Find the permutation


Problem Statement :


Consider a permutation, , of integers from  to . Let's determine the  of  to be the minimum absolute difference between any  consecutive integers in :

Generate a lexicographically sorted list of all permutations of length  having a maximal distance between all permutations of the same length. Print the lexicographically  permutation.

Input Format

The first line contains an integer, t (the number of test cases).

The t subsequent lines each contain two space-separated integers,  (the permutation length) and  (the 1-based index in the list of permutations having a maximal distance), respectively. The  line corresponds to the  test case.

Note: It is guaranteed that the sum of all ni  does not exceed 10^6.

Constraints

1  <=  t  <=  10
1  <=  ni  <=  10^6
1  <=  ki  <= 10^18

Output Format

For each test case: if the list of permutations having maximal distance has at least k elements, print the kth permutation as sequential (i.e.: from 1 to n ) space-separated integers on a new line; otherwise, print -1.



Solution :



title-img


                            Solution in C :

In   C++  :






#include <algorithm>
#include <assert.h>
#include <iostream>
#include <map>
#include <string>
#include <vector>
#include <unordered_set>

using namespace std;

static const uint64_t one = 1;

void SolveSimple(uint64_t n, uint64_t k)
{
    assert((n & 1) == 0);
    if (k > 2)
    {
        cout << -1 << endl;
    }
    else
    {
        for (uint64_t i = 0; i < n / 2; ++i)
        {
            if (k == 1)
                cout << (n / 2 - i) << " " << (n - i) << " ";
            else
                cout << (n / 2 + i + 1) << " " << (i + 1) << " ";
        }
        cout << endl;
    }
}

class Solution
{
public:
    vector<uint64_t> solution;
    uint64_t n;
    uint64_t k;

    int AdjustPlace()
    {
        uint64_t t = k;
        uint64_t n0 = 2;
        if (t < n0)
            return 1;
        uint64_t s = n0;
        t -= s;
        for (uint64_t i = 0; i < n / 2 - 1; ++i)
        {
            uint64_t p2 = (one << i);
            if (t < (s + p2))
                return 1;
            t -= (s + p2);
            s = 2 * s + p2;
        }
        uint64_t p2 = (one << (n / 2));
        if (t < p2)
        {
            k = t;
            return 2;
        }
        t -= p2;
        if (t < s)
        {
            k = s - 1 - t;
            return 3;
        }
        return -1;
    }

    void Inverse()
    {
        for (size_t i = 0; i < solution.size(); ++i)
        {
            solution[i] = n + 1 - solution[i];
        }
    }

    void Solve1()
    {
        vector<int> used(n + 2, 0);
        uint64_t p = 0;
        uint64_t mi = n / 2;
        for (;;)
        {
            if (k == 0)
            {
                solution[p++] = 1;
                for (size_t i = n / 2 + 1; i > 1; --i)
                {
                    if (!used[i])
                    {
                        solution[p++] = i;
                        solution[p++] = i + n / 2;
                    }
                }
                return;
            }
            if (k == 1)
            {
                size_t i1 = 1;
                size_t i2 = n / 2 + 2;
                for (; p < n - 1; )
                {
                    for (; used[i1]; ++i1);
                    for (; used[i2]; ++i2);
                    solution[p++] = i1++;
                    solution[p++] = i2++;
                }
                solution[p++] = n / 2 + 1;
                return;
            }
            uint64_t s = 2;
            k -= s;
            for (uint64_t i = 0; ; ++i)
            {
                if (k < s)
                {
                    solution[p++] = i + 2;
                    solution[p++] = i + 2 + n / 2;
                    used[i + 2] = 1;
                    used[i + 2 + n / 2] = 1;
                    break;
                }
                k -= s;
                size_t p2 = (one << i);
                if (k < p2)
                {
                    size_t i1 = i + 2;
                    size_t i2 = n / 2 + i1 + 1;
                    for (; ;)
                    {
                        for (; used[i1]; ++i1);
                        if (i1 == (n / 2 + 1))
                        {
                            break;
                        }
                        for (; used[i2]; ++i2);
                        solution[p++] = i1++;
                        solution[p++] = i2++;
                    }
                    solution[p++] = i1;
                    solution[p++] = 1;
                    Solve2I(i);
                    return;
                }
                k -= p2;
                s = 2 * s + p2;
            }
        }
    }

