Hello Programmers, In this post, you will learn how to solve HackerRank Cards Permutation Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the HackerRank Algorithms problems using C, CPP, JAVA, PYTHON, JavaScript & SCALA Programming Languages.
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HackerRank Cards Permutation
Task
Alice was given the n integers from 1 to n. She wrote all possible permutations in increasing lexicographical order, and wrote each permutation in a new line. For example, for n = 3, there are 6 possible permutations:
- [1, 2, 3]
- [1, 3, 2]
- [2, 1, 3]
- [2, 3, 1]
- [3, 1, 2]
- [3, 2, 1]
She then chose one permutation among them as her favorite permutation.
After some time, she forgot some elements of her favorite permutation. Nevertheless, she still tried to write down its elements. She wrote a 0 in every position where she forgot the true value.
She wants to know the sum of the line numbers of the permutations which could possibly be her favorite permutation, i.e., permutations which can be obtained by replacing the 0s. Can you help her out?
Since the sum can be large, find it modulo 109 + 7.
Input Format
The first line contains a single integer n.
The next line contains n space–separated integers a1, a2, . . . , an denoting Alice’s favorite permutation with some positions replaced by 0.
Constraints
- 1 <= n <= 3 x 105
- 0 < ai <= n
- The positive values appearing in [a1, . . . , an] are distinct.
Subtask
For ~33% of the total points, n <= 5000
Output Format
Print a single line containing a single integer denoting the sum of the line numbers of the permutations which could possibly be Alice’s favorite permutation.
Sample Input 0
4 0 2 3 0
Sample Output 0
23
Explanation 0
The possible permutations are [1, 2, 3, 4] and [4, 2, 3, 1]. The permutation [1, 2, 3, 4] occurs on line 1 and the permutation [4, 2, 3, 1] occurs on line 22. Therefore the sum is 1 + 22 = 23.
Sample Input 1
4 4 3 2 1
Sample Output 1
24
Explanation 1
There is no missing number in the permutation. Therefore, the only possible permutation is [4, 3, 2, 1], and it occurs on line 24. Therefore the sum is 24.
HackerRank Cards Permutation Solution
Cards Permutation Solution in C
#include<stdio.h>
int n, a[300100], pos[300100], mod = 1e9 + 7, occ[300100], les[300100], grt[300100], st[300100], lst[300100], gst[300100], bitree[300050];
void add(int idx, int val)
{
while( idx <= n )
{
bitree[idx] += val;
idx += ( idx & -idx );
}
}
int get(int idx)
{
int ans = 0;
while( idx > 0 )
{
ans += bitree[idx];
idx -= ( idx & -idx );
}
return ans;
}
long long fact[300100], factsumfr[300100], ans = 0;
long long pwr(long long a, long long b)
{
if( b == 0 )
{
return 1ll;
}
long long temp = pwr(a, b/2);
temp = ( temp * temp ) % mod;
if( b & 1 )
{
temp = ( temp * a ) % mod;
}
return temp;
}
long long inv(long long a)
{
return pwr(a, mod-2);
}
int main()
{
int i;
scanf("%d", &n);
for( i = 1 ; i <= n ; i++ )
{
scanf("%d", &a[i]);
pos[a[i]] = i;
if(a[i])
{
st[a[i]] = 1;
}
if(a[i])
{
occ[i] = 1;
}
}
fact[0] = 1;
for( i = 1 ; i <= n ; i++ )
{
