# 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.

Constraints

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

Example 1

Input

k = 2

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

Output

3

Explanation

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

Example 2

Input

k = 1

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

Output

5

Example 3

Input

k = 1

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

Output

1```

### 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) {
++ct[nums[i]];
while (ct.size() > k) {
--ct[nums[j]];
if (ct[nums[j]] == 0) {
ct.erase(nums[j]);
}
++j;
}
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)
map.remove(st);
}

// 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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