Leetcode #74: Search a 2D Matrix
In this guide, we solve Leetcode #74 Search a 2D Matrix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an m x n integer matrix matrix with the following two properties: Each row is sorted in non-decreasing order. The first integer of each row is greater than the last integer of the previous row.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Binary Search, Matrix
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true
Python Solution
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
m, n = len(matrix), len(matrix[0])
left, right = 0, m * n - 1
while left < right:
mid = (left + right) >> 1
x, y = divmod(mid, n)
if matrix[x][y] >= target:
right = mid
else:
left = mid + 1
return matrix[left // n][left % n] == target
Complexity
The time complexity is , where and are the number of rows and columns of the matrix, respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.