    void Solve2I(uint64_t r)
    {
        assert(k < (one << r));
        uint64_t sf = n - 2 * r - 1;
        uint64_t sb = n - 1;
        uint64_t s2 = 2;
        for (uint64_t i = 0; i < r; ++i)
        {
            uint64_t p2 = (one << (r - i - 1));
            if (k < p2)
            {
                solution[sf] = n / 2 + s2 + i;
                solution[sf + 1] = s2 + i;
                sf += 2;
            }
            else
            {
                k -= p2;
                solution[sb] = n / 2 + s2 + i;
                solution[sb - 1] = s2 + i;
                sb -= 2;
            }
        }
        assert(sf == sb);
        solution[sf] = n / 2 + s2 + r;
    }

    void Solve2()
    {
        uint64_t m = (one << (n / 2 - 1));
        if (k >= m)
        {
            k = 2 * m - 1 - k;
            Solve2();
            Inverse();
            return;
        }
        solution[0] = (n / 2) + 1;
        solution[1] = 1;
        Solve2I(n/2 - 1);
    }

    bool Solve(uint64_t _n, uint64_t _k)
    {
        n = _n;
k = _k - 1;
int status = AdjustPlace();
if (status == -1)
return false;
solution.resize(n);
if (status == 1)
Solve1();
if (status == 2)
Solve2();
if (status == 3)
{
Solve1();
Inverse();
}
return true;
}
};

class STest
{
public:
uint64_t count;
Solution S;

void PrintAllI(vector<int>& v, int k, int min_diff, vector<bool>& av)
{
if (k == v.size())
{
++count;
S.Solve(v.size(), count);
{
for (int i = 0; i < k; ++i)
{
cout << v[i] + 1 << " ";
}
cout << endl;
for (int i = 0; i < k; ++i)
{
cout << S.solution[i] << " ";
}
cout << endl;
}
}
else
{
int j = (k > 0) ? v[k - 1] : -min_diff;
for (int i = 0; i < int(av.size()); ++i)
{
if (av[i] && (abs(i - j) >= min_diff))
{
v[k] = i;
av[i] = false;
PrintAllI(v, k + 1, min_diff, av);
av[i] = true;
}
}
}
}

void PrintAll(int n)
{
count = 0;
vector<int> v(n, -1);
vector<bool> av(n, true);

PrintAllI(v, 0, n / 2, av);
}
};

void SSolve(uint64_t n, uint64_t k)
{
if (n == 1)
{
if (k == 1)
{
cout << 1 << endl;
}
else
{
cout << -1 << endl;
}
return;
}
Solution S;
if (S.Solve(n, k))
{
for (size_t i = 0; i < S.solution.size(); ++i)
{
cout << S.solution[i] << " ";
}
cout << endl;
}
else
{
cout << -1 << endl;
}
}

int main()
{
//STest t;
//t.PrintAll(11);
int T;
cin >> T;
for (; T; --T)
{
uint64_t n, k;
cin >> n >> k;
if (n & 1) 
{
SSolve(n, k);
}
else
{
SolveSimple(n, k);
}
}
return 0;
}







In   Java  :






import java.io.*;
import java.util.*;

public class Solution {

static String s = " ";
static long[] countSumms = new long[] { 
0, 2, 5, 12, 28, 64, 144, 320, 704, 1536, 3328, 7168, 15360, 32768, 69632,
147456, 311296, 655360, 1376256, 2883584, 6029312, 12582912, 26214400,
54525952, 113246208, 234881024, 486539264, 1006632960, 2080374784,
4294967296l, 8858370048l, 18253611008l, 37580963840l, 77309411328l,
158913789952l, 326417514496l, 670014898176l, 1374389534720l, 2817498546176l,
5772436045824l, 11819749998592l, 24189255811072l, 49478023249920l,
101155069755392l, 206708186021888l, 422212465065984l, 862017116176384l,
1759218604441600l, 3588805953060864l, 7318349394477056l, 14918173765664768l,
30399297484750848l, 61924494876344320l, 126100789566373888l, 256705178760118272l,
522417556774977536l };
static int[] NOT_FOUND = new int[] { -1 };

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i = 0; i < t; i++) {
int n = sc.nextInt();
long k = sc.nextLong();
int[] res = solve(n, k);
StringBuilder b = new StringBuilder(countString(n));
for (int j = 0; j < res.length; j++) {
b.append(res[j]).append(s);
}
System.out.println(b.toString());
}
sc.close();
}
static int[] solve(int n, long k) {

if(n == 1) {
if(k == 1) {
return new int[] { 1 };
}
return NOT_FOUND;
}

int min = n >> 1;

