# Super Kth LIS

### Problem Statement :

```Given an array of N integers (a0,a1,...,aN-1), find all possible increasing subsequences of maximum length, L. Then print the lexicographically Kth longest increasing subsequence as a single line of space-separated integers; if there are less than K subsequences of length L, print -1.

Two subsequences [ap0,ap1,...,apL-1] and [aq0,aq1,...,aqL-1] are considered to be different if there exists at least one i such that pi != qi.

Input Format

The first line contains 2 space-separated integers, N and K, respectively.
The second line consists of N space-separated integers denoting a0,a1,...,aN-1 respectively.

Constraints

1 <= N <= 10^5
1 <= K <= 10^18
1 <= ai <= N```

### Solution :

```                            ```Solution in C :

In C++ :

#pragma comment(linker, "/STACK:512000000")
#define _CRT_SECURE_NO_WARNINGS
//#include "testlib.h"
#include <cstdio>
#include <cassert>
#include <algorithm>
#include <iostream>
#include <memory.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <bitset>
#include <deque>
#include <ctime>
#include <stack>
#include <queue>
#include <fstream>
#include <sstream>
//#include <unordered_map>
using namespace std;
//#define FILENAME ""
#define mp make_pair
#define all(a) a.begin(), a.end()
typedef long long li;
typedef long double ld;
void solve();
void precalc();
clock_t start;
//int timer = 1;

int testNumber = 1;

bool todo = true;

int main() {
#ifdef _DDEBUG
freopen("input.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#else
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
//freopen(FILENAME".in", "r", stdin);
//freopen(FILENAME ".out", "w", stdout);
#endif
start = clock();
int t = 1;
cout.sync_with_stdio(0);
cin.tie(0);
precalc();
cout.precision(10);
cout << fixed;
//cin >> t;
int testNum = 1;
while (t--) {
//cerr << testNum << endl;
//cout << "Case #" << testNum++ << ": ";
solve();
++testNumber;
//++timer;
}

#ifdef room111
cerr << "\n\n" << (clock() - start) / 1.0 / CLOCKS_PER_SEC << "\n\n";
#endif

return 0;
}

//BE CAREFUL: IS INT REALLY INT?

//#define int li

/*int pr[] = { 97, 2011 };
int mods[] = { 1000000007, 1000000009 };

const int C = 300500;
int powers[2][C];*/

//int MOD = 1000000007;

//int c[5010][5010];

template<typename T>
T binpow(T q, T w, T mod) {
if (!w)
return 1 % mod;
if (w & 1)
return q * 1LL * binpow(q, w - 1, mod) % mod;
return binpow(q * 1LL * q % mod, w / 2, mod);
}

/*int curMod = 1000000009;

int fact[100500], revfact[100500];

int getC(int n, int k) {
int res = fact[n] * revfact[n - k] % curMod * revfact[k] % curMod;
return res;
}*/

/*const int C = 7000500;

int least_prime[C];*/

void precalc() {

/*for (int i = 2; i < C; ++i) {
if (!least_prime[i]) {
least_prime[i] = i;
for (li j = i * 1LL * i; j < C; j += i) {
least_prime[j] = i;
}
}
}*/

/*fact[0] = revfact[0] = 1;
for (int i = 1; i < 100500; ++i) {
fact[i] = fact[i - 1] * i % curMod;
revfact[i] = binpow(fact[i], curMod - 2, curMod);
}*/

/*for (int w = 0; w < 2; ++w) {
powers[w][0] = 1;
for (int j = 1; j < C; ++j) {
powers[w][j] = (powers[w][j - 1] * 1LL * pr[w]) % mods[w];
}
}*/
/*for (int i = 0; i < 5010; ++i) {
c[i][i] = c[i][0] = 1;
for (int j = 1; j < i; ++j) {
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
}
}*/
}

template<typename T>
T gcd(T q, T w) {
while (w) {
q %= w;
swap(q, w);
}
return q;
}
template<typename T>
T lcm(T q, T w) {
return q / gcd(q, w) * w;
}

