Imprima todos los elementos repetitivos en una array dada dentro del rango de valores [A, B] para consultas Q

Dada una array arr[] de tamaño N y Q consultas de la forma [A, B] , la tarea es encontrar todos los elementos únicos de la array que son repetitivos y sus valores se encuentran entre A y B (ambos inclusive) para cada una de las consultas Q.

Ejemplos:

Entrada: arr[] = { 1, 5, 1, 2, 3, 3, 4, 0, 0 }, Q = 2, consultas={ { 1, 3 }, { 0, 0 } }
Salida: {3, 1}, { 0 }
Explicación: Para la primera consulta, los elementos 1 y 3 se encuentran en el rango dado y ocurren más de una vez.
Para la segunda consulta, solo 0 se encuentra en el rango y es de naturaleza repetitiva.

Entrada: arr[] = {1, 5, 1, 2, 3, 4, 0, 0}, Q = 1, consultas={ { 1, 2 } }
Salida: 1

Enfoque ingenuo: el enfoque básico para resolver este problema es usar bucles anidados . El bucle externo realiza un recorrido de todos los elementos, y el bucle interno realiza una verificación de si el elemento seleccionado por el bucle externo aparece en cualquier otro lugar. Luego, si aparece en otro lugar, verifique si se encuentra dentro del rango de [A, B] . Si es así, imprímelo.

Complejidad de Tiempo: O(Q * N ² )
Espacio Auxiliar: O(1)

Enfoque eficiente:  un enfoque eficiente se basa en la idea de un mapa hash para almacenar la frecuencia de todos los elementos. Siga los pasos que se mencionan a continuación:

• Recorra el mapa hash y verifique si la frecuencia de un elemento es mayor que 1.
• En caso afirmativo, compruebe si el elemento está presente en el rango de [A, B] o no.
• En caso afirmativo, imprímalo; de lo contrario, omita el elemento.
• Continúe con el procedimiento anterior hasta que se atraviesen todos los elementos del mapa hash .

A continuación se muestra la implementación del enfoque anterior.

// C++ code for the above approach.
#include <bits/stdc++.h>
using namespace std;
 
// Initializing hashmap to store
// Frequency of elements
unordered_map<int, int> freq;
 
// Function to store frequency of elements in hash map
 
void storeFrequency(int arr[], int n)
{
    // Iterating the array
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
}
 
// Function to print elements
void printElements(int a, int b)
{
 
    // Traversing the hash map
    for (auto it = freq.begin(); it != freq.end(); it++) {
 
        // Checking 1st condition if frequency > 1, i.e,
        // Element is repetitive
 
        if ((it->second) > 1) {
 
            int value = it->first;
 
            // Checking 2nd condition if element
            // Is in range of a and b or not
            if (value >= a && value <= b) {
                // Printing the value
                cout << value << " ";
            }
        }
    }
}
 
// Function to find the elements
// satisfying given condition for each query
void findElements(int arr[], int N, int Q,
                  vector<pair<int, int> >& queries)
{
    storeFrequency(arr, N);
    for (int i = 0; i < Q; i++) {
        int A = queries[i].first;
        int B = queries[i].second;
        printElements(A, B);
        cout << endl;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 5, 1, 2, 3, 3, 4, 0, 0 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
    int Q = 2;
    vector<pair<int, int> > queries = { { 1, 3 }, { 0, 0 } };
 
    // Function call
    findElements(arr, N, Q, queries);
    return 0;
}

// Java code for the above approach.
 
import java.util.*;
 
class GFG{
 
// Initializing hashmap to store
// Frequency of elements
static HashMap<Integer,Integer> freq=new HashMap<Integer,Integer> ();
 
// Function to store frequency of elements in hash map
 
static void storeFrequency(int arr[], int n)
{
    // Iterating the array
    for (int i = 0 ; i < n; i++){
        if(freq.containsKey(arr[i])){
            freq.put(arr[i], freq.get(arr[i])+1);
        }
        else{
            freq.put(arr[i], 1);
        }
    }
}
 
// Function to print elements
static void printElements(int a, int b)
{
 
    // Traversing the hash map
    for (Map.Entry<Integer,Integer> it : freq.entrySet()) {
 
        // Checking 1st condition if frequency > 1, i.e,
        // Element is repetitive
 
        if ((it.getValue()) > 1) {
 
            int value = it.getKey();
 
            // Checking 2nd condition if element
            // Is in range of a and b or not
            if (value >= a && value <= b) {
                // Printing the value
                System.out.print(value+ " ");
            }
        }
    }
}
 
// Function to find the elements
// satisfying given condition for each query
static void findElements(int arr[], int N, int Q,
                  int[][] queries)
{
    storeFrequency(arr, N);
    for (int i = 0; i < Q; i++) {
        int A = queries[i][0];
        int B = queries[i][1];
        printElements(A, B);
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 5, 1, 2, 3, 3, 4, 0, 0 };
 
    // Size of array
    int N = arr.length;
    int Q = 2;
    int [][]queries = { { 1, 3 }, { 0, 0 } };
 
    // Function call
    findElements(arr, N, Q, queries);
}
}
 
// This code contributed by shikhasingrajput

# Python 3 code for the above approach.
from collections import defaultdict
 
# Initializing hashmap to store
# Frequency of elements
freq = defaultdict(int)
 
# Function to store frequency of elements in hash map
def storeFrequency(arr, n):
 
    # Iterating the array
    for i in range(n):
        freq[arr[i]] += 1
 
# Function to print elements
def printElements(a, b):
 
    # Traversing the hash map
    for it in freq:
 
        # Checking 1st condition if frequency > 1, i.e,
        # Element is repetitive
        # print("it = ",it)
        if ((freq[it]) > 1):
            value = it
 
            # Checking 2nd condition if element
            # Is in range of a and b or not
            if (value >= a and value <= b):
               
                # Printing the value
                print(value, end=" ")
 
# Function to find the elements
# satisfying given condition for each query
def findElements(arr, N,  Q,
                 queries):
 
    storeFrequency(arr, N)
    for i in range(Q):
        A = queries[i][0]
        B = queries[i][1]
        printElements(A, B)
        print()
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 5, 1, 2, 3, 3, 4, 0, 0]
 
    # Size of array
    N = len(arr)
    Q = 2
    queries = [[1, 3], [0, 0]]
 
    # Function call
    findElements(arr, N, Q, queries)
 
    # This code is contributed by ukasp.

// C# code for the above approach.
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Initializing hashmap to store
  // Frequency of elements
  static Dictionary<int,int> freq=new Dictionary<int,int> ();
 
  // Function to store frequency of elements in hash map
 
  static void storeFrequency(int []arr, int n)
  {
    // Iterating the array
    for (int i = 0 ; i < n; i++){
      if(freq.ContainsKey(arr[i])){
        freq[arr[i]]= freq[arr[i]]+1;
      }
      else{
        freq.Add(arr[i], 1);
      }
    }
  }
 
  // Function to print elements
  static void printElements(int a, int b)
  {
 
    // Traversing the hash map
    foreach (KeyValuePair<int,int> it in freq) {
 
      // Checking 1st condition if frequency > 1, i.e,
      // Element is repetitive
 
      if ((it.Value) > 1) {
 
        int value = it.Key;
 
        // Checking 2nd condition if element
        // Is in range of a and b or not
        if (value >= a && value <= b) {
          // Printing the value
          Console.Write(value+ " ");
        }
      }
    }
  }
 
  // Function to find the elements
  // satisfying given condition for each query
  static void findElements(int []arr, int N, int Q,
                           int[,] queries)
  {
    storeFrequency(arr, N);
    for (int i = 0; i < Q; i++) {
      int A = queries[i,0];
      int B = queries[i,1];
      printElements(A, B);
      Console.WriteLine();
    }
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int []arr = { 1, 5, 1, 2, 3, 3, 4, 0, 0 };
 
    // Size of array
    int N = arr.Length;
    int Q = 2;
    int [,]queries = { { 1, 3 }, { 0, 0 } };
 
    // Function call
    findElements(arr, N, Q, queries);
  }
}
 
// This code contributed by shikhasingrajput

<script>
// JavaScript code for the above approach.
 
// Initializing hashmap to store
// Frequency of elements
let freq = new Map();
 
// Function to store frequency of elements in hash map
function storeFrequency(arr,n)
{
    // Iterating the array
    for (let i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
}
 
// Function to print elements
function printElements(a, b)
{
 
    // Traversing the hash map
    for (let [key, val] of freq.values()) {
 
        // Checking 1st condition if frequency > 1, i.e,
        // Element is repetitive
 
        if (val > 1) {
 
            let value =key;
 
            // Checking 2nd condition if element
            // Is in range of a and b or not
            if (value >= a && value <= b) {
                // Printing the value
                 document.write(value + " ");
            }
        }
    }
}
 
// Function to find the elements
// satisfying given condition for each query
function findElements(arr, N, Q, queries)
{
    storeFrequency(arr, N);
    for (let i = 0; i < Q; i++) {
        let A = queries[i][0];
        let B = queries[i][1];
        printElements(A, B);
        document.write("<br>");
    }
}
 
// Driver code
 
    let arr = [ 1, 5, 1, 2, 3, 3, 4, 0, 0 ];
 
    // Size of array
    let N = arr.length;
    let Q = 2;
    let queries = [ [ 1, 3 ], [ 0, 0 ] ];
 
    // Function call
    findElements(arr, N, Q, queries);
     
    // This code is contributed by satwik4409.
    </script>

3 1 
0 

Complejidad temporal: O(Q*N + N)
Espacio auxiliar: O(N)