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2261 - K Divisible Elements Subarrays (Medium)

https://leetcode.com/problems/k-divisible-elements-subarrays/

Problem Statement​

Given an integer array nums and two integers k and p, return the number of distinct subarrays which have at most k elements divisible by p.

Two arrays nums1 and nums2 are said to be distinct if:

  • They are of different lengths, or
  • There exists at least one index i where nums1[i] != nums2[i].

A subarray is defined as a non-empty contiguous sequence of elements in an array.

Example 1:

Input: nums = [2,3,3,2,2], k = 2, p = 2
Output: 11
Explanation:
The elements at indices 0, 3, and 4 are divisible by p = 2.
The 11 distinct subarrays which have at most k = 2 elements divisible by 2 are:
[2], [2,3], [2,3,3], [2,3,3,2], [3], [3,3], [3,3,2], [3,3,2,2], [3,2], [3,2,2], and [2,2].
Note that the subarrays [2] and [3] occur more than once in nums, but they should each be counted only once.
The subarray [2,3,3,2,2] should not be counted because it has 3 elements that are divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 4, p = 1
Output: 10
Explanation:
All element of nums are divisible by p = 1.
Also, every subarray of nums will have at most 4 elements that are divisible by 1.
Since all subarrays are distinct, the total number of subarrays satisfying all the constraints is 10.

Constraints:

  • 1 <= nums.length <= 200
  • 1 <= nums[i], p <= 200
  • 1 <= k <= nums.length

Approach 1: Brute Force​

Build all subarrays and insert them to set. The answer would be the size of the set.

Written by @wingkwong
class Solution {
public:
    int countDistinct(vector<int>& nums, int k, int p) {
        int ans = 0, n = nums.size();
        set<string> s;
        for (int i = 0; i < n; i++) {
            int cnt = 0;
            string t;
            for (int j = i; j < n; j++) {
                // below line will cause TLE
                // cnt += nums[j] % p == 0;
                // use below instead
                cnt += nums[j] % p == 0 ? 1 : 0;
                // or 
                // if (nums[j] % p == 0) cnt += 1;
                if (cnt <= k) {
                    t = t + to_string(nums[j]) + "|";
                    s.insert(t);
                } else {
                    break;
                }
            }
        }
        return (int) s.size();
    }
};