2261-KDivisibleElementsSubarrays.go

package main

// 2261. K Divisible Elements Subarrays
// 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.

// 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

// Follow up:
// Can you solve this problem in O(n2) time complexity?

import "fmt"

func countDistinct(nums []int, k int, p int) int {
    count, n, key := 0, len(nums), ""
    mp := make(map[string]bool)
    for i := 0; i < n; i++ {
        count, key = 0, ""
        for j := i; j < n; j++ {
            if nums[j] % p == 0 && count == k {
                break
            }
            if nums[j] % p == 0 {
                count++
            }
            key += string(nums[j]) + "-"
            mp[key] = true
        }
    }
    return len(mp)
}

func countDistinct1(nums []int, k int, p int) int {
    res, n, mp := 0, len(nums), make(map[string]bool)
    for i := 0; i < n; i++ {
        count, key := 0, []byte{}
        for j := i ;j < n; j++{
            if nums[j] % p == 0 {
                count++
            }
            if count > k {
                break
            }
            key = append(key, byte(nums[j]) + '0')
            if !mp[string(key)] {
                mp[string(key)]= true
                res++
            }
        }
    }
    return res
}

func main() {
    // 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.
    fmt.Println(countDistinct([]int{2,3,3,2,2}, 2, 2)) // 11
    // 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.
    fmt.Println(countDistinct([]int{1,2,3,4}, 4, 1)) // 10

    fmt.Println(countDistinct1([]int{2,3,3,2,2}, 2, 2)) // 11
    fmt.Println(countDistinct1([]int{1,2,3,4}, 4, 1)) // 10
}