Extremum Permutations
Problem Statement :
Let's consider a permutation P = {p1, p2, ..., pN} of the set of N = {1, 2, 3, ..., N} elements .
P is called a magic set if it satisfies both of the following constraints:
Given a set of K integers, the elements in positions a1, a2, ..., aK are less than their adjacent elements, i.e., pai-1 > pai < pai+1
Given a set of L integers, elements in positions b1, b2, ..., bL are greater than their adjacent elements, i.e., pbi-1 < pbi > pbi+1
How many such magic sets are there?
Input Format
The first line of input contains three integers N, K, L separated by a single space.
The second line contains K integers, a1, a2, ... aK each separated by single space.
the third line contains L integers, b1, b2, ... bL each separated by single space.
Output Format
Output the answer modulo 1000000007 (109+7).
Constraints
3 <= N <= 5000
1 <= K, L <= 5000
2 <= ai, bj <= N-1, where i ∈ [1, K] AND j ∈ [1, L]
Solution :
Solution in C :
In C++ :
#include<math.h>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<stdio.h>
#include<map>
#include<ext/hash_map>
#include<ext/hash_set>
#include<set>
#include<string>
#include<assert.h>
#include<vector>
#include<time.h>
#include<queue>
#include<deque>
#include<sstream>
#include<stack>
#include<sstream>
#define MA(a,b) ((a)>(b)?(a):(b))
#define MI(a,b) ((a)<(b)?(a):(b))
#define AB(a) (-(a)<(a)?(a):-(a))
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define pob pop_back
#define ep 0.0000000001
#define pi 3.1415926535897932384626433832795
using namespace std;
using namespace __gnu_cxx;
const long long P=1000000000+7;
const int N=5001;
int n,m,i,j,kk,k,l,r;
int a[N];
long long d[N],dd[N];
int main()
{
cin>>n>>k>>l;
for (i=1;i<=k;i++)
{
cin>>j;
a[j]|=1;
a[j+1]|=2;
}
for (i=1;i<=l;i++)
{
cin>>j;
a[j]|=2;
a[j+1]|=1;
}
for (i=2;i<=n;i++)
if (a[i]==3) {cout<<0<<endl; return 0;}
d[1]=1;
for (i=2;i<=n;i++)
{
for (j=1;j<=n;j++)
d[j]=(d[j]+d[j-1])%P;
if (a[i]==2)
{
for (j=1;j<=i;j++)
dd[j]=d[j-1];
} else
if (a[i]==1)
{
for (j=1;j<=i;j++)
dd[j]=(d[n]-d[j-1]+P)%P;
}else
for (j=1;j<=i;j++)
dd[j]=d[n];
for (j=1;j<=n;j++) d[j]=0;
for (j=1;j<=i;j++)
d[j]=dd[j];
// for (j=1;j<=i;j++)
// cout<<d[j]<<" ";cout<<endl;
}
long long ans=0;
for (i=1;i<=n;i++) ans=(ans+d[i])%P;
cout<<ans<<endl;
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int l = in.nextInt();
int[] a = new int[5005];
int[] b = new int[5005];
long[][] dp = new long[5005][5005];
for (int i = 0; i < k; i++) {
a[in.nextInt()] = 1;
}
for (int i = 0; i < l; i++) {
b[in.nextInt()] = 1;
}
for (int i = 1; i < n; i++) {
if (a[i] == 1 && b[i] == 1){
System.out.println("0");
return;
}
if ((a[i-1] == 1 && a[i] == 1) || (b[i-1]==1 && b[i] == 1)){
System.out.println("0");
return;
}
}
dp[1][1] = 1;
for (int i = 2; i <= n; i++){
if (!(a[i-1] == 1 || b[i] == 1)){
long sum = 0;
for (int j=1; j <= i; j++){
dp[i][j] = add(dp[i][j], sum);
sum = add(sum, dp[i-1][j]);
}
}
if (!(b[i-1] == 1 || a[i] == 1)){
long sum = 0;
for (int j=i; j>=1; j--){
dp[i][j] = add(dp[i][j], sum);
sum = add(sum, dp[i-1][j-1]);
}
}
}
long ans = 0;
for (int i = 1; i <= n; i++){
ans = add(ans, dp[n][i]);
}
System.out.println(ans);
}
private static long add(long x, long v){
return (x+v) % 1000000007;
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
int o[5000]={0};
long long dp[5000][5000]={0};
int main(){
int N,K,L,x,i,j;
long long sum;
scanf("%d%d%d",&N,&K,&L);
for(i=0;i<K;i++){
scanf("%d",&x);
if(o[x-1]==1){
printf("0");
exit(0);
}
o[x-1]=-1;
if(o[x]==-1){
printf("0");
exit(0);
}
o[x]=1;
}
for(i=0;i<L;i++){
scanf("%d",&x);
if(o[x-1]==-1){
printf("0");
exit(0);
}
o[x-1]=1;
if(o[x]==1){
printf("0");
exit(0);
}
o[x]=-1;
}
dp[0][0]=1;
for(i=1;i<N;i++){
sum=0;
switch(o[i]){
case 0:
for(j=0,sum=0;j<i;j++)
sum=(sum+dp[i-1][j])%MOD;
for(j=0;j<=i;j++)
dp[i][j]=sum;
break;
case -1:
for(j=i-1,sum=0;j>=0;j--){
sum=(sum+dp[i-1][j])%MOD;
dp[i][j]=sum;
}
break;
default:
for(j=0,sum=0;j<i;j++){
sum=(sum+dp[i-1][j])%MOD;
dp[i][j+1]=sum;
}
}
}
for(i=0,sum=0;i<N;i++)
sum=(sum+dp[N-1][i])%MOD;
printf("%lld",sum);
return 0;
}
In Python3 :
from itertools import islice, accumulate
MOD = 10**9 + 7
def permcount(permlen, a, b):
if any(x+1 == y for c in map(sorted, (a, b)) for x, y in zip(c, c[1:])):
return 0
if set(a) & set(b):
return 0
goingup = [None] * permlen
for c, low in ((a, True), (b, False)):
for elt in c:
elt -= 1
if elt > 0:
goingup[elt] = not low
if elt < permlen - 1:
goingup[elt+1] = low
count = [1]
for i, inc in islice(enumerate(goingup), 1, permlen):
if inc is None:
count = [sum(count)] * (i+1)
elif inc:
count = [0] + list(accumulate(count))
else:
count = list(reversed(list(accumulate(reversed(count))))) + [0]
count = [elt % MOD for elt in count]
return sum(count) % MOD
def readints():
return [int(f) for f in input().split()]
permlen, alen, blen = readints()
a = readints()
b = readints()
assert len(a) == alen and len(b) == blen
print(permcount(permlen, a, b))
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