Search a 2D Matrix II LeetCode Solution | Easy Approach

Search a 2D Matrix II 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

Search a 2D Matrix II Solutions

✅Time: O(m+n)
✅Space: O(1)

C++

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

    while (r < matrix.size() && c >= 0) {
      if (matrix[r][c] == target)
        return true;
      if (matrix[r][c] > target)
        --c;
      else
        ++r;
    }

    return false;
  }
};

Java

 class Solution {
  public boolean searchMatrix(int[][] matrix, int target) {
    int r = 0;
    int c = matrix[0].length - 1;

    while (r <= matrix.length && c >= 0) {
      if (matrix[r][c] == target)
        return true;
      if (matrix[r][c] > target)
        --c;
      else
        ++r;
    }

    return false;
  }
}

Python

 class Solution:
  def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
    r = 0
    c = len(matrix[0]) - 1

    while r < len(matrix) and c >= 0:
      if matrix[r][c] == target:
        return True
      if target < matrix[r][c]:
        c -= 1
      else:
        r += 1

    return False

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