/* even n */
if ((n & 1) == 0) {
if (k == 1) {
int[] ret = new int[n];
int ix = 0;
for (int i = 0; i < min; i++) {
ret[ix++] = min - i;
ret[ix++] = n - i;
}
return ret;
} else if (k == 2) {
int[] ret = new int[n];
int ix = 0;
for (int i = 0; i < min; i++) {
ret[ix++] = min + i + 1;
ret[ix++] = i + 1;
}
return ret;
} else {
return NOT_FOUND;
}
}
/* odd n */
long midCount = 1L << min;
boolean flip = false;
boolean middle = false;
int countSummIx = min;
long countsSumm;

k--;
/* k is before mid section */
if (countSumms.length < countSummIx || k < (
    countsSumm = countSumms[countSummIx])) {
}
/* k is inside of mid section */
else if (k < (
countsSumm = countSumms[countSummIx]) + midCount) {
k = k - countsSumm;
middle = true;
}
/* k is after mid section but before end of last side */
else if (k < (countsSumm << 1) + midCount) {
k = Math.abs(k - (countsSumm << 1) - midCount + 1);
flip = true;
}
/* k out of range */
else {
return NOT_FOUND;
}

int[] arr = new int[n];
if (middle) {
arr[0] = min + 1;
arr[1] = 1;
if (k >= midCount >> 1) {
k = midCount - 1 - k;
flip = true;
}
solveRadius(n, k, min - 1, arr, min);
if (flip) {
int n_1 = n + 1;
for (int i = 0; i < arr.length; i++) {
arr[i] = n_1 - arr[i];
}
}
return arr;
}
solveSide(arr, n, k, min);
if (flip) {
int n_1 = n + 1;
for (int i = 0; i < arr.length; i++) {
arr[i] = n_1 - arr[i];
}
}
return arr;
}
static void solveSide(int[] arr, int n,
 long k, int min) {
boolean[] cache = new boolean[n + 1];
int ix = 0;
outer:while (true) {
if (k == 0) {
arr[ix++] = 1;
for (int i = min + 1; i > 1; i--) {
if (!cache[i]) {
arr[ix++] = i;
arr[ix++] = i + min;
}
}
break;
}
if (k == 1) {
for (int left = 1, right = min + 2, n_1 = n - 1; ix < n_1;) {
while (cache[left])    left++;
while (cache[right]) right++;
arr[ix++] = left++;
arr[ix++] = right++;
}
arr[ix++] = min + 1;
break;
}
k -= countSumms[1];
long next = 1L;
for (int i = 0, j = 2;; ++i, j++, next <<= 1) {
if (k < countSumms[i + 1]) {
arr[ix++] = j;
arr[ix++] = j + min;
cache[j] = cache[j + min] = true;
break;
}
k -= countSumms[i + 1];
if (k < next) {
int left = j;
int right = min + left + 1;
while (true) {
while (cache[left])    left++;
if (left == min + 1) {
break;
}
while (cache[right]) right++;
arr[ix++] = left++;
arr[ix++] = right++;
}
arr[ix++] = left;
arr[ix++] = 1;
solveRadius(n, k, i, arr, min);
break outer;
}
k -= next;
}
}
}
static int countString(int n) {
int ret = 0;
if(n < 10) { return n << 1; }
ret += 18;
if(n < 100) { ret += (n - 9) * 3; return ret;}
ret += 270;
if(n < 1000) { ret += (n - 99) << 2; return ret;}
ret += 3600;
if(n < 10000) { ret += (n - 999) * 5; return ret;}
ret += 45000;
if(n < 100000) { ret += (n - 9999) * 6; return ret;}
ret += 540000;
ret += (n - 99999) * 7;
return ret;
}
static void solveRadius(int n, long k, 
int radius, int[] arr, int min) {
int left = n - (radius << 1) - 1;
int right = n - 1;
int min_2 = min + 2;
for (int i = 0; i < radius; ++i) {
long next = (1L << (radius - (i + 1)));
if (k < next) {
arr[left] = min_2 + i;
arr[left + 1] = 2 + i;
left += 2;
} else {
arr[right] = min_2 + i;
arr[right - 1] = 2 + i;
right -= 2;
k -= next;
}
}
arr[left] = min_2 + radius;
}
}








