#74

Search a 2D Matrix

medium· Binary Searchruns: 0

You are given an m x n integer matrix with two properties: each row is sorted in non-decreasing order, and the first integer of each row is greater than the last integer of the previous row. Given an integer target, return true if target is in the matrix. The algorithm must run in O(log(m * n)) time.

wpm 0acc 100%time 0:000 / 492

class Solution:
    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
        rows, cols = len(matrix), len(matrix[0])
        left, right = 0, rows * cols - 1
        while left <= right:
            mid = (left + right) // 2
            value = matrix[mid // cols][mid % cols]
            if value == target:
                return True
            if value < target:
                left = mid + 1
            else:
                right = mid - 1
        return False

click the box to focus · tab inserts 4 spaces · backspace to correct · esc to pause

desktop only

codedrill is a typing game and needs a real keyboard. open this on a laptop or desktop to practice.

you can still browse problems and sections from your phone.