//#define int li

const int mod = 1000000007;

//#define double ld

const int shift = 1 << 17;

struct Node {
int mx;
li sum_mx;
Node():mx(0), sum_mx(0) {}
Node(int mx, li sum_mx):mx(mx), sum_mx(sum_mx) {}
};

const li INF = 2e18;

Node merge(const Node& l, const Node& r) {
Node res = l;
if (res.mx < r.mx) {
res = r;
}
else if (res.mx == r.mx) {
res.sum_mx += r.sum_mx;
if (res.sum_mx > INF) {
res.sum_mx = INF;
}
}
return res;
}

Node tree[2 * shift + 5];

void update(int v, Node new_val) {
v += shift;
tree[v] = new_val;
v /= 2;
while (v) {
tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
v /= 2;
}
}

Node rmq(int l, int r) {
if (l >= r) {
return Node(0, 0);
}
if (l & 1) {
return merge(tree[l], rmq(l + 1, r));
}
if (r & 1) {
return merge(tree[r - 1], rmq(l, r - 1));
}
return rmq(l / 2, r / 2);
}

Node get_rmq(int l, int r) {
return rmq(l + shift, r + shift);
}

li mult(li a, li b) {
if (b == 0) {
return 0;
}
if (a > INF / b) {
return INF;
}
return a * b;
}

li sum(li a, li b) {
li res = a + b;
if (res > INF) {
return INF;
}
return res;
}

void solve() {
int n;
li K;
cin >> n >> K;
vector<pair<int, int>> a(n);
vector<int> vals(n);
for (int i = 0; i < n; ++i) {
cin >> a[i].first;
vals[i] = a[i].first;
a[i].second = -i;
}
sort(all(a));
reverse(all(a));
for (int i = 0; i < n; ++i) {
a[i].second = -a[i].second;
}

vector<vector<int>> best_positions(n + 1);

vector<li> nums(n);

for (int i = 0; i < n; ++i) {
int pos = a[i].second;
Node cur_res = get_rmq(pos, n);
cur_res.mx += 1;
if (cur_res.mx == 1) {
cur_res.sum_mx = 1;
}
update(pos, cur_res);
best_positions[cur_res.mx].push_back(pos);
nums[pos] = cur_res.sum_mx;
}
for (int i = 0; i <= n; ++i) {
sort(all(best_positions[i]));
/*cout << "i = " << i << "\n";
for (int z : best_positions[i]) {
cout << z << ' ';
}
cout << "\n";*/
}

int cur_long = 0;
int last_value = 0;

li sum_all_ways = 0;
for (int i = n; i >= 0; --i) {
if (!best_positions[i].empty()) {
cur_long = i;
for (auto cur : best_positions[i]) {
sum_all_ways = sum(sum_all_ways, nums[cur]);
}
break;
}
}
if (sum_all_ways < K) {
cout << "-1\n";
return;
}

vector<int> ans;

vector<int> last_vec;
last_vec.push_back(-1);
vector<li> last_ways;
last_ways.push_back(1);

for (; cur_long > 0; --cur_long) {
vector<pair<pair<int, int>, pair<li, li>>> candies;
for (int pos : best_positions[cur_long]) {
if (vals[pos] > last_value) {
int id = upper_bound(all(last_vec), pos) - last_vec.begin();
if (id == 0) {
continue;
}
li new_sum_mx = mult(nums[pos], last_ways[id - 1]);
candies.push_back(mp(mp(vals[pos], pos), mp(new_sum_mx, last_ways[id - 1])));
}
}
sort(all(candies));
for (int i = 0; i < candies.size();) {
int j = i;
li sum_ways = 0;
while (j < candies.size() && candies[j].first.first == candies[i].first.first) {
sum_ways = sum(sum_ways, candies[j].second.first);
++j;
}
if (sum_ways >= K) {
ans.push_back(candies[i].first.first);
last_value = ans.back();
last_vec.clear();
last_ways.clear();
for (int r = i; r < j; ++r) {
last_vec.push_back(candies[r].first.second);
last_ways.push_back(candies[r].second.second);
}
for (int r = 1; r < last_ways.size(); ++r) {
last_ways[r] = sum(last_ways[r - 1], last_ways[r]);
}
break;
}
K -= sum_ways;
i = j;
}
}

for (int c : ans) {
cout << c << ' ';
}

}

In Java :

