Search a 2D Matrix LeetCode Solution

Problem – Search a 2D Matrix LeetCode Solution

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 LeetCode Solution in C++

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

Search a 2D Matrix LeetCode Solution in Java

  public boolean searchMatrix(int[][] matrix, int target) {
            int i = 0, j = matrix[0].length - 1;
            while (i < matrix.length && j >= 0) {
                    if (matrix[i][j] == target) {
                        return true;
                    } else if (matrix[i][j] > target) {
                        j--;
                    } else {
                        i++;
                    }
                }
            
            return false;
        }

Search a 2D Matrix LeetCode Solution in Python

class Solution:
    # @param matrix, a list of lists of integers
    # @param target, an integer
    # @return a boolean
    # 8:21
    def searchMatrix(self, matrix, target):
        if not matrix or target is None:
            return False

        rows, cols = len(matrix), len(matrix[0])
        low, high = 0, rows * cols - 1
        
        while low <= high:
            mid = (low + high) / 2
            num = matrix[mid / cols][mid % cols]

            if num == target:
                return True
            elif num < target:
                low = mid + 1
            else:
                high = mid - 1
        
        return False
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