**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 :

` ````
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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