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Find any one of the multiple repeating elements in read only array

Given a read only array of size ( n+1 ), find one of the multiple repeating elements in the array where the array contains integers only between 1 and n.
Read only array means that the contents of the array can’t be modified.

Examples:

Input : n = 5
        arr[] = {1, 1, 2, 3, 5, 4}
Output : One of the numbers repeated in the array is: 1

Input : n = 10
        arr[] = {10, 1, 2, 3, 5, 4, 9, 8, 5, 6, 4}
Output : One of the numbers repeated in the array is: 4 OR 5



Since, the size of the array is n+1 and elements ranges from 1 to n then it is confirmed that there will be at least one repeating element.

A simple solution is to create a count array and store counts of all elements. As soon as we encounter an element with count more than 1, we return it. This solution works in O(n) time and requires O(n) extra space.

A space optimized solution is to break the given range (from 1 to n) into blocks of size equal to sqrt(n). We maintain the count of elements belonging to each block for every block. Now as the size of array is (n+1) and blocks are of size sqrt(n), then there will be one such block whose size will be more than sqrt(n). For the block whose count is greater than sqrt(n), we can use hashing for the elements of this block to find which element appears more than once.

Explanation:
The method described above works because of the following two reasons:

  1. There would always be a block which has count greater than sqrt(n) because of one extra element. Even when one extra element has been added it will occupy a position in one of the blocks only, making that block to be selected.
  2. The selected block definitely has a repeating element. Consider that ith block is selected. Size of the block is greater than sqrt(n) (Hence, it is selected) Maximum distinct elements in this block = sqrt(n). Thus, size can be greater than sqrt(n) only if there is a repeating element in range ( i*sqrt(n), (i+1)*sqrt(n) ].

Note: The last block formed may or may not have range equal to sqrt(n). Thus, checking if this block has a repeating element will be different than other blocks. However, this difficulty can be overcome from implementation point of view by initialising the selected block with the last block. This is safe because at least one block has to get selected.

Below is the step by step algorithm to solve this problem:

  1. Divide the array in blocks of size sqrt(n).
  2. Make a count array which stores the count of element for each block.
  3. Pick up the block which has count more than sqrt(n), setting the last block
    as default.
  4. For the elements belonging to the selected block, use the method of hashing(explained in next step) to find the repeating element in that block.
  5. We can create a hash array of key value pair, where key is the element in the block and value is the count of number of times the given key is appearing. This can be easily implemented using unordered_map in C++ STL.

Below is the implementation of above idea:

C++

// C++ program to find one of the repeating
// elements in a read only array
#include <iostream>
#include <cmath>
#include <unordered_map>
using namespace std;
  
// Function to find one of the repeating
// elements
int findRepeatingNumber(const int arr[], int n)
{
    // Size of blocks except the
    // last block is sq
    int sq = sqrt(n);
  
    // Number of blocks to incorporate 1 to
    // n values blocks are numbered from 0
    // to range-1 (both included)
    int range = (n / sq) + 1;
  
    // Count array maintains the count for
    // all blocks
    int count[range] = {0};
  
    // Traversing the read only array and
    // updating count
    for (int i = 0; i <= n; i++)
    {
        // arr[i] belongs to block number
        // (arr[i]-1)/sq i is considered
        // to start from 0
        count[(arr[i] - 1) / sq]++;
    }
  
    // The selected_block is set to last
    // block by default. Rest of the blocks
    // are checked
    int selected_block = range - 1;
    for (int i = 0; i < range - 1; i++)
    {
        if (count[i] > sq)
        {
            selected_block = i;
            break;
        }
    }
  
    // after finding block with size > sq
    // method of hashing is used to find
    // the element repeating in this block
    unordered_map<int, int> m;
    for (int i = 0; i <= n; i++)
    {
        // checks if the element belongs to the
        // selected_block
        if ( ((selected_block * sq) < arr[i]) &&
                (arr[i] <= ((selected_block + 1) * sq)))
        {
            m[arr[i]]++;
  
            // repeating element found
            if (m[arr[i]] > 1)
                return arr[i];
        }
    }
  
    // return -1 if no repeating element exists
    return -1;
}
  
// Driver Program
int main()
{
    // read only array, not to be modified
    const int arr[] = { 1, 1, 2, 3, 5, 4 };
  
