Search a 2D Matrix LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Search a 2D Matrix in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Search a 2D Matrix– LeetCode Problem

Search a 2D Matrix– LeetCode Problem

Problem:

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

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

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

Search a 2D Matrix– LeetCode Solutions

Search a 2D Matrix in C++:

class Solution {
 public:
  bool searchMatrix(vector<vector<int>>& matrix, int target) {
    if (matrix.empty())
      return false;

    const int m = matrix.size();
    const int n = matrix[0].size();

    int l = 0;
    int r = m * n;

    while (l < r) {
      const int mid = l + (r - l) / 2;
      const int i = mid / n;
      const int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
      else
        r = mid;
    }

    return false;
  }
};

Search a 2D Matrix in Java:

class Solution {
  public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix.length == 0)
      return false;

    final int m = matrix.length;
    final int n = matrix[0].length;

    int l = 0;
    int r = m * n;

    while (l < r) {
      final int mid = l + (r - l) / 2;
      final int i = mid / n;
      final int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
      else
        r = mid;
    }

    return false;
  }
}

Search a 2D Matrix in Python:

class Solution:
  def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
    if not matrix:
      return False

    m = len(matrix)
    n = len(matrix[0])

    l = 0
    r = m * n

    while l < r:
      mid = l + (r - l) // 2
      i = mid // n
      j = mid % n
      if matrix[i][j] == target:
        return True
      if matrix[i][j] < target:
        l = mid + 1
      else:
        r = mid

    return False