K-Distinct Sublists - Google Top Interview Questions


Problem Statement :


Given a list of integers nums and an integer k, return the number of sublists such that there's exactly k distinct numbers in the sublist.

Constraints

k ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 1, 2, 3]

k = 2

Output

3

Explanation

We have the following sublists which have exactly 2 distinct numbers: [1, 1, 2], [1, 2], [2, 3]



Solution :



title-img




                        Solution in C++ :

int solve(vector<int>& nums, int k) {
    unordered_map<int, int> fr;
    nums.push_back(0);
    int i = 0, j = 0, n = nums.size(), res = 0, cnt = 1;
    while (j < n && k > 0) {
        fr[nums[j]]++;
        if (fr[nums[j]] == 1) k--;
        j++;
    }
    while (j < n) {
        if (k == 0) {
            while (fr[nums[i]] > 1) {
                fr[nums[i]]--;
                i++;
                cnt++;
            }
            res += cnt;
            fr[nums[j]]++;
            if (fr[nums[j]] == 1) k--;
            j++;
        } else {
            cnt = 1;
            fr[nums[i]]--;
            if (fr[nums[i]] == 0) k++;
            i++;
        }
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int k) {
        final int N = nums.length;
        Map<Integer, Integer> counts = new HashMap<>(N);
        int res = 0;
        for (int a = 0, b = -1, c = -1, kinds = 0; a != N; a++) {
            while (b + 1 != N && kinds < k) {
                b++;
                int prev = counts.getOrDefault(nums[b], 0);
                if (0 == prev)
                    kinds++;
                counts.put(nums[b], prev + 1);
            }
            c = Math.max(c, b);
            while (c + 1 != N && counts.containsKey(nums[c + 1])) c++;

            if (kinds == k)
                res += (c + 1 - b);
            if (kinds < k)
                break;
            {
                int prev = counts.get(nums[a]) - 1;
                if (prev == 0) {
                    counts.remove(nums[a]);
                    kinds--;
                } else
                    counts.put(nums[a], prev);
            }
        }
        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def windows(self, nums, k):
        i = j = dis = ans = 0
        d = defaultdict(int)
        while j < len(nums):
            if d[nums[j]] == 0:
                dis += 1
            d[nums[j]] += 1
            while i <= j and dis > k:
                if d[nums[i]] == 1:
                    dis -= 1
                d[nums[i]] -= 1
                i += 1
            if dis <= k:
                ans += j - i + 1
            j += 1
        return ans

    def solve(self, nums, k):
        return self.windows(nums, k) - self.windows(nums, k - 1)
                    


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