Longest Sublist with K Distinct Numbers - Google Top Interview Questions

Problem Statement :

Given an integer k and a list of integers nums, return the length of the longest sublist that contains at most k distinct integers.


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

Example 1


k = 2

nums = [0, 1, 2, 1, 0]




The longest substring with 2 distinct integers is [1,2,1], which has length of 3.

Example 2


k = 1

nums = [0, 0, 0, 0, 0]



Example 3


k = 1

nums = [0, 1, 2, 3, 4]



Solution :


                        Solution in C++ :

int solve(int k, vector<int>& nums) {
    unordered_map<int, int> ct;
    int j = 0;
    int ans = 0;
    for (int i = 0; i < nums.size(); ++i) {
        while (ct.size() > k) {
            if (ct[nums[j]] == 0) {
        ans = max(ans, i - j + 1);
    return ans;

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int k, int[] nums) {
        int start = 0, res = 0;
        Map<Integer, Integer> map = new HashMap<>();

        for (int end = 0; end < nums.length; end++) {
            // Add current number into frequency map
            map.put(nums[end], map.getOrDefault(nums[end], 0) + 1);

            // While we have too many distinct elements, close the window
            while (start <= end && map.size() > k) {
                // Remove from the map using the starting index
                int st = nums[start++];
                map.put(st, map.get(st) - 1);
                if (map.get(st) == 0)

            // Compute maximum result after shrinking the window
            res = Math.max(res, end - start + 1);
        return res;

                        Solution in Python : 
class Solution:
    def solve(self, k, nums):
        if k == 0:
            return 0
        if not nums:
            return 0
        curr_set = collections.defaultdict(int)
        l = 0
        max_len = 0
        curr_len = 0

        for r in range(0, len(nums)):
            while len(curr_set) >= k and nums[r] not in curr_set:
                curr_set[nums[l]] -= 1
                if curr_set[nums[l]] == 0:
                    del curr_set[nums[l]]
                l += 1
            curr_set[nums[r]] += 1
            max_len = max(max_len, r - l + 1)
        return max_len

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