Problem Link : http://www.spoj.com/problems/FREQUENT/
Solution Idea:
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In this problem an array of non-decreasing order is given. In each query a range l and r is given. You have to tell the maximum frequency of a number in the range l to r.
Now as the given array is in non-decreasing order. So you can replace array a = {1 1 1 2 2 3 3 3 3} by another array b = {3 2 4}. After this compression the problem is converted to a Range Maximum query problem which can be easily solved by segment tree or sparse table. Now for each query determine the index in the compress array and perform range maximum query on that range.
Think about the first and index of the query range. You need to handle them separately.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cerr<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/
#define mx 100005
pii block[mx];
int dp[mx][20];
int ara[mx],id[mx],cnt=0,table[mx];
int func(int idx, int p)
{
if(idx>cnt) return 0;
if(p==0)
{
return dp[idx][p]=table[idx];
}
int &ret=dp[idx][p];
if(ret!=-1) return ret;
ret=max(func(idx,p-1),func(idx+(1<<(p-1)),p-1));
return ret;
}
int query(int l, int r)
{
if(r<l) return 0;
if(l==r) return table[l];
int log=(int)log2(r-l);
return max(func(l,log),func(r-(1<<log)+1,log));
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,q;
while(sf(n))
{
if(n==0) break;
sf(q);
for(int i=1; i<=n; i++)
{
sf(ara[i]);
}
cnt=0;
ms(table,0);
ms(dp,-1);
// ms(block,0);
// ms(id,0);
for(int i=1; i<=n; i++)
{
int j=i;
cnt++;
block[cnt].ff=i;
while(ara[j]==ara[i] && j<=n)
{
id[j]=cnt;
table[cnt]++;
j++;
}
j--;
block[cnt].ss=j;
i=j;
}
while(q--)
{
int a,b;
sff(a,b);
if(id[a]==id[b])
{
printf("%d\n",b-a+1);
}
else
{
int l=id[a];
int r=id[b];
int ans=0;
ans=max(ans,block[l].ss-a+1);
l++;
ans=max(ans,b-block[r].ff+1);
r--;
ans=max(ans,query(l,r));
printf("%d\n",ans);
}
}
}
return 0;
}