Solution:
This problem can be efficiently solved by the Segment tree.
Since it is stated that, the input will be strictly in increasing order, a natural technique can be applied.
For this we require 3 arrays :
1. array input stores the input numbers,
2. array count stores the frequency of each input number,
3. and array start stores the index of the input list where a particular input number appeared first.
For example,
input = { -1, -1, 1, 1, 1, 1, 3, 10, 10, 10}
count = { 2, 2, 4, 4, 4, 1, 3, 3, 3 }
start = { 1, 1, 3, 3, 3, 7, 8, 8, 8 }
At first, a segment tree is constructed where each node will store the value of the maximum count (from the count array) of its respective range [ a,b ].
Let, the query range be [ i,j ].
Now 2 cases can occur The value at index i and j are –
1. same i.e input[ i ] = input[ j ].
2. different i.e input[ i ] ≠ input[ j ].
#Case 1:
Solving this case is the easiest. Since input[ i ] = input[ j ] , all the numbers in the range [ i,j ] are same ( since the numbers are non-descending ). So the answer for this case 1 is j – i + 1.
#Case 2:
In this case, there exists an index x where input[ i ] = input[ x ] and input[ i ] ≠ input[ x + 1 ]. Let, k = x + 1. So, the value of k = start [ i ] + count [ i ] .
So, the frequency of the value input[ i ] in the range [ i,k ] is cnt1 = k – i .
The frequency of input[ j ] in the range [ i,j ] is cnt2 = j –start [ j ] + 1 .
The maximum frequency of the values in range [ i,j ] may also exists in the range [ k, start[ j ] – 1 ]. This can be found by querying the segment tree for the maximum value in the range [ k, start[ j ] – 1 ]. Let the maximum count returned by the query be cnt3.
Therefore the answer for case 2 is max ( cnt1, cnt2 , cnt3 ).
#include <bits/stdc++.h>
#define pf printf
#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 sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define pii pair<int,int>
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define pbp(a,b) push_back({a,b})
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define ff first
#define ss second
#define sz(x) x.size()
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define for0(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forrev(i,n) for(int i=n-1; i>=0; i--)
#define forab(i,a,b) for(int i=a;i<=b;i++)
#define forba(i,b,a) for(int i=b;i>=a;i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define mx 100005
#define inf 100000008
/*------------------------------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
/*---------------------------------------------------------------------*/
using namespace std;
int input[mx],tree[3*mx],cnt[mx],start[mx];
map<int,int> mp;
void build (int node,int low,int high)
{
if(low==high)
{
tree[node]=cnt[low];
return;
}
int mid=(low+high)/2;
int left=2*node;
int right=2*node+1;
build(left,low,mid);
build(right,mid+1,high);
tree[node]=max(tree[left],tree[right]);
}
int query(int node,int low,int high,int qlow,int qhigh)
{
if(low>qhigh || high<qlow) return -inf;
else if(low>=qlow && high<=qhigh)
{
return tree[node];
}
int mid=(low+high)/2;
int left=2*node;
int right=2*node+1;
int l=query(left,low,mid,qlow,qhigh);
int r=query(right,mid+1,high,qlow,qhigh);
return max(l,r);
}
int main()
{
//CIN;
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,q;
while(sf(n)==1 && n!=0)
{
sf(q);
for1(i,n)
{
sf(input[i]);
mp[input[i]]++;
}
int y=-inf,k;
for1(i,n)
{
int x=mp[input[i]];
cnt[i]=x;
if(input[i]!=y)
{
k=i;
y=input[i];
}
start[i]=k;
}
build(1,1,n);
for1(i,q)
{
int qlow,qhigh,cnt1,cnt2,cnt3;
sff(qlow,qhigh);
if(input[qlow]!=input[qhigh])
{
int k=start[qlow]+cnt[qlow];
cnt1=k-qlow;
cnt2=qhigh-start[qhigh]+1;
cnt3=query(1,1,n,k,start[qhigh]-1);
pf("%d\n",max3(cnt1,cnt2,cnt3));
}
else pf("%d\n",qhigh-qlow+1);
}
ms(tree,0);
ms(cnt,0);
ms(start,0);
mp.clear();
}
return 0;
}
Fall out, Codes! 