Encuentre la longitud máxima del prefijo – Part 1

Dada una array arr[] de N enteros donde todos los elementos de la array pertenecen al rango [0, 9], es decir, un solo dígito, la tarea es encontrar la longitud máxima del prefijo de esta array de modo que se elimine exactamente un elemento de el prefijo hará que la ocurrencia de los elementos restantes en el prefijo sea la misma.
Ejemplos: 

Entrada: arr[] = {1, 1, 1, 2, 2, 2} 
Salida: 5  El
prefijo requerido es {1, 1, 1, 2, 2} 
Después de eliminar 1, cada elemento tendrá la misma frecuencia, es decir, {1, 1, 2, 2}

Entrada: arr[] = {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5} 
Salida: 13

Entrada: arr[] = {10, 2, 5, 4, 1} 
Salida: 5 

Enfoque: iterar sobre todos los prefijos y verificar cada prefijo si podemos eliminar un elemento para que cada elemento tenga la misma ocurrencia. Para satisfacer esta condición, una de las siguientes condiciones debe cumplirse: 

• Solo hay un elemento en el prefijo.
• Todos los elementos del prefijo tienen la ocurrencia de 1.
• Cada elemento tiene la misma ocurrencia, excepto exactamente un elemento que tiene la ocurrencia de 1.
• Cada elemento tiene la misma ocurrencia, excepto exactamente un elemento que tiene exactamente 1 ocurrencia más que cualquier otro elemento.

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum
// length of the required prefix
int Maximum_Length(vector<int> a)
{
     
    // Array to store the frequency
    // of each element of the array
    int counts[11] = {0};
 
    // Iterating for all the elements
    int ans = 0;
    for(int index = 0;
            index < a.size();
            index++)
    {
         
        // Update the frequency of the
        // current element i.e. v
        counts[a[index]] += 1;
 
        // Sorted positive values
        // from counts array
        vector<int> k;
        for(auto i : counts)
            if (i != 0)
                k.push_back(i);
 
        sort(k.begin(), k.end());
 
        // If current prefix satisfies
        // the given conditions
        if (k.size() == 1 ||
           (k[0] == k[k.size() - 2] &&
            k.back() - k[k.size() - 2] == 1) ||
           (k[0] == 1 and k[1] == k.back()))
            ans = index;
    }
     
    // Return the maximum length
    return ans + 1;
}
 
// Driver code
int main()
{
    vector<int> a = { 1, 1, 1, 2, 2, 2 };
 
    cout << (Maximum_Length(a));
}
 
// This code is contributed by grand_master

// Java implementation of the approach
import java.util.*;
public class Main
{
    // Function to return the maximum
    // length of the required prefix
    public static int Maximum_Length(Vector<Integer> a)
    {
           
        // Array to store the frequency
        // of each element of the array
        int[] counts = new int[11];
       
        // Iterating for all the elements
        int ans = 0;
        for(int index = 0;
                index < a.size();
                index++)
        {
               
            // Update the frequency of the
            // current element i.e. v
            counts[a.get(index)] += 1;
       
            // Sorted positive values
            // from counts array
            Vector<Integer> k = new Vector<Integer>();
            for(int i : counts)
                if (i != 0)
                    k.add(i);
       
            Collections.sort(k); 
       
            // If current prefix satisfies
            // the given conditions
            if (k.size() == 1 ||
               (k.get(0) == k.get(k.size() - 2) &&
                k.get(k.size() - 1) - k.get(k.size() - 2) == 1) ||
               (k.get(0) == 1 && k.get(1) == k.get(k.size() - 1)))
                ans = index;
        }
           
        // Return the maximum length
        return ans + 1;
    }
     
    // Driver code
    public static void main(String[] args) {
        Vector<Integer> a = new Vector<Integer>();
        a.add(1);
        a.add(1);
        a.add(1);
        a.add(2);
        a.add(2);
        a.add(2);
        
        System.out.println(Maximum_Length(a));
    }
}
 
// This code is contributed by divyeshrabadiya07

# Python3 implementation of the approach
 
# Function to return the maximum
# length of the required prefix
def Maximum_Length(a):
 
    # Array to store the frequency
    # of each element of the array
    counts =[0]*11
 
    # Iterating for all the elements
    for index, v in enumerate(a):
 
        # Update the frequency of the
        # current element i.e. v
        counts[v] += 1
 
        # Sorted positive values from counts array
        k = sorted([i for i in counts if i])
 
        # If current prefix satisfies
        # the given conditions
        if len(k)== 1 or (k[0]== k[-2] and k[-1]-k[-2]== 1) or (k[0]== 1 and k[1]== k[-1]):
            ans = index
 
    # Return the maximum length
    return ans + 1
 
# Driver code
if __name__=="__main__":
    a = [1, 1, 1, 2, 2, 2]
    n = len(a)
    print(Maximum_Length(a))

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to return the maximum
    // length of the required prefix
    static int Maximum_Length(List<int> a)
    {
          
        // Array to store the frequency
        // of each element of the array
        int[] counts = new int[11];
      
        // Iterating for all the elements
        int ans = 0;
        for(int index = 0;
                index < a.Count;
                index++)
        {
              
            // Update the frequency of the
            // current element i.e. v
            counts[a[index]] += 1;
      
            // Sorted positive values
            // from counts array
            List<int> k = new List<int>();
            foreach(int i in counts)
                if (i != 0)
                    k.Add(i);
      
            k.Sort();
      
            // If current prefix satisfies
            // the given conditions
            if (k.Count == 1 ||
               (k[0] == k[k.Count - 2] &&
                k[k.Count - 1] - k[k.Count - 2] == 1) ||
               (k[0] == 1 && k[1] == k[k.Count - 1]))
                ans = index;
        }
          
        // Return the maximum length
        return ans + 1;
    }
 
  static void Main() {
    List<int> a = new List<int>(new int[]{ 1, 1, 1, 2, 2, 2 });
    Console.Write(Maximum_Length(a));
  }
}
 
// This code is contributed by divyesh072019

<script>
    // Javascript implementation of the approach
     
    // Function to return the maximum
    // length of the required prefix
    function Maximum_Length(a)
    {
           
        // Array to store the frequency
        // of each element of the array
        let counts = new Array(11);
        counts.fill(0);
       
        // Iterating for all the elements
        let ans = 0;
        for(let index = 0; index < a.length; index++)
        {
               
            // Update the frequency of the
            // current element i.e. v
            counts[a[index]] += 1;
       
            // Sorted positive values
            // from counts array
            let k = [];
            for(let i = 0; i < counts.length; i++)
            {
                if (counts[i] != 0)
                {
                    k.push(i);
                }
            }
       
            k.sort(function(a, b){return a - b});
       
            // If current prefix satisfies
            // the given conditions
            if (k.length == 1 ||
               (k[0] == k[k.length - 2] &&
                k[k.length - 1] - k[k.length - 2] == 1) ||
               (k[0] == 1 && k[1] == k[k.length - 1]))
                ans = index;
        }
           
        // Return the maximum length
        return (ans);
    }
       
    let a = [ 1, 1, 1, 2, 2, 2 ];
    document.write(Maximum_Length(a));
        
       // This code is contributed by suresh07.
</script>

5

Complejidad temporal: O(aloga * a) donde a es la longitud del arreglo

Espacio auxiliar: O(a) donde a es la longitud del arreglo