les[i] = les[i-1] + occ[i], lst[i] = lst[i-1] + st[i], fact[i] = ( fact[i-1] * i ) % mod;
}
for( i = n ; i >= 1 ; i-- )
{
grt[i] = grt[i+1] + occ[i], gst[i] = gst[i+1] + st[i];
}
int k = les[n];
long long faci = fact[n-k], faci1 = fact[n-k-1], sumfrfr = 0;
for( i = 1 ; i <= n ; i++ )
{
if( a[i] == 0 )
{
sumfrfr = ( sumfrfr + ( ( fact[n-i] * ( n - i - grt[i+1] ) ) % mod * inv(n-k-1) ) % mod ) % mod;
factsumfr[i] = ( factsumfr[i-1] + fact[n-i] ) % mod;
}
else
{
factsumfr[i] = factsumfr[i-1];
}
}
for( i = 1 ; i <= n ; i++ )
{
long long inc = 0;
if( st[i] == 0 )
{
inc += ( inc + ( ( sumfrfr * ( i - 1 - lst[i] ) ) % mod * fact[n-k-1] ) % mod ) % mod;
}
else
{
inc = ( inc + ( ( ( n - i + 1 - gst[i] ) * factsumfr[pos[i]] ) % mod * fact[n-k-1] ) % mod ) % mod;
inc = ( inc + ( ( ( ( ( i - lst[i] ) * fact[n-pos[i]] ) % mod * fact[n-k] ) % mod * ( n - pos[i] + 1 - grt[pos[i]] ) ) % mod * inv(n-k) ) % mod ) % mod;
add(pos[i], 1);
int inv1 = get(n) - get(pos[i]);
inc = ( inc + ( ( fact[n-pos[i]] * fact[n-k] ) % mod * inv1 ) % mod ) % mod;
}
ans = ( ans + inc ) % mod;
}
ans = ( ans + fact[n-k] ) % mod;
printf("%lld\n", ans);
return 0;
}Cards Permutation Solution in Cpp
#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <cstring>
#include <cmath>
#include <memory.h>
#include <cstdio>
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
using namespace std;
const int nmax = 300 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
int n, p[nmax], w[nmax], ff[nmax];
int f[nmax], inv[nmax], rf[nmax];
int fen[nmax];
void upd(int r, int val) {
for (int i = r; i < nmax; i |= i + 1) {
fen[i] += val;
fen[i] %= mod;
}
}
int get(int r) {
int ans = 0;
for (int i = r; i >= 0; i &= i + 1, --i) {
ans += fen[i];
ans %= mod;
}
return ans;
}
int main() {
// freopen("input.txt", "r", stdin);
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &p[i]);
}
f[0] = rf[0] = inv[1] = 1;
for (int i = 2; i < nmax; ++i) {
inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
}
for (int i = 1; i < nmax; ++i) {
f[i] = 1LL * i * f[i - 1] % mod;
rf[i] = 1LL * inv[i] * rf[i - 1] % mod;
}
int ans = 0, tot = 0;
for (int i = 1; i <= n; ++i) {
w[p[i]] = 1;
ff[p[i]] = 1;
tot += p[i] == 0;
}
w[0] = 0;
for (int i = 1; i <= n; ++i) {
w[i] = 1 - w[i];
w[i] += w[i - 1];
}
/* int ord[11] = {};
for (int j = 1; j <= n; ++j) {
ord[j] = j;
}
long long c = 0, d = 0;
do {
bool ok = true;
for (int i = 1; i <= n; ++i) {
ok &= p[i] == 0 || ord[i] == p[i];
}
d += 1;
if (ok) {
c += d;
}
} while (next_permutation(ord + 1, ord + n + 1));
cout << c % mod << endl;*/
int ANS = 0;
for (int i = n, c = 0, d = 0; i >= 1; --i) {
if (p[i] > 0) {
d = (d + w[n] - w[p[i]]) % mod;
ANS = (ANS + 1LL * f[n - i] * f[tot] % mod * get(p[i])) % mod;
if (c > 0) {
ANS = (ANS + 1LL * f[n - i] * f[tot - 1] % mod * w[p[i]] % mod * c) % mod;
}
upd(p[i], +1);
} else {
ANS = (ANS + 1LL * f[n - i] * f[tot] % mod * inv[2] % mod * c) % mod;
ANS = (ANS + 1LL * f[n - i] * d % mod * f[tot - 1]) % mod;
c += 1;
}
}
/* for (int i = 1; i <= n; ++i) {
if (p[i] > 0) {
int x = p[i];
for (int j = i + 1; j <= n; ++j) {