In   C  :






#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

static const unsigned long long int one = 1;
static unsigned long long int* solution;
unsigned long long int n, k;

void solve2I(unsigned long long int r)
{
assert(k < (one << r));
unsigned long long int sf = n - 2 * r - 1;
unsigned long long int sb = n - 1;
unsigned long long int s2 = 2;
for (unsigned long long int i = 0; i < r; ++i)
{
unsigned long long int p2 = (one << (r - i - 1));
if (k < p2)
{
solution[sf] = n / 2 + s2 + i;
solution[sf + 1] = s2 + i;
sf += 2;
}
else
{
k -= p2;
solution[sb] = n / 2 + s2 + i;
solution[sb - 1] = s2 + i;
sb -= 2;
}
}
assert(sf == sb);
solution[sf] = n / 2 + s2 + r;
}

void solveTwo()
{
unsigned long long int m = (one << (n / 2 - 1));
if (k >= m)
{
k = 2 * m - 1 - k;
solveTwo();
for (size_t i = 0; i < n; ++i)
{
solution[i] = n + 1 - solution[i];
}
return;
}
solution[0] = (n / 2) + 1;
solution[1] = 1;
solve2I(n/2 - 1);
}

void SolveOne()
{
unsigned long long int* used = (
    unsigned long long int*)malloc((n + 2) * sizeof(
        unsigned long long int));
for(unsigned long long int i=0;i<n+2;i++){
used[i] = 0;
}
unsigned long long int p = 0;
for (;;)
{
if (k == 0)
{
solution[p++] = 1;
for (size_t i = n / 2 + 1; i > 1; --i)
{
if (!used[i])
{
solution[p++] = i;
solution[p++] = i + n / 2;
}
}
return;
}
if (k == 1)
{
size_t i1 = 1;
size_t i2 = n / 2 + 2;
for (; p < n - 1; )
{
for (; used[i1]; ++i1);
for (; used[i2]; ++i2);
solution[p++] = i1++;
solution[p++] = i2++;
}
solution[p++] = n / 2 + 1;
return;
}
unsigned long long int s = 2;
k -= s;
for (unsigned long long int i = 0; ; ++i)
{
if (k < s)
{
solution[p++] = i + 2;
solution[p++] = i + 2 + n / 2;
used[i + 2] = 1;
used[i + 2 + n / 2] = 1;
break;
}
k -= s;
size_t p2 = (one << i);
if (k < p2)
{
size_t i1 = i + 2;
size_t i2 = n / 2 + i1 + 1;
for (; ;)
{
for (; used[i1]; ++i1);
if (i1 == (n / 2 + 1))
{
break;
}
for (; used[i2]; ++i2);
solution[p++] = i1++;
solution[p++] = i2++;
}
solution[p++] = i1;
solution[p++] = 1;
solve2I(i);
return;
}
k -= p2;
s = 2 * s + p2;
}
}
}

int AdjustPlace()
{
unsigned long long int t = k;
unsigned long long int n0 = 2;
if (t < n0)
return 1;
unsigned long long int s = n0;
t -= s;
for (unsigned long long int i = 0; i < n / 2 - 1; ++i)
{
unsigned long long int p2 = (one << i);
if (t < (s + p2))
return 1;
t -= (s + p2);
s = 2 * s + p2;
}
unsigned long long int p2 = (one << (n / 2));
if (t < p2)
{
k = t;
return 2;
}
t -= p2;
if (t < s)
{
k = s - 1 - t;
return 3;
}
return -1;
}

unsigned long long int* solve(){
unsigned long long int* v;
if(n%2==1){
if(n==1) {
if (k == 1) {
v = (unsigned long long int*)malloc(
    1 * sizeof(unsigned long long int));
v[0] = 1;
return v;
} else {
v = NULL;
return v;
}
}
k--;
int status = AdjustPlace();
if (status == -1) {
v = NULL;
return v;
}
solution = (unsigned long long int*)malloc(
    n * sizeof(unsigned long long int));
if (status == 1)
SolveOne();
if (status == 2)
solveTwo();
if (status == 3)
{
SolveOne();
for (unsigned long long int i = 0; i < n; ++i)
{
solution[i] = n + 1 - solution[i];
}
}
return solution;
}
else {
assert((n & 1) == 0);
if (k > 2)
{
v = NULL;
return v;
}
else
{
unsigned long long int j = 0;
v = (unsigned long long int*)malloc(
    n * sizeof(unsigned long long int));
for (unsigned long long int i = 0; i < n / 2; ++i)
{
if (k == 1) {
v[j++] = (n/2-i);
v[j++] = (n-i);
}
else {
v[j++] = (n / 2 + i + 1);
v[j++] = (i + 1);
}
}
}
}
return v;
}

