2200 - 2299
2261 - K Divisible Elements Subarrays (Medium)
Problem Link
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
iwherenums1[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 <= 2001 <= nums[i], p <= 2001 <= k <= nums.length
Approach 1: Brute Force
Build all subarrays and insert them to set. The answer would be the size of the set.
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();
}
};