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Print all Repetitive elements in given Array within value range [A, B] for Q queries

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  • Last Updated : 07 Jul, 2022

Given an array arr[] of size N, and Q queries of the form [A, B], the task is to find all the unique elements from the array which are repetitive and their values lie between A and B (both inclusive) for each of the Q queries.

Examples:

Input: arr[] = { 1, 5, 1, 2, 3, 3, 4, 0, 0 }, Q = 2, queries={ { 1, 3 }, { 0, 0 } }
Output: {3, 1}, { 0 }
Explanation: For the first query the elements 1 and 3 lie in the given range and occurs more than once.
For the second queryonly 0 lies in the range and is repetitive in nature.

Input: arr[] = {1, 5, 1, 2, 3, 4, 0, 0}, Q = 1, queries={ { 1, 2 } }
Output: 1

 

Naive Approach: The basic approach to solve this problem is to use nested loops. A traversal of all elements is performed by the outer loop, and check whether the element picked by the outer loop appears anywhere else is performed by the inner loop. Then, if it appears elsewhere, check if it is lying within the range of [A, B]. If yes, then print it.

Time Complexity: O(Q * N2)
Auxiliary Space: O(1)

Efficient Approach:  An efficient approach is based on the idea of a hash map to store the frequency of all the elements. Follow the steps mentioned below:

  • Traverse the hash map, and check if the frequency of an element is greater than 1.
  • If yes, then check if the element is present in the range of [A, B] or not.
  • If yes, then print it else skip the element.
  • Continue the above procedure till all the elements of the hash map are traversed.

Below is the implementation of the above approach.

C++




// 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




// 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


Python3




# 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#




// 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


Javascript




<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>


Output

3 1 
0 

Time Complexity: O(Q*N + N)
Auxiliary Space: O(N)


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