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.
- Start travsering array from left and maintain two pointers.
- One pointer (lets say i) is used to iterate the array.
- 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
// 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
// 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
# 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#
// 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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