Search a 2D Matrix LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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 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 Solutions

Time: O(mnlogmn)
Space: O(1)

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;
  }
};

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;
  }
}

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

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