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;

public class E {
InputStream is;
PrintWriter out;
String INPUT = "";

static long I = 2000000000_000000000L;

void solve()
{
int n = ni();
long K = nl();
int[] a = na(n);
int[] b = Arrays.copyOf(a, n);
for(int i = 0;i < n;i++)b[i] = n-a[n-1-i];
b = shrink(b);
//        for(int i = 0;i < n;i++)b[i]++;

SegmentTreeRMQSumWhenMax st = new SegmentTreeRMQSumWhenMax(n+5);
st.updateOrAdd(0, 0, 1); // shifted by 1
int[] maxs = new int[n+1];
long[] counts = new long[n+1];
double[] cx = new double[n+1];
for(int i = 0;i < n;i++){
int max = st.maxx(0, b[i]+1); // <=a[i]
maxs[i] = max;
if(st.gwd >= I)st.gw = I;
counts[i] = st.gw;
cx[i] = st.gwd;
st.updateOrAdd(b[i]+1, max+1, st.gw);
}
int max = st.maxx(0, n+1); // <=a[i]
maxs[n] = max;
counts[n] = st.gw;
cx[n] = st.gwd;
if(cx[n] <= 2E18 && K > counts[n]){
out.println(-1);
return;
}
int lis = maxs[n];
int[][] g = makeBuckets(maxs, lis);
for(int i = 0;i < n+1;i++){
if(cx[i] >= 2E18){
counts[i] = I;
}
}

long[] ft = new long[n+3];
double[] ftd = new double[n+3];
addFenwick(ft, 0, 1);
addFenwick(ftd, 0, 1);
int[] ret = new int[lis];
int[] prevs = new int[n];
long[] pvs = new long[n];
int pp = 0;
prevs[pp] = 0; pvs[pp] = 1; pp++;
for(int h = lis-1;h >= 0;h--){
long[][] temps = new long[g[h].length][];
int p = 0;
for(int i : g[h]){
int ind = n-1-i;
if(h < lis-1 && a[ind] <= ret[lis-(h+1)-1])continue;
long sum = sumFenwick(ft, ind+1);
double sumd = sumFenwick(ftd, ind+1);
if(sumd >= I)sum = I;
long cc = sum*counts[i];
double cd = (double)counts[i]*sum;
if(cd > 2E18){
cc = I;
}
temps[p++] = new long[]{a[ind], cc, sum, ind+1};
}
for(int j = 0;j < pp;j++){
addFenwick(ft, prevs[j], -pvs[j]);
addFenwick(ftd, prevs[j], -pvs[j]);
}

// 1 4 5 3 5
Arrays.sort(temps, 0, p, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
return Long.compare(a[0], b[0]);
}
});
//            tr("temps", temps);
for(int i = 0;i < p;){
int j = i;
while(j < p && temps[j][0] == temps[i][0])j++;
long lnum = 0;
for(int k = i;k < j;k++){
lnum += temps[k][1];
if(lnum >= I)lnum = I;
}
if(K - lnum <= 0){
ret[lis-h-1] = (int)temps[i][0];
break;
}else{
K -= lnum;
}
i = j;
}
pp = 0;
for(int i = 0;i < p;i++){
long[] t = temps[i];
if(t[0] == ret[lis-h-1]){
//                    tr("add", t);
addFenwick(ft, (int)t[3], t[2]);
addFenwick(ftd, (int)t[3], t[2]);
prevs[pp] = (int)t[3]; pvs[pp] = t[2]; pp++;
}
}
}
//        tr(g);