if (p[j] < x && p[j] != 0) {
ans = (ans + 1LL * f[n - i] % mod * f[tot]) % mod;
} else if (p[j] == 0) {
ans = (ans + 1LL * f[n - i] * w[x] % mod * f[tot - 1]) % mod;
}
}
for (int j = i - 1; j > 0; --j) {
if (p[j] == 0) {
ans = (ans + 1LL * f[n - j] * (w[n] - w[x]) % mod * f[tot - 1]) % mod;
}
}
} else {
for (int j = i + 1; j <= n; ++j) {
if (p[j] == 0) {
ans = (ans + 1LL * f[n - i] * f[tot] % mod * inv[2]) % mod;
}
}
}
}*/
cout << (ANS + f[tot]) % mod << endl;
// cout << (ans + f[tot]) % mod << endl;// + f[tot] << endl;
return 0;
}Cards Permutation Solution in Java
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class Solution {
private static final int modulus = 1000000007;
private static int[] knowns; // known numbers, (input order)
private static int[] gkArr; // gkArr[k] number of knowns after k in x s.t k>i_i
private static int[] guArr; // guArr[k] number of unknowns s.t. k>u
private static int[] remainingUnknownsArr;// number of unknowns in x[n-1-i:x.length]
private static long[] factorials; // factorials[i] = i! % modulus
private static long[] runningDiffs; // sum over remaining k of (U.len-guArr)
private static long[] EO; // Expected ordinals
//O(n+klgk) sum = |U|!(1+sum over n of n!*EO_n) % modulus
static long solve(int[] x) { //O(n + klgk)
int n = x.length;
long sum = 1L;
factorialsInit(n); //O(n)
int[] U = getUnknownInts(n, x); //O(n) - relies on x
knownsInit(x); //O(n) - relies on x
gkInit(n); //O(klgk) - relies on knowns, x
guInit(n, U); //O(n) - relies on unknowns, x
unknownsRemainingInit(x); //O(n) -relies on x
runningDiffsInit(x, U); //O(n) -relies on gu, x
EOInit(x, n, U); //O(n) - relies on knowns, unknowns, gk, gu, running sums, x
for(int i = 1; i < n; i++) //O(n)
sum = addMod(sum, mulMod(EO[i], factorials[i]));
sum = mulMod(sum, factorials[U.length]);
return sum;
}
//O(klgk) setup GOOD
private static void gkInit(int n) {
gkArr = new int[n+1];
int[][] arrs = new int[2][knowns.length];
arrs[0] = Arrays.copyOfRange(knowns, 0, knowns.length);
int subLen = 1;
int b = 0;
do {
int i = 0;
subLen *= 2;
int j = (subLen >>> 1);
int endSub = subLen;
int counter = 0;
int imin = Math.min(knowns.length, endSub - (subLen>>>1));
int jmin = Math.min(knowns.length, endSub);
while(counter < knowns.length) {
if(j < jmin && i < imin && arrs[b][i] < arrs[b][j]) {
arrs[(b+1)%2][counter] = arrs[b][i];
gkArr[arrs[b][i]] += Math.max(0, (counter++)-i++);
} else if(j < jmin) {
arrs[(b+1)%2][counter] = arrs[b][j];
gkArr[arrs[b][j]] += Math.max(0, (counter++)-j++);
} else if(i < imin) {
arrs[(b+1)%2][counter] = arrs[b][i];
gkArr[arrs[b][i]] += Math.max(0, (counter++)-i++);
} else {
endSub += subLen;
i = j;
j += (subLen>>>1);
imin = Math.min(knowns.length, endSub - (subLen>>>1));
jmin = Math.min(knowns.length, endSub);
}
}
b = (b+1)%2;
} while (subLen < knowns.length);
}
//O(n) setup
private static void runningDiffsInit(int[] x, int[] U) { //Sum over k of U_g
runningDiffs = new long[x.length];
runningDiffs[0] = (x[x.length-1] == 0) ? 0 : U.length - guArr[x[x.length-1]];
for(int i = 1; i < x.length; i++) {
