Reduce the array such that each element appears at most 2 times

Given a sorted array arr of size N, the task is to reduce the array such that each element can appear at most two times.

Examples:

Input: arr[] = {1, 2, 2, 2, 3}
Output: {1, 2, 2, 3}
Explanation:
Remove 2 once, as it occurs more than 2 times.

Input: arr[] = {3, 3, 3}
Output: {3, 3}
Explanation:
Remove 3 once, as it occurs more than 2 times.

Approach: This can be solved with the help of two pointer algorithm.



  1. Start travsering array from left and maintain two pointers.
  2. One pointer (lets say i) is used to iterate the array.
  3. And the second pointer (lets say st) moves forward to find the next unique element, the element at i appears more than twice.

Below is the implementation of the above approach:

CPP

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// C++ program to reduce the array
// such that each element appears
// at most 2 times
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to remove duplicates
void removeDuplicates(int arr[], int n)
{
    // Initalise 2nd pointer
    int st = 0;
  
    // Itereate over the array
    for (int i = 0; i < n; i++) {
  
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
  
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
  
    cout << "{";
    for (int i = 0; i < st; i++) {
        cout << arr[i];
  
        if (i != st - 1)
            cout << ", ";
    }
    cout << "}";
}
  
// Driver code
int main()
{
    int arr[]
        = { 1, 1, 1, 2,
            2, 2, 3, 3,
            3, 3, 3, 3,
            4, 5 };
  
    int n = sizeof(arr)
            / sizeof(arr[0]);
  
    // Function call
    removeDuplicates(arr, n);
  
    return 0;
}

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Java

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// Java program to reduce the array
// such that each element appears
// at most 2 times
class GFG
{
  
// Function to remove duplicates
static void removeDuplicates(int arr[], int n)
{
    // Initalise 2nd pointer
    int st = 0;
  
    // Itereate over the array
    for (int i = 0; i < n; i++) {
  
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
  
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
  
    System.out.print("{");
    for (int i = 0; i < st; i++) {
        System.out.print(arr[i]);
  
        if (i != st - 1)
            System.out.print(", ");
    }
    System.out.print("}");
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 1, 2,
                  2, 2, 3, 3,
                  3, 3, 3, 3,
                  4, 5 };
  
    int n = arr.length;
  
    // Function call
    removeDuplicates(arr, n);
}
}
  
// This code is contributed by sapnasingh4991

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Python3

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# Python3 program to reduce the array 
# such that each element appears 
# at most 2 times 
  
# Function to remove duplicates 
def removeDuplicates(arr, n) : 
  
    # Initalise 2nd pointer 
    st = 0
  
    # Itereate over the array 
    for i in range(n) :
  
        if (i < n - 2 and arr[i] == arr[i + 1
            and arr[i] == arr[i + 2]) :
            continue
  
        # Updating the 2nd pointer 
        else :
            arr[st] = arr[i]; 
            st += 1
  
    print("{",end="") 
    for i in range(st) :
        print(arr[i],end="");
          
        if (i != st - 1) :
            print(", ",end=""); 
      
    print("}",end=""); 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 1, 1, 2
            2, 2, 3, 3
            3, 3, 3, 3
            4, 5 ]; 
  
    n = len(arr);
      
    # Function call 
    removeDuplicates(arr, n); 
  
# This code is contributed by Yash_R

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C#

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// C# program to reduce the array
// such that each element appears
// at most 2 times
using System;
  
class GFG
{
   
// Function to remove duplicates
static void removeDuplicates(int []arr, int n)
{
    // Initalise 2nd pointer
    int st = 0;
   
    // Itereate over the array
    for (int i = 0; i < n; i++) {
   
        if (i < n - 2
            && arr[i] == arr[i + 1]
            && arr[i] == arr[i + 2])
            continue;
   
        // Updating the 2nd pointer
        else {
            arr[st] = arr[i];
            st++;
        }
    }
   
    Console.Write("{");
    for (int i = 0; i < st; i++) {
        Console.Write(arr[i]);
   
        if (i != st - 1)
            Console.Write(", ");
    }
    Console.Write("}");
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 1, 1, 2,
                  2, 2, 3, 3,
                  3, 3, 3, 3,
                  4, 5 };
   
    int n = arr.Length;
   
    // Function call
    removeDuplicates(arr, n);
}
}
  
// This code is contributed by sapnasingh4991

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Output:

{1, 1, 2, 2, 3, 3, 4, 5}

Time complexity: O(N)
Space complexity: O(1)

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Improved By : sapnasingh4991, Yash_R