//        tr(maxs);
//        tr(counts);
for(int i = 0;i < lis;i++){
out.print(ret[i] + " ");
}
out.println();
//        tr(K);
}

public static long sumFenwick(long[] ft, int i)
{
long sum = 0;
for(i++;i > 0;i -= i&-i){
sum += ft[i];
}
return sum;
}

public static void addFenwick(long[] ft, int i, long v)
{
if(v == 0)return;
int n = ft.length;
for(i++;i < n;i += i&-i)ft[i] += v;
}

public static double sumFenwick(double[] ft, int i)
{
double sum = 0;
for(i++;i > 0;i -= i&-i){
sum += ft[i];
}
return sum;
}

public static void addFenwick(double[] ft, int i, double v)
{
if(v == 0)return;
int n = ft.length;
for(i++;i < n;i += i&-i)ft[i] += v;
}

public static int[][] makeBuckets(int[] a, int sup)
{
int n = a.length;
int[][] bucket = new int[sup+1][];
int[] bp = new int[sup+1];
for(int i = 0;i < n;i++)bp[a[i]]++;
for(int i = 0;i <= sup;i++)bucket[i] = new int[bp[i]];
for(int i = n-1;i >= 0;i--)bucket[a[i]][--bp[a[i]]] = i;
return bucket;
}

public static int[] shrink(int[] a) {
int n = a.length;
long[] b = new long[n];
for (int i = 0; i < n; i++)
b[i] = (long) a[i] << 32 | i;
Arrays.sort(b);
int[] ret = new int[n];
int p = 0;
for (int i = 0; i < n; i++) {
if (i > 0 && (b[i] ^ b[i - 1]) >> 32 != 0)
p++;
ret[(int) b[i]] = p;
}
return ret;
}

private static class SegmentTreeRMQSumWhenMax {
public int M, H, N;
public int[] st;
public long[] w;
public double[] wd;

public SegmentTreeRMQSumWhenMax(int n)
{
N = n;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new int[M];
w = new long[M];
wd = new double[M];
Arrays.fill(st, 0, M, Integer.MIN_VALUE);
}

// if x equals to before st[H+pos], +y, else update y
public void updateOrAdd(int pos, int x, long y)
{
if(x < st[H+pos])throw new RuntimeException("x < st[H+pos]");
if(x == st[H+pos]){
w[H+pos] += y;
wd[H+pos] += y;
}else{
st[H+pos] = x;
w[H+pos] = y; // y % mod
wd[H+pos] = y; // y % mod
}
for(int i = (H+pos)>>>1;i >= 1;i >>>= 1)propagate(i);
}

private void propagate(int i)
{
if(st[2*i] < st[2*i+1]){
st[i] = st[2*i+1];
w[i] = w[2*i+1];
wd[i] = wd[2*i+1];
}else if(st[2*i] > st[2*i+1]){
st[i] = st[2*i];
w[i] = w[2*i];
wd[i] = wd[2*i];
}else{
st[i] = st[2*i];
w[i] = w[2*i] + w[2*i+1];
wd[i] = wd[2*i] + wd[2*i+1];
}
}

public long gw;
public double gwd;

public int maxx(int l, int r){
gw = 0;
gwd = 0.;
if(l >= r)return 0;
int max = Integer.MIN_VALUE;
while(l != 0){
int f = l&-l;
if(l+f > r)break;
int v = st[(H+l)/f];
if(v > max){
max = v;
gw = w[(H+l)/f];
gwd = wd[(H+l)/f];
}else if(v == max){
gw += w[(H+l)/f];
gwd += wd[(H+l)/f];
}
l += f;
}

while(l < r){
int f = r&-r;
int v = st[(H+r)/f-1];
if(v > max){
max = v;
gw = w[(H+r)/f-1];
gwd = wd[(H+r)/f-1];
}else if(v == max){
gw += w[(H+r)/f-1];
gwd += wd[(H+r)/f-1];
}
r -= f;
}
return max;
}
}

void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);

long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}

public static void main(String[] args) throws Exception { new E().run(); }

private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;

private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}

private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }

private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}

private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}

private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}

private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}

private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}

while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}

private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}

while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}

private static void tr(Object... o) {
System.out.println(Arrays.deepToString(o)); }
}

