**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);
}
return total / k >= 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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