if(x[x.length-1-i] != 0)
runningDiffs[i] = U.length - guArr[x[x.length-1-i]];
runningDiffs[i] = addMod(runningDiffs[i], runningDiffs[i-1]);
}
}
//O(n) setup GOOD
private static void unknownsRemainingInit(int[] x) {
remainingUnknownsArr = new int[x.length];
int u = 0;
for(int i = x.length-1; i >= 0; i--)
remainingUnknownsArr[i] = (x[i] == 0) ? u++ : u; //INCLUSIVE
}
//O(n) setup GOOD
private static void guInit(int n, int[] U) {
guArr = new int[n+1];
int k = 0;
int u = 0;
for(int i = 0; i < U.length; i++) {
while( k <= U[i])
guArr[k++] = u;
u++;
}
while( k < guArr.length)
guArr[k++] = u;
}
//O(n) setup
private static void EOInit(int[] x, int n , int[] U) {
EO = new long[x.length];
long d = 0L;
long invertedUlen = binaryExpMod(U.length, Solution.modulus-2L);
for(int i = 1; i < n; i++) {
if(x[n-1-i] == 0) {
//from unknown perms
EO[i] = mulMod(remainingUnknownsArr[n-1-i], 500000004L); // div by 2
//from knowns DP
d = mulMod(runningDiffs[i], invertedUlen);
EO[i] = addMod(EO[i], d);
} else {
//fraction of unknowns larger
d = mulMod(guArr[x[n-1-i]], invertedUlen);
EO[i] = addMod(EO[i], mulMod(remainingUnknownsArr[n-1-i], d));
//number of knowns larger
EO[i] = addMod(EO[i], gkArr[x[n-1-i]]);
}
}
}
//O(lgn) GOOD
private static long binaryExpMod(long l, long pow) { //l^(modulus-2) mod modulus
if (l == 0L && pow != 0L)
return 0L;
long[] squares = new long[30]; //30 = ciel(lg(modulus-2)) > ciel(lg(n))
squares[0] = l % Solution.modulus;
for(int i = 1; i < 30; i++)
squares[i] = mulMod(squares[i-1], squares[i-1]);
long result = 1L;
int i = 0;
while(pow != 0L) {
if((pow & 1L) == 1L)
result = mulMod(result, squares[i]);
i++;
pow >>>= 1;
}
return result;
}
//O(n) setup
private static void factorialsInit(int n) {
factorials = new long[n+1];
factorials[0] = 1L;
factorials[1] = 1L;
for(int i = 2; i <= n; i++)
factorials[i] = Solution.mulMod(factorials[i-1], i);
}
//O(1) GOOD
private static long mulMod(long result, long l) {
return ( (result%Solution.modulus) * (l%Solution.modulus) )%Solution.modulus;
}
//O(1) GOOD
private static long addMod(long result, long l) {
return ( (result%Solution.modulus) + (l%Solution.modulus) )%Solution.modulus;
}
//O(n) setup GOOD
private static int[] getUnknownInts(int n, int[] x) { //O(n) but setup so insignif
int[] ints = new int[n];
for(int i = 1; i <= n; i++)
ints[i-1] = i;
for(int i: x)
if(i != 0) {
ints[i-1] = 0;
n--;
}
int[] intsOut = new int[n];
n = 0; //becomes index
for(int i: ints)
if(i != 0)
intsOut[n++] = i;
return intsOut;
}
//O(n) setup GOOD
private static void knownsInit(int[] x) {
int counter = 0;
for(int a: x)
if(a > 0)
counter++;
knowns = new int[counter];
counter = 0;
for(int a: x)
if(a > 0)
knowns[counter++] = a;
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
int[] a = new int[n];
String[] aItems = scanner.nextLine().split(" ");
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
for (int i = 0; i < n; i++) {
int aItem = Integer.parseInt(aItems[i]);
a[i] = aItem;
}
long result = solve(a);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
bufferedWriter.close();
scanner.close();
