### Zhunussov's blog

By Zhunussov, history, 2 months ago,

Help pls!

• +4

 » 2 months ago, # | ← Rev. 2 →   +32 You can keep an array of Boolean values, with each index representing whether or not you have seen this number in the array. It will be of size n+1, since mex of n elements is at most n+1. Fill this array in with a O(n) loop over the array, and then go over the Boolean array and find the first false value, that index will be your mex.
•  » » 2 months ago, # ^ |   -222 please provide him code you dumb assss
•  » » 2 months ago, # ^ |   0 u can simply use lower bound on the array instead of iterating all over. With a complexity of O(logn) insetad of O(n).
•  » » » 2 months ago, # ^ |   +31 lower_bound won't work because the array will not be of the form [1,1,1, 0, 0, 0, ...] it can be something like [1, 0, 1, 1, 1, 1, 1]
•  » » » » 2 months ago, # ^ |   +15 Oh yeah u are right :)) Thats why i am newbie.
 » 2 months ago, # |   0 You can use cycle sort to put elements to it's corresponding index. Then traverse again and first element not equal to index is your mex.Remember to ignore elements out of bound.Time: O(N) Space: O(1)
 » 2 months ago, # |   +20 If there are $Q$ mex queries of ranges in an array lengthened $N$, then it can be solved in $O(Q\sqrt N+N)$ with Mo's algorithm.
 » 2 months ago, # |   +1 in case of the given array is sorted, you can binary search on the first index that is not equal to it's value in the array, the answer will be that index. timeO(logN). codeint mex(int *a, int n) { int l = 0, r = n - 1, ans = n; while (l <= r) { int mid = (l + r) / 2; if (a[mid] == mid) { l = mid + 1; } else { r = mid - 1; ans = mid; } } return ans; } 
•  » » 2 months ago, # ^ |   +10 Besides being sorted, the array should only have unique values.
•  » » » 2 months ago, # ^ |   0 this can be easily handled while taking the initial array. codevoid solve() { int n; cin >> n; int a[n], newN = 0, last = -1; for (int i = 0; i < n; i++) { int x; cin >> x; if (x != last) { a[newN++] = x; } last = x; } cout << mex(a, newN); } 
 » 2 months ago, # |   +1 If you want to just find out the mex of the whole array, write a simple loop. int mex(vector vc) { vector cnt; int n=vc.size(); cnt.resize(n+2); for(auto id:vc) if(vc<=n) cnt[id]=1; for(int i=0;i<=n;i++) if(cnt[i]==0) return i; } If you want to query the mex of an interval for many times, this problem may help. https://www.luogu.com.cn/problem/P4137Sorry I don't know "可持久化线段树"'s English translation.
•  » » 2 months ago, # ^ |   +13 It is a persistent segment tree.
 » 2 months ago, # |   -7 I think this video may be helpful. It is about different ways to find MEX of an array. From the simplest one to the most efficienthttps://youtu.be/JDuVLyKn7Yw?si=CRqVObXm7JGdCMmg
 » 2 months ago, # |   0 if array is permutation of lenght N, and one value is missing, you can just find total sum = n*(n+1)//2 and then subtract each element of array from sum. Final value of sum is going to be your MEX.
 » 2 months ago, # | ← Rev. 2 →   0 Construct a hash array. Use Binary search and range query data structure such as minimum Sparse Table, check if in given range the minimum value is zero and accordingly update pointers.