In Python3 :

#!/bin/python3

import os
import sys

class Node:
prev = []
marked = False

def __init__(self, num: int):
self.num = num

def _set_all_prefixed(new_heap, patience_heaps, curr_ind):
prefix_list = list()
prev_patience_heap = patience_heaps[curr_ind - 1]
patience_heap_size = len(prev_patience_heap)
for i in range(0, patience_heap_size):
elem = prev_patience_heap[i]
if elem.num < new_heap.num:
prefix_list.append(elem)
else:
pass
return prefix_list

def patience_sorting(A: list):
size = len(A)
patience_heaps: list(list(Node)) = [[Node(A[0])]]
for i in range(1, size):
curr = A[i]
smallest_number_bigger_than_curr = sys.maxsize
snbtc_ind = -1
patience_heaps_size = len(patience_heaps)
for j in range(patience_heaps_size):
patience_heap_top = patience_heaps[j][-1]
if smallest_number_bigger_than_curr > patience_heap_top.num >= curr:
smallest_number_bigger_than_curr = patience_heap_top.num
snbtc_ind = j
set_prev = False
if snbtc_ind < 0:
new_heap = Node(curr)
set_prev = True
patience_heaps.append([new_heap])
else:
new_heap = Node(curr)
if snbtc_ind > 0:
set_prev = True
patience_heaps[snbtc_ind].append(new_heap)
# point to the top of the previous patience heap
if set_prev:
# new_heap.prev = patience_heaps[snbtc_ind-1][-1]
new_heap.prev = _set_all_prefixed(new_heap, patience_heaps, snbtc_ind)
return patience_heaps

# def print_all_LIS_wrapper(patience_heaps, k):
#     sequence_tails = patience_heaps[-1]
#     sequence_tails.reverse()
#     count = 1
#     for tail in sequence_tails:
#         count = print_all_LIS(list(), tail, k, count)
#
#
# def print_all_LIS(subsequence, node, k, count):
#     if count > k:
#         return count
#     temp = [node] + subsequence
#     if len(node.prev) == 0:
#         if count == k:
#             print(' '.join([str(n.num) for n in temp]))
#         return count+1
#     else:
#         node.prev.reverse()
#         for prev in node.prev:
#             count = print_all_LIS(temp, prev, k, count)
#     return count

def print_all_LIS_wrapper(patience_heaps, k):
sequence_tails = patience_heaps[-1]
sequence_tails.reverse()
count = 0
subsequence = list()
for node in sequence_tails:
stack = [node]
while len(stack) != 0:
curr = stack.pop(0)
while len(subsequence) > 0 and curr.num >= subsequence[0]:
subsequence.pop(0)
subsequence.insert(0, curr.num)
prevs = curr.prev
if len(prevs) == 0:
count += 1
if count == k:
return subsequence
else:
for prev in prevs:
stack.insert(0, prev)
return -1

def superKth(k, a):
sorted_heaps = patience_sorting(a)
subsequence = print_all_LIS_wrapper(sorted_heaps, k)
if isinstance(subsequence, list):
retval = [str(a) for a in subsequence]
print(' '.join(retval))
else:
print(subsequence)

if __name__ == '__main__':

nk = input().split()

n = int(nk[0])

k = int(nk[1])

a = list(map(int, input().rstrip().split()))

result = superKth(k, a)```
```

## Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

## Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

## AND xor OR

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

## 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

## 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

## 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):