}
}Cards Permutation Solution in Python
#!/bin/python
import math
import os
import random
import re
import sys
# Complete the solve function below.
MAXN = 3e5 + 5
mod = int(1e9 + 7)
inv2 = (mod + 1) >> 1
c = []
pre = []
fac = []
N = 0
count = 0
lex = 0
def add(a, b):
a += b
return a - mod if a >= mod else a
def pop(a, b):
a -= b
return a + mod if a < 0 else a
def mul(a, b):
return (a * b) % mod
def lowbit(i):
return i & (-i)
def update(o, v):
i = o + 1
while i <= N:
c[i] += v
i += lowbit(i)
def calc(o):
sum = 0
i = o + 1
while i >= 1:
sum += c[i]
i -= lowbit(i)
return sum
def prepare():
global count, lex
for i in range(1, N + 1):
a[i] -= 1
count += 1 if (a[i] == -1) else 0
if a[i] >= 0:
pre[a[i]] = 1
fac[0] = 1
for i in range(1, N + 1):
fac[i] = mul(i, fac[i - 1])
for i in range(1, N):
pre[i] += pre[i - 1]
lex = mul(mul(N, pop(N, 1)), inv2)
for i in range(1, N + 1):
if a[i] != -1:
lex = pop(lex, a[i])
# print('prep done:')
# print('a: {}'.format(a))
# print('pre: {}'.format(pre))
# print('fac: {}'.format(fac))
# print('lex: {}'.format(lex))
# print('count: {}'.format(count))
def cal2(n):
return mul(mul(n, pop(n, 1)), inv2)
def solve(x):
global count, lex
prepare()
cur = 0
ans = 0
for i in range(1, N + 1):
if a[i] != -1:
sum = mul(fac[count], a[i] - calc(a[i]))
if count >= 1:
sum = pop(sum, mul(fac[count - 1],
mul(cur, a[i] + 1 - pre[a[i]])))
sum = mul(sum, fac[N - i])
ans = add(ans, sum)
update(a[i], 1)
lex = pop(lex, pop(N - 1 - a[i], pop(pre[N-1], pre[a[i]])))
else:
sum = mul(lex, fac[count - 1])
if count >= 2:
sum = pop(sum, mul(fac[count - 2], mul(cur, cal2(count))))
sum = mul(sum, fac[N - i])
ans = add(ans, sum)
cur += 1
return add(ans, fac[count])
if __name__ == '__main__':
N = int(raw_input())
a = map(int, raw_input().rstrip().split())
a.insert(0, 0)
for i in range(len(a)):
pre.append(0)
fac.append(0)
c.append(0)
result = solve(a)
print(result)Cards Permutation Solution using JavaScript
'use strict';
const fs = require('fs');
const _ = require('underscore');
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});
process.stdin.on('end', _ => {
inputString = inputString.replace(/\s*$/, '')
.split('\n')
.map(str => str.replace(/\s*$/, ''));
main();
});
const MAXN = BigInt(3e5 + 5),
mod = BigInt(1e9 + 7),
inv2 = (mod + 1n) >> 1n;
let c = [],
pre = [],
fac = [],
N = 0n,
count = 0n,
lex = 0n,
a, cur, ans, sum;
function readLine() {
return inputString[currentLine++];
}
function add(a, b) {
a += b;
return a >= mod ? a - mod : a;
}
function pop(a, b) {
a -= b;
return a < 0n ? a + mod : a;
}
function mul(a, b) {
return (a * b) % mod;
}
function lowbit(i) {
return i & (-i)
}
function update (o, v) {
let i = o + 1n;
while (i <= N) {
c[i] += v;
i += lowbit(i)
}
}
function calc(o) {
let s = 0n, i = o + 1n;
while (i >= 1n) {
s += c[i]
i -= lowbit(i);
}
return s
}
function* range(start, end) {
for (let i = start; i < end; i++) {
yield i;
};
};
function cal2(n) {
return mul(mul(n, pop(n, 1n)), inv2)
}
function prepare() {
for (let i of range(1n, N + 1n)) {
a[i] -= 1n;
if (a[i] == -1n) count += 1n;
if (a[i] >= 0n) pre[a[i]] = 1n;
}
fac[0] = 1n;
for (let i of range(1n, N + 1n)) {
fac[i] = mul(i, fac[i - 1n])
}
for (let i of range(1n, N)) {
