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MEX - Editorial



Author: Hruday Pabbisetty
Tester: Alexey Zayakin
Editorialist: Jakub Safin




sorting, greedy algorithms


You are given a multiset of non-negative integers of size $N$; maximise its minimum excluded non-negative integer by adding at most $K$ integers to it.


Sort $S_1,\dots,S_N$, remove duplicate elements. Greedily add numbers between elements $S_i$, the answer is the smallest number you can't add; it's possible to add all numbers between two consecutive elements at once.


If we want the mex to be $\ge k$, we need to have all integers $0$ through $k-1$ in the multiset.

The most straightforward algorithm would be to go through non-negative integers in increasing order. We can ignore integers that are already in the multiset. Whenever encountering an integer $x$ that's not in the multiset, we need to add it - if we didn't do it, the mex would be at most $x$. The first such integer which we can't add to our multiset (because we already added $K$ of them) is the maximum possible mex.

If we simply mark the numbers from the initial multiset in an array of size $\mathrm{max}S$, we can determine if a number is in the initial multiset in $O(1)$ and the time complexity of this greedy algorithm is $O(\mathrm{max}S+K+N)$, since we only need to process at most $K+N$ integers before stopping; the memory complexity is $O(\mathrm{max}S)$.

The constraints are low enough that this is enough to solve the problem, but we can do better.

A faster solution

Let's make this algorithm independent on $\mathrm{max}S$.

We can sort the array $S_{1..N}$ in non-decreasing order and discard duplicate elements, since the mex doesn't depend on how many times each number occurs in the multiset, only which numbers; as we go through all non-negative integers, we can remember the index of the last element $S_i$ we encountered (or $i=0$ if we haven't encountered any yet). If $i < N$, then we know that the next number we could encounter that's in the multiset is $S_{i+1}$; when that happens, we can just increment $i$ by 1.

With this approach, we only need $O(1)$ time with $O(N)$ memory to check if a number is in the initial multiset and sorting takes $O(N\log{N})$, so the whole algorithm takes $O(N\log{N}+K)$ time and $O(N)$ memory.

An even faster solution

Let's make it independent on $K$, too.

Look at what we're doing so far after sorting $S$ and removing duplicates: add all numbers from $0$ up to $S_1-1$ if we can, skip $S_1$, add all numbers from $S_1+1$ up to $S_2-1$ if we can, skip $S_2$, etc. Instead of adding numbers in a range $[x,y]$ one by one, why not just subtract $y-x+1$ from $K$, since it has the same effect?

So, we can simply go through the numbers $S_{1..N}$ and for each $S_i$, subtract $S_i-1-S_{i-1}$ from $K$ (we can consider $S_0=-1$ so that we're subtracting $S_1$ for $i=1$). As soon as we can't do this, because $K < S_i-1-S_{i-1}$, the only numbers we can add are $S_{i-1}+1$ through $S_{i-1}+K$ and the answer will be $S_{i-1}+K+1$.

Naturally, if this didn't happen for any $i$, then we should be adding numbers starting from $S_N+1$; the first number we couldn't add is $S_N+1+K$, so that will be the answer.

This solution takes just $O(N\log{N})$ time and $O(N)$ memory. There's an additional $\log$ factor, but we can solve the problem even for very large values of $K$ and $S_{1..N}$.


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asked 06 Oct '17, 19:41

xellos0's gravatar image

accept rate: 10%

edited 17 Oct '17, 10:44

admin's gravatar image

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i have implemented similar kind of solution but it's failing can any one please tell at what test case it's failing. The below is my solution


answered 17 Oct '17, 22:58

prasadram126's gravatar image

accept rate: 0%

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question asked: 06 Oct '17, 19:41

question was seen: 1,196 times

last updated: 17 Oct '17, 22:58