Skip to content
Related Articles

Related Articles

Find the maximum length of the prefix
  • Last Updated : 13 Apr, 2021

Given an array arr[] of N integers where all elements of the array are from the range [0, 9] i.e. a single digit, the task is to find the maximum length of the prefix of this array such that removing exactly one element from the prefix will make the occurrence of the remaining elements in the prefix same.
Examples: 

Input: arr[] = {1, 1, 1, 2, 2, 2} 
Output:
Required prefix is {1, 1, 1, 2, 2} 
After removing 1, every element will have equal frequency i.e. {1, 1, 2, 2}

Input: arr[] = {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5} 
Output: 13

Input: arr[] = {10, 2, 5, 4, 1} 
Output:

Approach: Iterate over all the prefixes and check for each prefix if we can remove an element so that each element has same occurrence. In order to satisfy this condition, one of the following conditions must hold true: 



  • There is only one element in the prefix.
  • All the elements in the prefix have the occurrence of 1.
  • Every element has the same occurrence, except for exactly one element which has occurrence of 1.
  • Every element has the same occurrence, except for exactly one element which has the occurrence exactly 1 more than any other elements.

Below is the implementation of the above approach:

C++14




// 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




// 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




# 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#




// 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

Javascript




<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>

 
 

Output: 
5

 

 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :