Hey everyone, I was trying ADAUNIQ problem on SPOJ https://www.spoj.com/problems/ADAUNIQ/. I want to solve it using Mo's algorithm with updates, I have written the code. It gives the correct output for the default test case also but gives wrong answer on submission. I tried debugging it but couldn't find any fault. I would be extremely thankful if someone can look at my code. Please note that the question asks for "number of unique elements in a segment", so if a segment is [1, 2, 3, 3] the answer will be 2 (because 1 and 2 occur once) and not 3 (not 1, 2, 3 because 3 occurs twice).
#include<bits/stdc++.h>
using namespace std;
const int N = 2*1e5;
class query {
public:
int id, l, r, res, time;
// int block;
};
vector<query> qv;
vector<pair<int, int> > updateQueries;
int distinct = 0; // distinct elements in current window
int cnt[N+5]; // stores the frequency
void add(int val) {
cnt[val]++;
if(cnt[val]==1) distinct++;
if(cnt[val]==2) distinct--;
// we are finding unique values in a range
// if we find duplicate, the kind will no longer be distinct
// not same distinct as in DQUERY
}
void rem(int val) {
cnt[val]--;
if(cnt[val]==0) distinct--;
if(cnt[val]==1) distinct++;
}
void update(vector<int>& v, int i, int x, int y) {
int idx = updateQueries[i].first;
int val = updateQueries[i].second;
int old = v[idx];
v[idx] = val;
updateQueries[i].second = old;
if(idx>=x && idx<=y) {
rem(old);
add(val);
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(NULL);
int n, num_q; cin>>n>>num_q;
vector<int> v(n);
for(int i=0; i<n; i++) cin>>v[i];
int sz = pow(n, 2/3)+1;
for(int i=0; i<num_q; i++) {
int c; cin>>c;
if(c==1) {
int idx, val; cin>>idx>>val;
updateQueries.push_back(make_pair(idx, val));
} else if(c==2) {
query q;
cin>>q.l; cin>>q.r;
q.id = i; q.time = updateQueries.size();
// q.block = (q.l)/sz;
qv.push_back(q);
}
}
// sort queries
sort(qv.begin(), qv.end(), [&](query a, query b){
if((a.l/sz)==(b.l)/sz){
if((a.r/sz)==(b.r/sz)) return a.time<b.time;
return (a.r/sz)<(b.r/sz);
}
return (a.l/sz)<(b.l/sz);
});
int x = 0, y = 0; // maintain current window, [x, y)
int current_time = 0;
for(int i=0; i<qv.size(); i++) {
// update time
int time = qv[i].time;
while(current_time<time) {
update(v, current_time, x, y);
current_time++;
}
while(current_time>time) {
current_time--;
update(v, current_time, x, y);
}
// add y
while(y<=qv[i].r) {
add(v[y]);
y++;
}
// removing y
while(y>(qv[i].r+1)) {
y--;
rem(v[y]);
}
// removing x
while(x<qv[i].l) {
rem(v[x]);
x++;
}
// add x
while(x>qv[i].l) {
x--;
add(v[x]);
}
qv[i].res = distinct;
}
sort(qv.begin(), qv.end(), [&](query a, query b){
return a.id<b.id;
});
for(int i=0; i<qv.size(); i++) {
cout<<qv[i].res<<'\n';
}
return 0;
}
I don't know what's wrong with MO but
Seems you mean
pow(n, 2./3.)
.Yeah, but that won't make difference right. It's for optimal time complexity.
It will make difference since 2/3 is integer division, which makes pow(n, 2/3) = pow(n, 0) = 1. What you mean is floating point division just as feodorv said.