Segunda palabra más repetida en una secuencia – Part 1

Dada una secuencia de strings, la tarea es encontrar la segunda string más repetida (o frecuente) en la secuencia dada. (Teniendo en cuenta que no hay dos palabras que sean las segundas más repetidas, siempre habrá una sola palabra).

Ejemplos: 

Input : {"aaa", "bbb", "ccc", "bbb", 
         "aaa", "aaa"}
Output : bbb

Input : {"geeks", "for", "geeks", "for", 
          "geeks", "aaa"}
Output : for

Preguntado en: Amazon

Implementación:

// C++ program to find out the second
// most repeated word
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the word
string secMostRepeated(vector<string> seq)
{
 
    // Store all the words with its occurrence
    unordered_map<string, int> occ;
    for (int i = 0; i < seq.size(); i++)
        occ[seq[i]]++;
 
    // find the second largest occurrence
    int first_max = INT_MIN, sec_max = INT_MIN;
    for (auto it = occ.begin(); it != occ.end(); it++) {
        if (it->second > first_max) {
            sec_max = first_max;
            first_max = it->second;
        }
 
        else if (it->second > sec_max &&
                 it->second != first_max)
            sec_max = it->second;
    }
 
    // Return string with occurrence equals
    // to sec_max
    for (auto it = occ.begin(); it != occ.end(); it++)
        if (it->second == sec_max)
            return it->first;
}
 
// Driver program
int main()
{
    vector<string> seq = { "ccc", "aaa", "ccc",
                          "ddd", "aaa", "aaa" };
    cout << secMostRepeated(seq);
    return 0;
}

// Java program to find out the second
// most repeated word
 
import java.util.*;
 
class GFG
{
    // Method to find the word
    static String secMostRepeated(Vector<String> seq)
    {
        // Store all the words with its occurrence
        HashMap<String, Integer> occ = new HashMap<String,Integer>(seq.size()){
            @Override
            public Integer get(Object key) {
                 return containsKey(key) ? super.get(key) : 0;
            }
        };
        
        for (int i = 0; i < seq.size(); i++)
            occ.put(seq.get(i), occ.get(seq.get(i))+1);
      
        // find the second largest occurrence
       int first_max = Integer.MIN_VALUE, sec_max = Integer.MIN_VALUE;
         
       Iterator<Map.Entry<String, Integer>> itr = occ.entrySet().iterator();
       while (itr.hasNext())
       {
           Map.Entry<String, Integer> entry = itr.next();
           int v = entry.getValue();
           if( v > first_max) {
                sec_max = first_max;
                first_max = v;
            }
      
            else if (v > sec_max &&
                     v != first_max)
                sec_max = v;
       }
        
       // Return string with occurrence equals
        // to sec_max
       itr = occ.entrySet().iterator();
       while (itr.hasNext())
       {
           Map.Entry<String, Integer> entry = itr.next();
           int v = entry.getValue();
           if (v == sec_max)
                return entry.getKey();
       }
        
       return null;
    }
     
    // Driver method
    public static void main(String[] args)
    {
        String arr[] = { "ccc", "aaa", "ccc",
                         "ddd", "aaa", "aaa" };
        List<String> seq =  Arrays.asList(arr);
         
        System.out.println(secMostRepeated(new Vector<>(seq)));
    }   
}
// This program is contributed by Gaurav Miglani

# Python3 program to find out the second
# most repeated word
 
# Function to find the word
def secMostRepeated(seq):
     
    # Store all the words with its occurrence
    occ = {}
    for i in range(len(seq)):
        occ[seq[i]] = occ.get(seq[i], 0) + 1
 
    # Find the second largest occurrence
    first_max = -10**8
    sec_max = -10**8
 
    for it in occ:
        if (occ[it] > first_max):
            sec_max = first_max
            first_max = occ[it]
             
        elif (occ[it] > sec_max and
              occ[it] != first_max):
            sec_max = occ[it]
 
    # Return with occurrence equals
    # to sec_max
    for it in occ:
        if (occ[it] == sec_max):
            return it
 
# Driver code
if __name__ == '__main__':
     
    seq = [ "ccc", "aaa", "ccc",
            "ddd", "aaa", "aaa" ]
    print(secMostRepeated(seq))
 
# This code is contributed by mohit kumar 29

// C# program to find out the second
// most repeated word
using System;
using System.Collections.Generic;
 
class GFG
{
    // Method to find the word
    static String secMostRepeated(List<String> seq)
    {
        // Store all the words with its occurrence
        Dictionary<String, int> occ =
        new Dictionary<String, int>();
                 
        for (int i = 0; i < seq.Count; i++)
            if(occ.ContainsKey(seq[i]))
                occ[seq[i]] = occ[seq[i]] + 1;
            else
                occ.Add(seq[i], 1);
     
        // find the second largest occurrence
        int first_max = int.MinValue,
            sec_max = int.MinValue;
         
        foreach(KeyValuePair<String, int> entry in occ)
        {
            int v = entry.Value;
            if( v > first_max)
            {
                sec_max = first_max;
                first_max = v;
            }
         
            else if (v > sec_max &&
                    v != first_max)
                sec_max = v;
        }
         
        // Return string with occurrence equals
        // to sec_max
        foreach(KeyValuePair<String, int> entry in occ)
        {
            int v = entry.Value;
            if (v == sec_max)
                return entry.Key;
        }
             
        return null;
    }
     
    // Driver method
    public static void Main(String[] args)
    {
        String []arr = { "ccc", "aaa", "ccc",
                        "ddd", "aaa", "aaa" };
        List<String> seq = new List<String>(arr);
         
        Console.WriteLine(secMostRepeated(seq));
    }
}
 
// This code is contributed by Rajput-Ji

<script>
 
// JavaScript program to find out the second
// most repeated word
 
// Function to find the word
function secMostRepeated(seq)
{
 
    // Store all the words with its occurrence
    let occ = new Map();
    for (let i = 0; i < seq.length; i++)
    {
        if(occ.has(seq[i])){
            occ.set(seq[i], occ.get(seq[i])+1);
        }
        else occ.set(seq[i], 1);
    }
 
    // find the second largest occurrence
    let first_max = Number.MIN_VALUE, sec_max = Number.MIN_VALUE;
    for (let [key,value] of occ) {
        if (value > first_max) {
            sec_max = first_max;
            first_max = value;
        }
 
        else if (value > sec_max && value != first_max)
            sec_max = value;
    }
 
    // Return string with occurrence equals
    // to sec_max
    for (let [key,value] of occ)
        if (value == sec_max)
            return key;
}
 
// Driver program
let seq = [ "ccc", "aaa", "ccc", "ddd", "aaa", "aaa" ];
document.write(secMostRepeated(seq));
 
// This code is contributed by shinjanpatra
</script>

ccc

Complejidad temporal: O(N), donde N representa el tamaño del vector dado.
Espacio auxiliar: O(N) , donde N representa el tamaño del vector dado.

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