    // array of size 6(n + 1) having
    // elements between 1 and 5
    int n = 5;
  
    cout << "One of the numbers repeated in"
         " the array is: "
         << findRepeatingNumber(arr, n) << endl;
}

Java

// Java program to find one of the repeating 
// elements in a read only array 
import java.io.*;
import java.util.*;
  
class GFG
{
  
    // Function to find one of the repeating 
    // elements 
    static int findRepeatingNumber(int[] arr, int n)
    {
        // Size of blocks except the 
        // last block is sq 
        int sq = (int) Math.sqrt(n);
  
        // Number of blocks to incorporate 1 to 
        // n values blocks are numbered from 0 
        // to range-1 (both included) 
        int range = (n / sq) + 1;
  
        // Count array maintains the count for 
        // all blocks 
        int[] count = new int[range];
  
        // Traversing the read only array and 
        // updating count 
        for (int i = 0; i <= n; i++)
        {
            // arr[i] belongs to block number 
            // (arr[i]-1)/sq i is considered 
            // to start from 0 
            count[(arr[i] - 1) / sq]++;
        }
  
        // The selected_block is set to last 
        // block by default. Rest of the blocks 
        // are checked 
        int selected_block = range - 1;
        for (int i = 0; i < range - 1; i++) 
        
            if (count[i] > sq) 
            
                selected_block = i; 
                break
            
        }
  
        // after finding block with size > sq 
        // method of hashing is used to find 
        // the element repeating in this block 
        HashMap<Integer, Integer> m = new HashMap<>();
        for (int i = 0; i <= n; i++)
        {
            // checks if the element belongs to the 
            // selected_block 
            if ( ((selected_block * sq) < arr[i]) && 
                    (arr[i] <= ((selected_block + 1) * sq))) 
            
                m.put(arr[i], 1);
  
                // repeating element found
                if (m.get(arr[i]) == 1
                    return arr[i]; 
            
        }
  
        // return -1 if no repeating element exists 
        return -1
}
  
// Driver code
public static void main(String args[])
{
    // read only array, not to be modified 
    int[] arr = { 1, 1, 2, 3, 5, 4 };
  
    // array of size 6(n + 1) having 
    // elements between 1 and 5 
    int n = 5;
  
    System.out.println("One of the numbers repeated in the array is: "
                                    findRepeatingNumber(arr, n));
}
}
  
// This code is contributed by rachana soma

Python3

# Python 3program to find one of the repeating
# elements in a read only array
from math import sqrt
  
# Function to find one of the repeating
# elements
def findRepeatingNumber(arr, n):
      
    # Size of blocks except the
    # last block is sq
    sq = sqrt(n)
  
    # Number of blocks to incorporate 1 to
    # n values blocks are numbered from 0
    # to range-1 (both included)
    range__= int((n / sq) + 1)
  
    # Count array maintains the count for
    # all blocks
    count = [0 for i in range(range__)] 
  
    # Traversing the read only array and
    # updating count
    for i in range(0, n + 1, 1):
          
        # arr[i] belongs to block number
        # (arr[i]-1)/sq i is considered
        # to start from 0
        count[int((arr[i] - 1) / sq)] += 1
  
    # The selected_block is set to last
    # block by default. Rest of the blocks
    # are checked
    selected_block = range__ - 1
    for i in range(0, range__ - 1, 1):
        if (count[i] > sq):
            selected_block = i
            break
          
    # after finding block with size > sq
    # method of hashing is used to find
    # the element repeating in this block
    m = {i:0 for i in range(n)}
    for i in range(0, n + 1, 1):
          
        # checks if the element belongs 
        # to the selected_block
        if (((selected_block * sq) < arr[i]) and 
             (arr[i] <= ((selected_block + 1) * sq))):
            m[arr[i]] += 1
  
            # repeating element found
            if (m[arr[i]] > 1):
                return arr[i]
  
    # return -1 if no repeating element exists
    return -1
  
# Driver Code
if __name__ == '__main__':
      
    # read only array, not to be modified
    arr = [1, 1, 2, 3, 5, 4]
  
    # array of size 6(n + 1) having
    # elements between 1 and 5
    n = 5
  
    print("One of the numbers repeated in the array is:",
                             findRepeatingNumber(arr, n))
      
# This code is contributed by
# Sahil_shelangia


Output:

One of the numbers repeated in the array is: 1

Time Complexity: O(N)
Auxiliary Space: sqrt(N)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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