5499. Detect Pattern of Length M Repeated K or More Times

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 = 3
Output: true
Explanation: 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 = 2
Output: true
Explanation: 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 = 3
Output: false
Explanation: 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 = 2
Output: false
Explanation: 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 = 3
Output: false
Explanation: 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 <= 100
• 1 <= arr[i] <= 100
• 1 <= m <= 100
• 2 <= k <= 100

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给定一个数组，为是否存在长度为m的子数组，这个子数组重复了至少k词。

观察数据范围，数据量很小，直接暴力求解，代码如下：

class Solution {
public:
    bool containsPattern(vector<int>& arr, int m, int k) {
        int n = arr.size();
        if(m * k > n) return false;
        for(int i = 0; i < n; ++i) {
            int rep = 0;
            for(int j = i; j < n; j += m) {
                bool good = true;
                for(int u = 0; u < m; ++u) {
                    if(u + j >= n || arr[u + j] != arr[i + u]) {
                        good = false;
                        break;
                    }
                }
                if(!good) break;
                else ++rep;
            }
            if(rep >= k) return true;
        }
        return false;
    }
};

本代码提交AC，用时4MS。