# Inverse RMQ

### Problem Statement :

```Range Minimum Query is a well-known problem: given an array of distinct integers with size n=2^k and m queries, find the minimum element on subsegment [Li,Ri].

One of the most efficient and famous solutions to this problem is a segment tree. A segment tree is a full binary tree with 2.n-1 nodes where the leaves contain the values of the original array and each non-leaf node contains the minimum value of its entire subtree.

Usually, a segment tree is represented as an array of integers with 2.n-1 elements. The left child of the ith node is in the (2.i+1)th cell, and the right child is in the (2.i+2)th cell. For example, A = [1,1,3,1,2,3,4] represents the following segment tree where the first number in a node describes the array index, i, in A and the second number denotes the value stored at index i (which corresponds to the minimum value in that node's subtree):

example1_graph.png

You've just used n distinct integers to construct your first segment tree and saved it as an array, A, of 2.n-1 values. Unfortunately, some evil guy came and either shuffled or altered the elements in your array. Can you use the altered data to restore the original array? If no, print NO on a new line; otherwise, print two lines where the first line contains the word YES and the second line contains 2.n-1 space-separated integers denoting the array's original values. If there are several possible original arrays, print the lexicographically smallest one.

Input Format

The first line contains a single integer, n, denoting the size of the array.
The second line contains 2.n-1 space-separated integers denoting the shuffled values of the segment tree.

Constraints
1 <= n <= 2^18
n is a power of two.
Each value in the segment tree is between -10^9 and 10^9.
Output Format

Print NO if this array could not be constructed by shuffling some segment tree. Otherwise, print YES on the first line, and 2.n-1 space-separated integers describing the respective values of the original array on the second line. If there are several possible answers, print the lexicographically smallest one.```

### Solution :

```                            ```Solution in C :

In C++ :

/*
*/

#define _CRT_SECURE_NO_WARNINGS

#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#include <assert.h>

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd

#define eps 1e-9
#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 350

using namespace std;

const int INF = 1e9;
const int N = 1500031;

int n;
map<int, int> cnt;
map<int, int>::iterator it;

int ANS[N];
set<int> set_sz[N];
int DEP[N * 2];
set<int>::iterator it2;

void run_builder(int v, int tl, int tr,int dep)
{
if (tl == tr)
{
set_sz[dep].insert(v);
DEP[v] = dep;
return;
}
int tm = tl + tr;
tm /= 2;
run_builder(v * 2, tl, tm, dep + 1);
run_builder(v * 2 + 1, tm + 1, tr, dep + 1);
}

int used[N * 2];

void remove_vertex(int v)
{
int here = DEP[v];
//used[v] = 1;
if (v>=n)
set_sz[here].erase(v);
}

{
DEP[v] = val;
if (v>=n)
set_sz[val].insert(v);
}

void DFS(int vert, int D)
{
if (used[vert])
return;
remove_vertex(vert);
if (vert<n)
{
DFS(vert * 2, D + 1);
DFS(vert * 2 + 1, D + 1);
}
}

{
while (vert > 1 && used[vert] == 0)
{
remove_vertex(vert);
used[vert] = 1;
if (vert < n)
{
DFS(vert * 2, 1);
DFS(vert * 2 + 1, 1);
}
vert /= 2;
}
if (vert == 1)
{
used[vert] = 1;
remove_vertex(vert);
}

if (vert < n)
{
DFS(vert*2, 1);
DFS(vert * 2 + 1, 1);
}
}

int T[N * 2];

void TRACE(int v, int l, int r)
{
if (l == r)
{
T[v] = ANS[v];
return;
}
int m = l + r;
m /= 2;
TRACE(v * 2, l, m);
TRACE(v * 2 + 1, m + 1, r);
T[v] = min(T[v * 2], T[v * 2 + 1]);
}

int main(){
//freopen("fabro.in","r",stdin);
//freopen("fabro.out","w",stdout);
//freopen("F:/in.txt", "r", stdin);
//freopen("F:/output.txt", "w", stdout);
ios_base::sync_with_stdio(0);
//cin.tie(0);

cin >> n;

for (int i = 1; i < 2 * n; i++)
{
int val;
cin >> val;
cnt[val]++;
}

run_builder(1,1,n,1);

for (it = cnt.begin(); it != cnt.end(); ++it)
{
int am = (*it).second;
int val = (*it).first;

//	cout << val << "%" << am << endl;

/*for (int i = 4; i <= 7; i++)
{
cout << DEP[i] << " ";
}
cout << endl;
*/	//cout << set_sz[1].size() << endl;

if (set_sz[am].size() == 0)
{
cout << "NO" << endl;
return 0;
}

it2 = set_sz[am].begin();
int vert = (*it2);
//cout << "\$\$" << vert << endl;

//cout << vert << "%%" << val << endl;

ANS[vert] = val;
}

TRACE(1,1,n);

cout << "YES" << endl;
for (int i = 1; i < n * 2; i++)
{
if (i>1)
cout << " ";
cout << T[i];
}
cout << endl;

cin.get(); cin.get();
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 n = sc.nextInt();
int faken = 1;
int d = 1;
while (faken < n) {
faken <<= 1;
d++;
}
sc.nextLine();
String[] nums = sc.nextLine().split(" ");
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
for (int i = 0; i < 2*n-1; i++) {
int j = Integer.parseInt(nums[i]);
if (!hm.containsKey(j))
hm.put(j, 0);
hm.put(j, hm.get(j)+1);
}
ArrayList<TreeSet<Integer>> tms = new ArrayList<TreeSet<Integer>>();
for (int i = 0; i <= d; i++)
for (int j : hm.keySet()) {
if (hm.get(j) > d) {
System.out.println("NO");
return;
}
}
int arrind = 0;
int[] tree = new int[2*n-1];
int expectation = 1;
for (int i = 0; i < d; i++) {
if (tms.get(d-i).size()!=expectation) {
System.out.println("NO");
return;
}
for (int j = 0; j < expectation; j++) {
Integer value;
if (i==0) {
value = tms.get(d-i).first();
} else {
value = tms.get(d-i).ceiling(tree[arrind-1]);
}
if (value == null) {
System.out.println("NO");
return;
}
tms.get(d-i).remove(value);
int val = value;
int temparrind = arrind;
for (int k = 0; k < d-i; k++) {
tree[temparrind] = val;
temparrind = temparrind*2+1;
}
arrind += 2;
}
if (i > 0)
expectation *= 2;
}
boolean valid = true;
for (int i = 0; i < n-1; i++) {
if (arrind<2*n-1||tree[i]!=tree[2*i+1]||tree[2*i+1]>=tree[2*i+2]) {
valid = false;
break;
}
}
if (valid) {
System.out.println("YES");
StringBuilder ans = new StringBuilder();
String space = "";
for (int i = 0; i < 2*n-1; i++) {
ans.append(space+tree[i]);
space = " ";
}
System.out.println(ans);
} else {
System.out.println("NO");
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
struct mul_t {
int el;
int mul;
};

int cmpmul(const void* a_in, const void* b_in) {
const struct mul_t* a = a_in;
const struct mul_t* b = b_in;
if (a->mul < b->mul) return 1;
if (a->mul > b->mul) return -1;

if (a->el < b->el) return -1;
if (a->el > b->el) return 1;
return 0;
}

int cmpint(const void* a_in, const void* b_in) {
const int* a = a_in;
const int* b = b_in;
if (*a < *b) return -1;
if (*a > *b) return 1;
return 0;
}

int main() {
int n;
scanf("%d\n", &n);
int tn = n + n - 1;
int els[tn];
for (int i = 0; i < tn; i++) {
scanf("%d", els + i);
}
struct mul_t muls[tn];
memset(muls, 0, sizeof(muls));
qsort(els, tn, sizeof(int), cmpint);
int nmuls = 0;
for (int i = 0; i < tn; i++) {
if (nmuls == 0 || muls[nmuls-1].el != els[i]) nmuls++;
muls[nmuls-1].el = els[i];
muls[nmuls-1].mul++;
}
qsort(muls, nmuls, sizeof(struct mul_t), cmpmul);
// Check if this is a segment tree -- multiplicities must work for that
int depth = log2(n) + 1;
int expel = 1;
int mp = 0;
for (int expmul = depth; expmul > 0; expmul--) {
for (int i = 0; i < expel; i++) {
if (muls[mp++].mul != expmul) {
printf("NO\n");
exit(0);
}
}
if (expmul < depth) expel *= 2;
}

int tl = 2;
mp = 1;
int st[tn];
int sp = 0;
st[sp++] = muls[0].el;
//   fprintf(stderr, "%d ", st[sp - 1]);
while (mp < nmuls) {
int tp = sp / 2;
int emp = mp + tl;
for (int tc = 0; tc < tl; tc += 2) {
st[sp++] = st[tp++];
for (int i = mp; i < emp; i++) {
if (muls[i].el > st[(sp-1)/2]) {
if (i != mp) {
struct mul_t tmp = muls[i];
memmove(muls + mp + 1, muls + mp, (i - mp) * sizeof(struct mul_t));
muls[mp] = tmp;
}
break;
}
}
st[sp++] = muls[mp++].el;
if (st[(sp - 2)/2] > st[sp - 1]) exit(-1);
//           fprintf(stderr, "%d %d ", st[sp - 2], st[sp - 1]);
//           if (st[(sp - 2)/2] != st[sp - 2]) exit(-1);
}
tl*=2;
}
printf("YES\n");
for (sp = 0; sp < tn; sp++) {
printf("%d ", st[sp]);
}
printf("\n");
return 0;
}

In Python3 :

import sys
from bisect import bisect_right

def z(a, b):
B = sorted(b)
r = []
for i in range(len(a)):
ind = bisect_right(B, a[i])
r += [a[i], B[ind]]
del B[ind]
#print(a, b, r)
return r

n = int(input())
nl = 0
x = n
while x:
nl += 1
x >>= 1

a = list(map(int,input().split()))
occ = {}
for e in a:
if not e in occ:
occ[e] = 0
occ[e] += 1

# test #############
cnt_occ = {}
for i, v in occ.items():
if not v in cnt_occ:
cnt_occ[v] = 0
cnt_occ[v] += 1
for i in range(1, nl):
if not i in cnt_occ or cnt_occ[i] != (1 << (nl - 1 - i)):
print('NO')
sys.exit(0)
#####################

ah = [[] for _ in range(nl + 2)]
for i, v in occ.items():
ah[v].append(i)

sofar = ah[nl]
res = list(sofar)
for i in range(1, nl):
sofar = z(sofar, ah[nl - i])
res += sofar
print('YES')
print(' '.join(map(str,res)))```
```

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