void print(unsigned long long int* v){
for(unsigned long long int i=0;i<n-1;i++)
printf("%lld ", v[i]);
printf("%lld\n", v[n-1]);
}
int main() {
unsigned long long int t;
scanf("%lld", &t);
while(t--){
scanf("%lld %lld", &n, &k);
unsigned long long int* t = solve();
if(t == NULL){
printf("%i\n", -1);
} else {
print(t);
}
}
return 0;
}








In Python3 :






from bisect import *
import collections
from time import time
import random

popdistr = collections.Counter()

def naive(n, k):
    def gen(perm, nums):
        if len(perm) == n:
            perms.append(perm)
        for i in sorted(nums):
            if abs(perm[-1] - i) >= mindist:
                gen(perm + [i], nums - {i})
    perms = []
    mindist = n // 2
    for i in range(n):
        gen([i], set(range(n)) - {i})
    return perms[k] if k < len(perms) else -1

if 0:
    for k in range(7):
        print(naive(3, k))
    input()


def smart(n, k):
    if n < 5:
        return naive(n, k)
    
    half = n // 2
    h = half
    H = half + 1

    # Even n cases
    if not n & 1:
        if k > 1:
            return -1
        perm = [None] * n
        if k == 0:
            # 4 9 3 8 2 7 1 6 0 5
            perm[::2] = range(h-1, -1, -1)
            perm[1::2] = range(n-1, h-1, -1)
        else:
            # 5 0 6 1 7 2 8 3 9 4
            perm[::2] = range(h, n)
            perm[1::2] = range(h)
        return perm

    low = (h + 3) << (h - 2)
    #low = 2 if n == 3 else (h + 3) << (h - 2)
    lowmid = 1 << (h - 1)
    #print(k, low, lowmid)
    if k >= (low + lowmid) * 2:
        return -1
    if k >= low + lowmid:
        merp = smart(n, (low + lowmid) * 2 - 1 - k)
        if merp == -2:
            return merp
        return [n-1 - m for m in merp]
    if k >= low:
        return binary(list(range(n)), k-low, h)
    
    offset = [2]
    for i in range(half - 1):
        offset.append(offset[-1] * 2 + (1 << i))
        if offset[-1] > 10**30:
            break
    offset.append(offset[-1] + (1 << (i + 1)))
    offset.append(0)  # offset[-1] = 0
    #print(offset)

    nums = list(range(n))
    perm = []
    pops = 0
    while True:

        # Cases k=0, k=1
        if k < 2:
            # n=11: 0 5 10 4 9 3 8 2 7 1 6
            #       0 6 1 7 2 8 3 9 4 10 5
            add = h + k
            return perm + [nums[i*add % n] for i in range(n)]

        i = bisect(offset, k)
        k -= offset[i-1]

        #print(offset, i, k, end=' ... ')
        
        # Binary cases
        if k >= offset[i-1]:# or i == h:
            return perm + binary(nums, k - offset[i-1], i)

        # Ugly cases
        perm += nums.pop(i), nums.pop(i+h-1)
        n -= 2
        half -= 1
        h -= 1
        H -= 1

        if pops:
            popdistr[pops] -= 1
        pops += 1
        popdistr[pops] += 1

def binary(nums, k, i):
    n = len(nums)
    half = n // 2
    H = half + 1
    perm = [None] * n
    ks, testbit = bin(k)[:1:-1], half - 1
    left, right = 0, n - 1
    for m in range(i, i+half):
        if testbit < len(ks) and ks[testbit] == '1':
            perm[right] = nums[m]
            perm[right-1] = nums[(m + H) % n]
            right -= 2
        else:
            perm[left] = nums[m]
            perm[left+1] = nums[(m + H) % n]
            left += 2
        testbit -= 1
    perm[left] = nums[i + half]
    return perm

if 1:
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        perm = smart(n, k-1)
        print(-1 if perm == -1 else ' '.join(str(p+1) for p in perm))
                        








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Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

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Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

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Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

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Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

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Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

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Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

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