# K-Distinct Groups - Google Top Interview Questions

### Problem Statement :

```You are given a list of integers counts where counts[i] represents the number of items that exist of type i.

You are also given an integer k.

Return the maximum number of groups of size k we can have given that each group must have items of distinct types.

Constraints

n ≤ 100,000 where n is the length of counts

k ≤ n

Example 1

Input

counts = [3, 3, 2, 5]

k = 2

Output

6

Explanation

Let's name the four item types [a, b, c, d] respectively. We can have the following groups of two where
all elements are distinct types:

(a, d)

(b, d)

(a, b)

(a, b)

(c, d)

(c, d)

Example 2

Input

counts = [3, 2, 4]

k = 3

Output

2

Explanation

Let's name the three item types [a, b, c] respectively. We can only have these groups of three where all
elements are distinct types:

(a, b, c)

(a, b, c)

We cannot use (a, c, c) since there are two cs.```

### Solution :

```                        ```Solution in C++ :

bool can(vector<int>& v, int take, int k) {
long long total = 0;
for (int val : v) {
total += min(val, take);
}
}

int solve(vector<int>& counts, int k) {
int lhs = 0;
int rhs = 2e9;
while (lhs < rhs) {
int mid = (lhs + rhs + 1) / 2;
if (can(counts, mid, k)) {
lhs = mid;
} else {
rhs = mid - 1;
}
}
return lhs;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
// returns whether we can make num_groups of size k
public boolean possible(int[] count, int k, long num_groups) {
long required = num_groups * k;
for (int i = 0; i < count.length; i++) {
long use_this_much = Math.min(count[i], num_groups);
use_this_much = Math.min(use_this_much, required);
required -= use_this_much;
if (required == 0)
return true;
}
return false;
}
public int solve(int[] counts, int k) {
long l = 1;
long r = 0;
for (int c : counts) r += c;

int res = 0;
while (l <= r) {
long num_groups = l + (r - l) / 2;

if (possible(counts, k, num_groups)) {
res = (int) Math.max(res, num_groups);
l = num_groups + 1;
} else {
r = num_groups - 1;
}
}
return (int) res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def possible(self, counts, groups, k):
required = groups * k
for i in range(len(counts)):
use_this_much = min(counts[i], groups, required)
required -= use_this_much
if required == 0:
return True
return False

def solve(self, counts, k):
res = 0
l = 0
r = sum(counts)
while l <= r:
m = l + (r - l) // 2
if self.possible(counts, m, k):
l = m + 1
res = max(res, m)
else:
r = m - 1
return res```
```

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