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.


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

k ≤ n

Example 1


counts = [3, 3, 2, 5]

k = 2




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


counts = [3, 2, 4]

k = 3




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)
                r = m - 1
        return res

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