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 :
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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For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a
View Solution →Square-Ten Tree
The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the
View Solution →Balanced Forest
Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a
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