### broly_1033's blog

By broly_1033, history, 6 months ago,

Hey guys, I was solving ADAUNIQ on SPOJ but couldn't find a proper understandable solution when I was stuck. I was trying to solve it using Mo's algorithm with updates but couldn't find a working code for the same. So I thought of sharing this, if it could help someone.

// first writing without coordinateCompression
// Note: We also have to coordinateCompress the update values in query

#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

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) { // as our range is [x, y)
rem(old);
}
}

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)<(b.l);

});

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);
}

while(y<=qv[i].r) {
y++;
}
// removing y
while(y>(qv[i].r+1)) {
y--;
rem(v[y]);
}
// removing x
while(x<qv[i].l) {
rem(v[x]);
x++;
}
while(x>qv[i].l) {
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;
}

/*
8 8
1 2 3 3 1 2 3 3
2 1 3
2 0 3
2 0 7
1 3 4
1 7 0
2 1 3
2 0 3
2 0 7
*/



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