从最后一个数组中删除 Triplet

Question

Given an integer array of nums, remove the triplet from last.
The relative order of elements should be kept the same.

Example 1
input: nums = [2,4,2,2,7,5,6,7,8,6,6,2,6,7,6]
output: nums = [2,4,5,6,8,6]

Example 2 input: nums = [2,2,3,2,3,2]
output: nums = [2,3,3]

I have this in java

int[] nums = {2,4,2,2,7,5,6,7,8,6,6,2,6,7,6};
int[] ans = new int[6];
int count=1; 
for(int i=0;i<nums.length;i++){
   for(int j=0;j<nums.length;j++){
     if(arr[i] == arr[j]){
        if(count < 3){
         count++;
         ans[i] = nums[i];
        }
     }
   }
}

Answer 1

I couldn't let this one go, so I did code up an example in c#. This is one approach. There are probably many ways to do this better:

private static int[] RemoveLastTriplet(int[] input)
{
    var toRemove = new bool[input.Length];
    var counts = new Dictionary<int, int>();
    
    //Count how many times each input value is found.
    foreach(var value in input)
    {
        int count;
        
        if (counts.TryGetValue(value, out count))
        {
            counts[value] = count + 1;
        }
        else
        {
            counts[value] = 1;
        }
    }
    
    foreach(var kvp in counts)
    {
        //Determine how many triplets we have for this value
        var tripletCount = kvp.Value / 3;
        
        //Keep track of where we're starting
        var currentIndex = input.Length - 1;
        
        //Remove each triplet
        for(var tripletIndex = 0; tripletIndex < tripletCount; tripletIndex++)
        {
            //counts the number of elements in this triplet
            var thisTripletCount = 0;
            
            //Mark each member of the triplet for deletion.
            for(var inputIndex = currentIndex; thisTripletCount < 3; inputIndex--)
            {
                if (input[inputIndex] == kvp.Key)
                {
                    //Mark this index for removal
                    toRemove[inputIndex] = true;
                    thisTripletCount++;
                }

                //Keep track of where we are in the overall input array
                currentIndex--;
            }
        }
    }
    
    //We could be more clever here and keep track of how many
    // items we'll have in the output list and just create an array.
    var output = new List<int>();
    
    for(int index = 0; index < input.Length; index++)
    {
        if (!toRemove[index])
        {
            output.Add(input[index]);
        }
    }
    
    return output.ToArray();
}

Answer 2

This is my n^2 solution for this problem. I hope it will help you. Just find the count of element first and then find the total number of elements to be removed. Let say it is 't' after that, just replace the t number of elements from the end with -1.

#include <iostream>
#include <unordered_set>

using namespace std;

int main() {
  int n;
  cout << "Enter the length of array : ";
  cin >> n;
  int arr[n];
  cout << "Enter items : ";
  for (int i = 0; i < n; ++i)
    cin >> arr[i];

  //   set for fast lookups
  unordered_set<int> uset;

  for (int i = n - 1; i > -1; --i) {
    int e = arr[i];
    if (uset.find(e) != uset.end())
      continue;

    // count the element frequiency
    int c = 0;

    for (int j = 0; j < n; ++j) {
      if (arr[j] == e)
        c++;
    }

    // number of elements would we remove from end
    c = c - c % 3;

    for (int j = n - 1; j > -1; --j) {
      if (c == 0)
        break;
      if (arr[j] == e) {
        arr[j] = -1; // put -1 at removed element location
        c--;
      }
    }
  }

  //   now just print only elemt those value is not equals to zero
  for (int i = 0; i < n; ++i) {
    if (arr[i] != (-1))
      cout << arr[i] << " ";
  }
  cout << endl;
  return 0;
}

Answer 3

arr = input().split()[::-1]
# print(arr)
i = 0


def count_same(num, ind):
    count = 0
    indexes = []
    for index in range(ind, len(arr)):
        if arr[index] == num:
            count += 1
            indexes.append(index)
            if count == 3:
                break
    if count == 3:
        for index in indexes[::-1]:
            arr.pop(index)
        return True
    return False


while i < len(arr):
    if not count_same(arr[i], i):
        i += 1
    # print(arr)
print(list(map(int, arr[::-1])))