Prefijo más largo en una string con la frecuencia más alta

Dada una string, encuentre un prefijo con la frecuencia más alta. Si dos prefijos tienen la misma frecuencia, considere el que tiene la longitud máxima.
Ejemplos: 
 

Input : str = "abc" 
Output : abc
Each prefix has same frequency(one) and the 
prefix with maximum length is "abc".

Input : str = "abcab"
Output : ab
Both prefix "a" and "ab" occur two times and the 
prefix with maximum length is "ab".

La idea es observar que cada prefijo de la string dada contendrá el primer carácter de la string y el primer carácter solo es también un prefijo de la string dada. Entonces, el prefijo con la ocurrencia más alta es el primer carácter. Ahora queda la tarea de maximizar la longitud del prefijo de frecuencia más alta.
Acercarse : 
 

1. Tome un vector que almacenará los índices del primer elemento de la string.
2. Si el primer elemento apareció solo una vez, el prefijo más largo será la string completa.
3. De lo contrario, haga un bucle hasta la segunda aparición del primer elemento y marque una letra después de cada índice almacenado.
4. Si no hay desajuste, avanzamos, de lo contrario, nos detenemos.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find Longest prefix string with the
// highest frequency
void prefix(string str)
{
    int k = 1, j;
    int n = str.length();
 
    vector<int> g;
    int flag = 0;
 
    // storing all indices where first element is found
    for (int i = 1; i < n; i++) {
        if (str[i] == str[0]) {
            g.push_back(i);
            flag = 1;
        }
    }
 
    // if the first letter in the string does not occur
    // again  then answer will be the whole string
    if (flag == 0) {
        cout << str << endl;
    }
    else {
        int len = g.size();
 
        // loop till second appearance of the first element
        while (k < g[0]) {
 
            int cnt = 0;
            for (j = 0; j < len; j++) {
 
                // check one letter after every stored index
                if (str[g[j] + k] == str[k]) {
                    cnt++;
                }
            }
 
            // If there is no mismatch we move forward
            if (cnt == len) {
                k++;
            }
            // otherwise we stop
            else {
                break;
            }
        }
 
        for (int i = 0; i < k; i++) {
            cout << str[i];
        }
 
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    string str = "abcab";
    prefix(str);
 
    return 0;
}

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    // Function to find Longest prefix string with the
    // highest frequency
    static void prefix(char[] str)
    {
        int k = 1, j;
        int n = str.length;
 
        Vector<Integer> g = new Vector<>();
        int flag = 0;
 
        // storing all indices where first element is found
        for (int i = 1; i < n; i++)
        {
            if (str[i] == str[0])
            {
                g.add(i);
                flag = 1;
            }
        }
 
        // if the first letter in the string does not occur
        // again then answer will be the whole string
        if (flag == 0)
        {
            System.out.println(String.valueOf(str));
        }
        else
        {
            int len = g.size();
 
            // loop till second appearance of the first element
            while (k < g.get(0))
            {
 
                int cnt = 0;
                for (j = 0; j < len; j++)
                {
 
                    // check one letter after every stored index
                    if ((g.get(j) + k) < n &&
                        str[g.get(j) + k] == str[k])
                    {
                        cnt++;
                    }
                }
 
                // If there is no mismatch we move forward
                if (cnt == len)
                {
                    k++;
                }
                // otherwise we stop
                else
                {
                    break;
                }
            }
 
            for (int i = 0; i < k; i++)
            {
                System.out.print(str[i]);
            }
 
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String str = "abcab";
        prefix(str.toCharArray());
    }
}
 
/* This code contributed by PrinciRaj1992 */

# Python3 implementation of the above approach
 
# Function to find Longest prefix string with the
# highest frequency
def prefix(string) :
 
    k = 1;
    n = len(string);
 
    g = [];
    flag = 0;
 
    # storing all indices where first element is found
    for i in range(1, n) :
        if (string[i] == string[0]) :
            g.append(i);
            flag = 1;
 
    # if the first letter in the string does not occur
    # again then answer will be the whole string
    if (flag == 0) :
        print(string);
      
    else :
        length = len(g);
         
        # loop till second appearance of the first element
        while (k < g[0]) :
            cnt = 0;
             
            for j in range(length) :
                 
                # check one letter after every stored index
                if (string[g[j] + k] == string[k]) :
                    cnt += 1;
                     
            # If there is no mismatch we move forward
            if (cnt == len) :
                k += 1;
            
            # otherwise we stop
            else :
                break;
        
        for i in range(k+1) :
            print(string[i],end="");
        
        print()
 
# Driver Code
if __name__ == "__main__" :
 
    string = "abcab";
    prefix(string);
 
# This code is contributed by AnkitRai01

// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find Longest prefix string with the
    // highest frequency
    static void prefix(char[] str)
    {
        int k = 1, j;
        int n = str.Length;
 
        List<int> g = new List<int>();
        int flag = 0;
 
        // storing all indices where first element is found
        for (int i = 1; i < n; i++)
        {
            if (str[i] == str[0])
            {
                g.Add(i);
                flag = 1;
            }
        }
 
        // if the first letter in the string does not occur
        // again then answer will be the whole string
        if (flag == 0)
        {
            Console.WriteLine(String.Join("",str));
        }
        else
        {
            int len = g.Count;
 
            // loop till second appearance of the first element
            while (k < g[0])
            {
 
                int cnt = 0;
                for (j = 0; j < len; j++)
                {
 
                    // check one letter after every stored index
                    if ((g[j] + k) < n &&
                        str[g[j] + k] == str[k])
                    {
                        cnt++;
                    }
                }
 
                // If there is no mismatch we move forward
                if (cnt == len)
                {
                    k++;
                }
                // otherwise we stop
                else
                {
                    break;
                }
            }
 
            for (int i = 0; i < k; i++)
            {
                Console.Write(str[i]);
            }
 
            Console.WriteLine();
        }
    }
 
    // Driver Code
    public static void Main()
    {
        String str = "abcab";
        prefix(str.ToCharArray());
    }
}
 
// This code contributed by Rajput-Ji

<script>
 
// Javascript program for the above approach
 
    // Function to find Longest prefix string with the
    // highest frequency
    function prefix(str)
    {
        let k = 1, j;
        let n = str.length;
 
        let g = [];
        let flag = 0;
 
        // storing all indices where first element is found
        for (let i = 1; i < n; i++)
        {
            if (str[i] == str[0])
            {
            
                g.push(i);
                flag = 1;
            }
        }
 
        // if the first letter in the string does not occur
        // again then answer will be the whole string
        if (flag == 0)
        {
            document.write((str.join("")));
        }
        else
        {
            let len = g.length;
 
            // loop till second appearance of the first element
            while (k < g[0])
            {
 
                let cnt = 0;
                for (j = 0; j < len; j++)
                {
 
                    // check one letter after every stored index
                    if ((g[j] + k) < n &&
                        str[g[j] + k] == str[k])
                    {
                        cnt++;
                    }
                }
 
                // If there is no mismatch we move forward
                if (cnt == len)
                {
                    k++;
                }
                // otherwise we stop
                else
                {
                    break;
                }
            }
 
            for (let i = 0; i < k; i++)
            {
                document.write(str[i]);
            }
 
            document.write("<br/>");
        }
    }
 
 
// Driver Code
 
     let str = "abcab";
     prefix(str);
  
 // This code is co tributed by sanjoy_62.
</script>

ab

Complejidad de tiempo : O(N)