Skip to content
Programmingoneonone
Programmingoneonone

LEARN EVERYTHING ABOUT PROGRAMMING

  • Home
  • CS Subjects
    • IoT ? Internet of Things
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone

LEARN EVERYTHING ABOUT PROGRAMMING

HackerRank Inverse RMQ problem solution

YASH PAL, 31 July 2024

In this HackerRank Inverse RMQ problem solution, we have given an array of distinct integers with size n = 2k and m queries. we need to find the minimum element on subsegment [Li, Ri].

Hackerrank inverse RMQ problem solution

Topics we are covering

Toggle
  • Problem solution in Python.
  • Problem solution in Java.
  • Problem solution in C++.
  • Problem solution in C.

Problem solution in Python.

import sys
import heapq

n = int(input())
a = list(map(int, input().split()))
freq = dict()
for i in a:
    if i not in freq:
        freq[i] = 0
    freq[i] += 1
if len(freq) < n:
    print("NO")
    sys.exit()
exp_freq = 1
depth = 1
while 2**(depth - 1) < n:
    depth += 1
prev = dict()
ans = [0] * (n + n - 1)
while exp_freq <= n:
    v = list()
    v1 = list()
    for key in prev:
        v1.append((key, prev[key]))
    for key in freq:
        if freq[key] == depth:
            v.append(key)
    if len(prev) == 0:
        ans[0] = v[0]
        prev[v[0]] = 0
        freq[v[0]] -= 1
        exp_freq *= 2
        depth -= 1
        continue
    v.sort()
    v1.sort()
    cur = exp_freq // 2 - 1
    pq = list()
    j = 0
    for i in v:
        if i in prev:
            ans[prev[i] * 2 + 1] = i
            prev[i] = prev[i] * 2 + 1
            freq[i] -= 1
        else:
            while j < len(v1):
                if v1[j][0] < i:
                    heapq.heappush(pq, v1[j][1])
                    j += 1
                else:
                    break
            if len(pq) == 0:
                print("NO")
                sys.exit()
            temp = heapq.heappop(pq)
            ans[temp * 2 + 2] = i
            prev[i] = temp * 2 + 2
            freq[i] -= 1
    exp_freq *= 2
    depth -= 1
print("YES")
for i in ans:
    print(i, end=" ")

{“mode”:”full”,”isActive”:false}

Problem solution in Java.

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.io.IOException;
import java.io.Reader;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.TreeMap;
import java.util.StringTokenizer;
import java.util.Map;
import java.util.Map.Entry;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Solution {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        InversedRMQ solver = new InversedRMQ();
        solver.solve(1, in, out);
        out.close();
    }

    static class InversedRMQ {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.readInt();
            int k = 2 * n - 1;
            final Map<Integer, Integer> count = new TreeMap<>();
            for (int i = 0; i < k; i++) {
                int x = in.readInt();
                count.put(x, count.getOrDefault(x, 0) + 1);
            }
            final TreeSet<Integer>[] types = new TreeSet[k + 1];
            final Map<Integer, Integer> result = new HashMap<>();
            for (int i = 0; i < types.length; i++) {
                types[i] = new TreeSet<>();
            }
            int log = 0, n_ = n;
            while (n_ > 0) {
                n_ >>= 1;
                log++;
            }
            types[log].add(1);
            for (Map.Entry<Integer, Integer> entry : count.entrySet()) {
                if (types[entry.getValue()].isEmpty()) {
                    out.printLine("NO");
                    return;
                }
                int wh = types[entry.getValue()].pollFirst();
                result.put(entry.getKey(), wh);
                for (int i = 0; i < entry.getValue() - 1; i++) {
                    types[entry.getValue() - i - 1].add(wh * 2 + 1);
                    wh <<= 1;
                }
            }
            final int[] ans = new int[k + 1];
            for (Map.Entry<Integer, Integer> entry : result.entrySet()) {
                int wh = entry.getValue(), value = entry.getKey();
                while (wh <= k) {
                    ans[wh] = value;
                    wh <<= 1;
                }
            }
            out.printLine("YES");
            for (int i = 1; i <= k; i++) {
                out.print(ans[i], ' ');
            }
            out.printLine();
        }

    }

    static class OutputWriter {
        private PrintWriter writer;

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public OutputWriter(OutputStream stream) {
            this(new OutputStreamWriter(stream));
        }

        public void print(Object... args) {
            for (Object arg : args) {
                writer.print(arg);
            }
        }

        public void printLine(Object... args) {
            print(args);
            writer.println();
        }

        void close() {
            writer.close();
        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(Reader reader) {
            this.reader = new BufferedReader(reader);
        }

        public InputReader(InputStream stream) {
            this(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public String readWord() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(nextLine());
            }
            return tokenizer.nextToken();
        }

        public int readInt() {
            return Integer.parseInt(readWord());
        }

    }
}

{“mode”:”full”,”isActive”:false}

Problem solution in C++.

