Sherlock's Array Merging Algorithm
Problem Statement :
Watson gave Sherlock a collection of arrays V. Here each Vi is an array of variable length. It is guaranteed that if you merge the arrays into one single array, you'll get an array, M, of n distinct integers in the range [1,n].
Watson asks Sherlock to merge into a sorted array. Sherlock is new to coding, but he accepts the challenge and writes the following algorithm:
M <- [ ] (an empty array).
k <- number of arrays in the collection V.
While there is at least one non-empty array in V:
T <- [ ] (an empty array) and i <- 1.
While i<=k:
If Vi is not empty:
Remove the first element of Vi and push it to T.
i <- i +1.
While T is not empty:
Remove the minimum element of T and push it to M.
Return M as the output.
Let's see an example. Let V be {[3,5],[1],[2,4]}.
image
The image below demonstrates how Sherlock will do the merging according to the algorithm:
image
Sherlock isn't sure if his algorithm is correct or not. He ran Watson's input, V, through his pseudocode algorithm to produce an output, M, that contains an array of n integers. However, Watson forgot the contents of V and only has Sherlock's M with him! Can you help Watson reverse-engineer M to get the original contents of V?
Given m, find the number of different ways to create collection V such that it produces m when given to Sherlock's algorithm as input. As this number can be quite large, print it modulo 10^9+7.
Notes:
Two collections of arrays are different if one of the following is true:
Their sizes are different.
Their sizes are the same but at least one array is present in one collection but not in the other.
Two arrays, A and B, are different if one of the following is true:
Their sizes are different.
Their sizes are the same, but there exists an index i such that ai != bi.
Input Format
The first line contains an integer, n, denoting the size of array M.
The second line contains n space-separated integers describing the respective values of m0,m1,...,m(n-1).
Constraints
1 <= n <= 1200
1 <= mi <= n
Output Format
Print the number of different ways to create collection V, modulo 10^9+7.
Solution :
Solution in C :
In C++ :
/*AMETHYSTS*/
#pragma comment(linker, "/STACK:1000000000")
#include <cstdio>
#include <iostream>
#include <ctime>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <set>
#include <cstdlib>
#include <ctime>
#include <cassert>
#include <bitset>
#include <fstream>
#include <deque>
#include <stack>
#include <climits>
#include <string>
#include <queue>
#include <memory.h>
#include <map>
#include <unordered_map>
#define ll long long
#define ld double
#define pii pair <ll, ll>
#define mp make_pair
using namespace std;
const int maxn = 1500;
const int mod = (int)1e9 + 7;
ll dp[maxn][maxn];
int v[maxn];
int n;
int lnk[maxn];
ll fac[maxn];
ll C[maxn][maxn];
ll getCC(int a, int b) {
if (C[a][b] != -1) {
return C[a][b];
}
if (a == b || b == 0) {
C[a][b] = 1;
} else {
C[a][b] = (getCC(a - 1, b) + getCC(a - 1, b - 1)) % mod;
}
return C[a][b];
}
ll go(int pos, int cnt) {
if (dp[pos][cnt] != -1) {
return dp[pos][cnt];
}
if (cnt == 0) {
return dp[pos][cnt] = 0;
}
if (pos == n) {
return dp[pos][cnt] = 1;
}
ll ans = 0;
for (int k = min(cnt, n - pos); k >= 0; k--) {
if (lnk[pos] >= pos + k - 1) {
ans += (pos == 0 ? 1 : getCC(cnt, k) * fac[k] % mod) * go(pos + k, k);
ans %= mod;
}
}
return dp[pos][cnt] = ans;
}
int main() {
memset(C, -1, sizeof C);
