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09 Mo's Algo (RANGE mode query Optimized).cpp
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09 Mo's Algo (RANGE mode query Optimized).cpp
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/**
Problem : Given a range [l,r], find the highest freq / the value with highest freq
**/
/**Which of the favors of your Lord will you deny ?**/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define MP make_pair
#define F first
#define S second
#define INF INT_MAX
#define ALL(x) (x).begin(), (x).end()
#define DBG(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << endl
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class TIn>
using indexed_set = tree<
TIn, null_type, less<TIn>,
rb_tree_tag, tree_order_statistics_node_update>;
/*
PBDS
-------------------------------------------------
1) insert(value)
2) erase(value)
3) order_of_key(value) // 0 based indexing
4) *find_by_order(position) // 0 based indexing
*/
inline void optimizeIO()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const int nmax = 2e5+7;
const LL LINF = 1e17;
string to_str(LL x)
{
stringstream ss;
ss<<x;
return ss.str();
}
//bool cmp(const PII &A,const PII &B)
//{
//
//}
int ara[nmax];
/** Mo's Algorithm **/
/*** 1 based indexing ***/
int block_size;
struct Query
{
int l, r, idx;
bool operator<(Query other) const
{
return make_pair(l / block_size, r) <
make_pair(other.l / block_size, other.r);
}
};
int cnt[nmax];
int max_cnt[nmax]; /** this part is needed because when a value is removed , we need to check
if this value was the only one with maximum freq **/
int mx = 0;
inline void init(int n)
{
mx = 0;
fill(cnt,cnt+nmax,0);
fill(max_cnt,max_cnt+nmax,0);
block_size = sqrt(n);
}
void remov(int idx); // TODO: remove value at idx from data structure
void add(int idx); // TODO: add value at idx from data structure
int get_answer(); // TODO: extract the current answer of the data structure
inline void add(int idx)
{
int val = ara[idx];
cnt[val]++;
max_cnt[cnt[val]]++;
mx = max(mx,cnt[val]);
}
inline void remov(int idx)
{
int val = ara[idx];
max_cnt[cnt[val]]--;
if(max_cnt[cnt[val]]==0) /** if this value was the NOT only one with maximum freq , then reduce mx **/
mx--;
cnt[val]--;
}
inline int get_answer()
{
return mx;
}
vector<int> mo_s_algorithm(vector<Query> queries)
{
vector<int> answers(queries.size()+1);
sort(queries.begin(), queries.end());
// TODO: initialize data structure
int cur_l = 0;
int cur_r = -1;
// invariant: data structure will always reflect the range [cur_l, cur_r]
for (Query q : queries)
{
while (cur_l > q.l)
{
cur_l--;
add(cur_l);
}
while (cur_r < q.r)
{
cur_r++;
add(cur_r);
}
while (cur_l < q.l)
{
remov(cur_l);
cur_l++;
}
while (cur_r > q.r)
{
remov(cur_r);
cur_r--;
}
answers[q.idx] = get_answer();
}
return answers;
}
int main()
{
int tc;
scanf("%d",&tc);
while(tc--)
{
int n,q;
scanf("%d %d",&n,&q);
for(int i=1; i<=n; i++)
{
scanf("%d",&ara[i]);
}
init(n); /** IMP **/
vector<Query>qv;
for(int i=1; i<=q; i++)
{
int l,r;
scanf("%d %d",&l,&r);
l++,r++;
qv.push_back({l,r,i});
}
auto ans = mo_s_algorithm(qv);
for(int i=1; i<=q; i++)
{
printf("%d\n",ans[i]);
}
}
return 0;
}
/**
100
10 4
1 2 2 2 6 6 7 2 2 6
6 9
3 5
4 8
1 4
10 4
1 2 2 2 6 6 7 2 2 6
4 6
8 10
1 10
5 10
**/