an m x n integer matrix. The 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 <= matix[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

Explanation

Start with the top right corner. If the value if greater than the target value move left else move down.
ALGORITHM:
  • Start at top right corner.
  • If the current value is greater than the target value move left.
  • Else move down.
  • Repeat uniti you go out of bounds.
  • Time complexity : O(n+m)
  • Space complexity : O(1)

code

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        
        int row = 0;
        int col = n-1;
        while(col>=0 && row<m){
            if(matrix[row][col] == target) return true;
            if(matrix[row][col] > target) --col;
            else ++row;
        }
        return false;
    }
};

PYTHON

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m = len(matrix)
        n = len(matrix[0])
        
        row = 0
        col = n-1
        
        while(col>=0 and row<m):
            if(matrix[row][col] == target):
                return True
            if(matrix[row][col] > target):
                col -= 1
            else:
                row += 1
        return False
            

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