#include<bits/stdc++.h>
using namespace std;
void Q(){
    cout << "NO" << endl;
    exit(0);
}
set<int> Adj[(1 << 21)];
int Ar[(1 << 20)] , X , nn;
vector<int> Ans[(1 << 20)];
void build(int lvl , int V , bool F){
    if(lvl == nn)return ;
    set<int> :: iterator no = Adj[lvl].upper_bound(V);
    if(F && (no == Adj[lvl].end() || *no <= V)){
        Q();
    }
    if(F){
        int x = *no;
        Ans[lvl].push_back(x);
        Adj[lvl].erase(no);
        build(lvl + 1 , x , 0);
        build(lvl + 1 , x , 1);
    }
    else {
        Ans[lvl].push_back(V);
        build(lvl + 1 , V , 0);
        build(lvl + 1 , V , 1);
    }
}
int main()
{
    int n;
    cin >> n;
    vector<int> A(2 * n - 1);
    map<int , int > mp ;
    map<int , set<int> > mp1;
    if(n == 1){
        cin >> n;
        cout << "YESn" << n << endl;;
        return 0;
    }
    int N = 2 * n - 1 , mx = 0 , idx = 0;
    for(int i = 0 ;i < N ; i ++){
        cin >> A[i] , mp[A[i]] ++;
        if(mp[A[i]] > mx)
            mx = mp[A[i]] , idx = A[i];
    }
    //if(mp.begin() -> first != idx)  Q();
    set<int> S;
    for(int i = 0 ;i < N ;i ++)
        mp1[mp[A[i]]].insert(A[i]),  S.insert(A[i]);
    for(int m = n / 2 , x = 1; m ; m >>= 1 , x ++){
        if(mp1[x].size() != m)
            Q();
    }
    for(auto no : S)
        Adj[mx - mp[no] + 1].insert(no);
    nn = mx + 1 ;
    int KK = *Adj[1].begin();
    Adj[1].erase(Adj[1].begin());
    build(1 , KK , 0);
    cout << "YESn";
    for(int i = 1; i <= mx ;i ++)
        for(int l = 0; l < Ans[i].size() ;l ++)
            cout << Ans[i][l] << " ";

}

{“mode”:”full”,”isActive”:false}

Problem solution in C.

#include <assert.h>
#include <stddef.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* readline();
char** split_string(char*);

int* canTree(int n, int* arr){
    int logn = 0;
    while(n >> logn > 1){
        logn++;
    }

    int sortarr[2*n - 1];
    for(int i = 0; i < 2*n - 1; i++){
        sortarr[i] = arr[i];
        int curr = i;
        int inplace = 0;
        while(curr > 0 && inplace == 0){
            int next = (curr - 1)/2;
            if(sortarr[next] < sortarr[curr]){
                int temp = sortarr[curr];
                sortarr[curr] = sortarr[next];
                sortarr[next] = temp;
                curr = next;
            }
            else{
                inplace = 1;
            }
        }
    }

    for(int i = 0; i < 2*n - 1; i++){
        int temp = sortarr[0];
        sortarr[0] = sortarr[2*n - i - 2];
        sortarr[2*n - i - 2] = temp;

        int curr = 0;
        int reheaped = 0;
        while(reheaped == 0){
            int next = curr;
            if(2*curr + 1 < 2*n - i - 2 && sortarr[2*curr + 1] > sortarr[next]){
                next = 2*curr + 1;
            }
            if(2*curr + 2 < 2*n - i - 2 && sortarr[2*curr + 2] > sortarr[next]){
                next = 2*curr + 2;
            }
            if(next != curr){
                temp = sortarr[curr];
                sortarr[curr] = sortarr[next];
                sortarr[next] = temp;
                curr = next;
            }
            else{
                reheaped = 1;
            }
        }
    }

    int *levellist[logn + 1];
    int numatlevel[logn + 1];
    int **numopen[logn + 1];
    for(int i = 0; i <= logn; i++){
        levellist[i] = malloc((1<<i)*sizeof(int));
        assert(levellist[i] != NULL);
        for(int j = 0; j < 1<<i; j++){
            levellist[i][j] = INT_MIN;
        }
        numopen[i] = malloc((1<<i)*sizeof(int*));
        assert(numopen[i] != NULL);
        for(int j = 0; j <= (1<<i); j++){
            numopen[i][j] = malloc((logn + 1)*sizeof(int));
            assert(numopen[i][j] != NULL);
            for(int k = 0; k <= logn; k++){
                numopen[i][j][k] = 0;
            }
        }
        numatlevel[i] = 0;
    }
    numopen[0][0][0] = 1;
    

