[leetcode] 1566. Detect Pattern of Length M Repeated K or More Times

[leetcode] 1566. Detect Pattern of Length M Repeated K or More Times

Description

Given an array of positive integers arr, find a pattern of length m that is repeated k or more times.

A pattern is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times consecutively without overlapping. A pattern is defined by its length and the number of repetitions.

Return true if there exists a pattern of length m that is repeated k or more times, otherwise return false.

Example 1:

Input: arr = [1,2,4,4,4,4], m = 1, k = 3Output: trueExplanation: The pattern (4) of length 1 is repeated 4 consecutive times. Notice that pattern can be repeated k or more times but not less.

Example 2:

Input: arr = [1,2,1,2,1,1,1,3], m = 2, k = 2Output: trueExplanation: The pattern (1,2) of length 2 is repeated 2 consecutive times. Another valid pattern (2,1) is also repeated 2 times.

Example 3:

Input: arr = [1,2,1,2,1,3], m = 2, k = 3Output: falseExplanation: The pattern (1,2) is of length 2 but is repeated only 2 times. There is no pattern of length 2 that is repeated 3 or more times.

Example 4:

Input: arr = [1,2,3,1,2], m = 2, k = 2Output: falseExplanation: Notice that the pattern (1,2) exists twice but not consecutively, so it doesn't count.

Example 5:

Input: arr = [2,2,2,2], m = 2, k = 3Output: falseExplanation: The only pattern of length 2 is (2,2) however it's repeated only twice. Notice that we do not count overlapping repetitions.

Constraints:

2 <= arr.length <= 1001 <= arr[i] <= 1001 <= m <= 1002 <= k <= 100

分析

题目的意思是：给你一个数组，找出重复超过K次的子数组，这道题目虽然是easy类型，我没啥思路，关键在于暴力破解写三个循环有点麻烦，参考了一下别人的实现，首先遍历数组，然后判断把一个子数组拓展K次是否等于相对应位置的值，这样就很容易判别出来了，python这种数组进行比较的做法还是挺巧妙的。

代码

class Solution: def containsPattern(self, arr: List[int], m: int, k: int) -> bool: n=len(arr) if(m*k>n): return False for i in range(n-m): if(arr[i:i+m]*k==arr[i:i+m*k]): return True return False

参考文献

​​[LeetCode] [Python 3] Short solution faster than 92%​​

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