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++ :
/*
*/
//#pragma comment(linker, "/STACK:16777216")
#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 yn askfhwqriuperikldjk
#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);
}
void add_vertex(int v, int val)
{
DEP[v] = val;
if (v>=n)
set_sz[val].insert(v);
}
void DFS(int vert, int D)
{
if (used[vert])
return;
remove_vertex(vert);
add_vertex(vert, D);
if (vert<n)
{
DFS(vert * 2, D + 1);
DFS(vert * 2 + 1, D + 1);
}
}
void run_updates(int vert)
{
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;
run_updates(vert);
//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) {
Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
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++)
tms.add(new TreeSet<Integer>());
for (int j : hm.keySet()) {
if (hm.get(j) > d) {
System.out.println("NO");
return;
}
tms.get(hm.get(j)).add(j);
}
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)))
View More Similar Problems
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
View Solution →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
View Solution →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
View Solution →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):
View Solution →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.
View Solution →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
View Solution →