    

    int index = 0;
    while(index < 2*n - 1){
        int currval = sortarr[index];
        int currindex = index;
        while(index < 2*n - 1 && sortarr[index] == currval){
            index++;
        }
        int numreps = index - currindex;
        int level = (logn + 1) - numreps;
        int pos = 0;
        if(level < 0 || numopen[0][0][level] == 0){
            return NULL;
        }
        for(int i = 0; i < level; i++){
            assert(numopen[i][pos][level] > 0);
            numopen[i][pos][level] -= 1;
            for(int j = level + 1; j <= logn; j++){
                numopen[i][pos][j] += 1;
            }
            if(i + 1 < level && numopen[i + 1][2*pos][level] > 0){
                pos = 2*pos;
            }
            else{
                pos = 2*pos + 1;
            }
        }
        for(int i = level; i <= logn; i++){
            levellist[i][pos] = currval;
            for(int j = i + 1; j <= logn; j++){
                numopen[i][pos][j] += 1;
            }
            pos = 2*pos;
        }
    }

    int *toreturn = malloc((2*n - 1)*sizeof(int));
    assert(toreturn != NULL);
    index = 0;
    for(int i = 0; i <= logn; i++){
        for(int j = 0; j < (1<<i); j++){
            toreturn[index] = levellist[i][j];
            index++;
        }
    }
    
    return toreturn;
}

int main() {
    FILE* fptr = fopen(getenv("OUTPUT_PATH"), "w");

    char* n_endptr;
    char* n_str = readline();
    int n = strtol(n_str, &n_endptr, 10);

    if(n_endptr == n_str || *n_endptr != ''){ exit(EXIT_FAILURE);}
    
    char** vals_str = split_string(readline());
    int *treelist = malloc((2*n - 1) * sizeof(int));

    for(int i = 0; i < 2*n - 1; i++){
        char* currval_str = vals_str[i];
        char* currval_endptr;
        int currval = strtol(currval_str, &currval_endptr, 10);

        if(currval_endptr == currval_str || *currval_endptr != ''){ exit(EXIT_FAILURE);}
        treelist[i] = currval;
    }
    int* arrlist = canTree(n, treelist);
    if(arrlist == NULL){
        fprintf(fptr, "NO");
    }
    else{
        fprintf(fptr, "YESn");
        for(int i = 0; i < 2*n - 2; i++){
            fprintf(fptr, "%d ", arrlist[i]);
        }
        fprintf(fptr, "%d", arrlist[2*n - 2]);
    }   
    return 0;
}


char* readline() {
    size_t alloc_length = 1024;
    size_t data_length = 0;
    char* data = malloc(alloc_length);

    while (true) {
        char* cursor = data + data_length;
        char* line = fgets(cursor, alloc_length - data_length, stdin);

        if (!line) { break; }

        data_length += strlen(cursor);

        if (data_length < alloc_length - 1 || data[data_length - 1] == 'n') { break; }

        size_t new_length = alloc_length << 1;
        data = realloc(data, new_length);

        if (!data) { break; }

        alloc_length = new_length;
    }

    if (data[data_length - 1] == 'n') {
        data[data_length - 1] = '';
    }
    if(data[data_length - 1] != ''){
        data_length++;
        data = realloc(data, data_length);
        data[data_length - 1] = '';
    }

    data = realloc(data, data_length);

    return data;
}

char** split_string(char* str) {
    char** splits = NULL;
    char* token = strtok(str, " ");

    int spaces = 0;

    while (token) {
        splits = realloc(splits, sizeof(char*) * ++spaces);
        if (!splits) {
            return splits;
        }

        splits[spaces - 1] = token;

        token = strtok(NULL, " ");
    }

    return splits;
}

{“mode”:”full”,”isActive”:false}

Algorithms coding problems solutions

Post navigation

Previous post
Next post
  • Automating Image Format Conversion with Python: A Complete Guide
  • HackerRank Separate the Numbers solution
  • How AI Is Revolutionizing Personalized Learning in Schools
  • GTA 5 is the Game of the Year for 2024 and 2025
  • Hackerrank Day 5 loops 30 days of code solution
How to download udemy paid courses for free

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes