K Divisible Elements Subarrays

Problem Statement

In this tutorial, we will solve the K Divisible Elements Subarrays problem . We will provide the implementation of the solution in Python, Java, and C++.

Problem Description

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

Examples

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

Solution of Given Problem

Intuition and Approach

The problem can be solved using a brute force approach or an optimized Technique.

• Brute Force
• Optimized approach

Approach 1:Brute Force (Naive)

Brute Force Approach: Generate all possible subarrays and count those that satisfy the condition (at most k elements divisible by p).

Codes in Different Languages

C++

Written by @AmruthaPariprolu

#include <vector>
#include <unordered_set>

using namespace std;

int countDistinctSubarrays(vector<int>& nums, int k, int p) {
    unordered_set<string> subarrays;
    int n = nums.size();
    
    for (int start = 0; start < n; ++start) {
        int count = 0;
        for (int end = start; end < n; ++end) {
            if (nums[end] % p == 0) count++;
            if (count > k) break;
            string subarray;
            for (int i = start; i <= end; ++i) {
                subarray += to_string(nums[i]) + ",";
            }
            subarrays.insert(subarray);
        }
    }
    return subarrays.size();
}

Java

Written by @AmruthaPariprolu

import java.util.HashSet;

public class Solution {
    public int countDistinctSubarrays(int[] nums, int k, int p) {
        HashSet<String> subarrays = new HashSet<>();
        int n = nums.length;
        
        for (int start = 0; start < n; ++start) {
            int count = 0;
            for (int end = start; end < n; ++end) {
                if (nums[end] % p == 0) count++;
                if (count > k) break;
                StringBuilder subarray = new StringBuilder();
                for (int i = start; i <= end; ++i) {
                    subarray.append(nums[i]).append(",");
                }
                subarrays.add(subarray.toString());
            }
        }
        return subarrays.size();
    }
}

Python

Written by @AmruthaPariprolu

def countDistinctSubarrays(nums, k, p):
    subarrays = set()
    n = len(nums)
    
    for start in range(n):
        count = 0
        for end in range(start, n):
            if nums[end] % p == 0:
                count += 1
            if count > k:
                break
            subarray = ','.join(map(str, nums[start:end + 1]))
            subarrays.add(subarray)
    
    return len(subarrays)

Complexity Analysis

• Time Complexity: $O(n^3)$
• where n is the length of the array due to nested loops.
• Space Complexity: $O(n^2)$
• Additional space may be required for storing for storing subarrays in a set.

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