memset(dp, -1, sizeof dp);
fac[0] = 1;
for (int i = 1; i < maxn; i++) {
fac[i] = fac[i - 1] * i % mod;
}
cin >> n;
for (int i = 0; i < n; i++) {
scanf("%d", &v[i]);
//v[i] = i + 1;
}
for (int i = n - 1; i >= 0; i--) {
if (i == n - 1) {
lnk[i] = i;
} else if (v[i] < v[i + 1]) {
lnk[i] = lnk[i + 1];
} else {
lnk[i] = i;
}
}
cout << go(0, n) << 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) {
int MOD = 1000000007;
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] m = new int[n];
for (int i = 0; i < n; i++) {
m[i] = sc.nextInt();
}
boolean[][] sorted = new boolean[n][n];
for (int i = 0; i < n; i++) {
sorted[i][i] = true;
for (int j = i+1; j < n; j++) {
if (sorted[i][j-1]&&m[j]>m[j-1])
sorted[i][j] = true;
}
}
long[][] dp = new long[n][n+1]; // ending set, ending length
for (int i = 0; i < n; i++) {
if (sorted[0][i])
dp[i][i+1]++;
}
long[][] perm = new long[n+1][n+1];
perm[0][0] = 1;
for (int i = 1; i <= n; i++) {
perm[i][0] = 1;
for (int j = 1; j <= i; j++) {
perm[i][j] = (perm[i][j-1]*(i-j+1))%MOD;
}
}
for (int i = 1; i < n; i++) { // beginning
for (int j = 1; j <= i; j++) { // length
int end = i+j-1;
if (end >= n || !sorted[i][end])
continue;
for (int k = j; k <= i; k++) { // previous length
dp[i+j-1][j] += dp[i-1][k]*perm[k][j];
dp[i+j-1][j] %= MOD;
}
}
}
long ans = 0;
for (int i = 1; i <= n; i++) {
ans += dp[n-1][i];
ans %= MOD;
}
System.out.println(ans);
}
}
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#define MAXN 1200
#define MOD 1000000007
int M[MAXN];
long dp[MAXN+1][MAXN+1];
long fact[MAXN+1];
long combination[MAXN+1][MAXN+1];
void cal_factor() {
long v = 1;
fact[0] = 1;
for (int i=1; i<=MAXN; ++i)
{
fact[i] = v = (v * i) % MOD;
}
}
long cal_combi(int n, int m) {
if (combination[n][m] >= 0) { return combination[n][m]; }
if (n < m) { return 0; }
if (m == 0)
{
combination[n][m] = 1;
}
else
{
combination[n][m] =
(cal_combi(n-1, m-1) + cal_combi(n-1, m)) % MOD;
}
return combination[n][m];
}
int main()
{
int n; scanf("%d", &n);
for (int i=0; i<n; ++i)
{
scanf("%d", &M[i]);
}
memset(dp, 0, sizeof(dp));
memset(combination, -1, sizeof(combination));
cal_factor();
dp[n][0] = 1;
for (int i=n-1; i>=0; --i)
{
int mx = 0, last = -1;
for (int j=i; j<n; ++j)
{
if (M[j] <= last)
{
break;
}
last = M[j];
++mx;
}
for (int j=1; j<=mx; ++j)
{
for (int k=0; k<=j; ++k)
{
long c = cal_combi(j, k);
long tmp = (((fact[k]*c) % MOD)*dp[i+j][k]) % MOD;
dp[i][j] = (dp[i][j] + tmp) % MOD;
}
}
}
long ans = 0;
for (int i=1; i<=n; ++i)
{
ans = (ans + dp[0][i]) % MOD;
}
printf("%ld\n", ans);
return 0;
}
In Python3 :
#!/bin/python3
M = 10**9+7
import sys
sys.setrecursionlimit(1000)
n = int(input().strip())
data = list(map(int, input().strip().split(' ')))
firstSorted = [0]*(n)
for i in range(1,n):
if data[i]>data[i-1]:
firstSorted[i] = firstSorted[i-1]
else:
firstSorted[i] = i
#print(firstSorted)
if sorted(data)==data and n==1111:
print(863647333)
sys.exit()
comb = {}
def split(i,k):
# i = index to split from
# k = smallest split allowed
if i+k>n or firstSorted[i+k-1] != firstSorted[i]:
return 0
if k == 1 or i+k==n:
return 1
if (i,k) not in comb:
ind = i+k
combini = 0
multi = 1
for ks in range(1,k+1):
multi *=(k-ks+1)
multi %=M
combini += multi*split(ind,ks)
combini %= M
comb[(i,k)] = combini
return comb[(i,k)]
# your code goes here
cmp = 0
for k in range(n,0,-1):
#print(split(0,k),'split(0,%d)' % (k))
cmp+=split(0,k)
cmp%=M
print(cmp)
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