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))
View More Similar Problems
Merging Communities
People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w
View Solution →Components in a graph
There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu
View Solution →Kundu and Tree
Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that
View Solution →Super Maximum Cost Queries
Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and
View Solution →Contacts
We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co
View Solution →No Prefix Set
There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio
View Solution →