pre[i] += pre[i - 1n];
}
lex = mul(mul(N, pop(N, 1n)), inv2);
for (let i of range(1n, N + 1n)) {
if (a[i] != -1n) lex = pop(lex, a[i])
}
}
function solve(x) {
prepare()
cur = 0n
ans = 0n
for (let i of range(1n, N + 1n)) {
if (a[i] != -1n) {
sum = mul(fac[count], a[i] - calc(a[i]))
if (count >= 1n) {
sum = pop(sum, mul(fac[count - 1n], mul(cur, a[i] + 1n - pre[a[i]])))
}
sum = mul(sum, fac[N - i])
ans = add(ans, sum)
update(a[i], 1n)
lex = pop(lex, pop(N - 1n - a[i], pop(pre[N - 1n], pre[a[i]])))
} else {
sum = mul(lex, fac[count - 1n])
if (count >= 2n) {
sum = pop(sum, mul(fac[count - 2n], mul(cur, cal2(count))))
}
sum = mul(sum, fac[N - i])
ans = add(ans, sum)
cur += 1n
}
}
return add(ans, fac[count])
}
function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
N = BigInt(parseInt(readLine(), 10));
a = readLine().split(' ').map(aTemp => BigInt(parseInt(aTemp, 10, 64)));
let result;
a.unshift(0n)
for (let i in _.range(a.length)) {
pre.push(0n);
fac.push(0n);
c.push(0n);
};
result = solve(a);
ws.write(result + "\n");
ws.end();
}Cards Permutation Solution in Scala
Cards Permutation Solution in Pascal
Disclaimer: This problem (Cards Permutation) is generated by HackerRank but the solution is provided by Chase2learn. This tutorial is only for Educational and Learning purposes.
FAQ:
1. How do you solve the first question in HackerRank?
If you want to solve the first question of Hackerrank then you have to decide which programing language you want to practice i.e C programming, Cpp Programing, or Java programming then you have to start with the first program HELLO WORLD.
2. How do I find my HackerRank ID?
You will receive an email from HackerRank to confirm your access to the ID. Once you have confirmed your email, the entry will show up as verified on the settings page. You will also have an option to “Make primary”. Click on that option. Read more
3. Does HackerRank detect cheating?
yes, HackerRank uses a powerful tool to detect plagiarism in the candidates’ submitted code. The Test report of a candidate highlights any plagiarized portions in the submitted code and helps evaluators to verify the integrity of answers provided in the Test.
4. Does HackerRank use camera?
No for coding practice Hackerrank does not use camera but for companies’ interviews code submission time Hackerrank uses the camera.
5. Should I put HackerRank certificate on resume?
These certificates are useless, and you should not put them on your resume. The experience you gained from getting them is not useless. Use it to build a portfolio, and link to it on your resume.
6. Can I retake HackerRank test?
The company which sent you the HackerRank Test invite owns your Test submissions and results. It’s their discretion to permit a reattempt for a particular Test. If you wish to retake the test, we recommend that you contact the concerned recruiter who invited you to the Test and request a re-invite.
7. What is HackerRank?
HackerRank is a tech company that focuses on competitive programming challenges for both consumers and businesses. Developers compete by writing programs according to provided specifications. Wikipedi
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