What I am trying to achieve is, when there is a given word, I need to find out the second most frequent character within that string.

In additions to that, there is some more options I want to achieve. For example,

1. If a string, "ababababd" is given, there are four 'a's, four 'b's, and one 'd', so in this case, I need to printout 'd'.

2. If a string is "ababababdc" is given, there are four 'a's, four 'b's, one 'd' and 'c', I need to printout 'dc'

   public class MemorizeWorld {
       public static void main(String[] args) {
       Solution solution = new Solution();
       String a = "ababababd";
       solution.solution(a);
    }
   }
   
   class Solution {
        public void solution(String a) {
            int[] array = new int[26];
            char[] char_array = a.toCharArray();
   
            for(int i = 0; i < char_array.length; i  ) {
                int index = char_array[i] - 'a';
                array[index]  ;
            }
            int max = 0;
            for(int i = 0; i < array.length; i  ) {
                if(array[i] > max) {
                    max = array[i];
                }
            }
        }
    }
   

This is what I have done so far and I am stuck on getting a solution. I know how to get the second most frequent character by following the post from geeksforgeeks, but I don't know how I should apply those two options above to that. Is there anyone who can help me with this problem?

CodePudding user response：

Find the second maximum just like you've found the maximum.

int max2 = 0;
for(int i = 0; i < array.length; i  ) {
    if(array[i] > max2 && array[i] < max) {
         max2 = array[i];
     }
}

Then get the result string whose frequency is same as the second maximum.

String result = "";
for(int i = 0; i < array.length; i  ) {
    if(array[i] == max2) {
         result  = (i   'a');
     }
}

CodePudding user response：

// most frequent character in a given string public class Example { static final int NO_OF_CHARS = 256;

static char getSecondMostFreq(String str)
{
    
    int[] count = new int[NO_OF_CHARS];
    int i;
    for (i=0; i< str.length(); i  )
        (count[str.charAt(i)])  ;
  
   
    int first = 0, second = 0;
    for (i = 0; i < NO_OF_CHARS; i  )
    {
       
        if (count[i] > count[first])
        {
            second = first;
            first = i;
        }
  
      
        else if (count[i] > count[second] &&
                 count[i] != count[first])
            second = i;
    }
  
    return (char)second;
}
  

public static void main(String args[])
{
  String str = "ababababd";
  char res = getSecondMostFreq(str);
  if (res != '\0')
     System.out.println("Second most frequent char" 
                                   " is "   res);
  else
     System.out.println("No second most frequent" 
                                   "character");
}

}

CodePudding user response：

Stream API may be used in this solution:

1. create a frequency map using Collectors.groupingBy Collectors.summingInt / Collectors.counting()
2. "invert" the frequency and character using Collectors.toMap or Collectors.groupingBy Collectors.joining to merge characters with the same frequency; use a TreeMap with the reversed key comparator
3. skip the first value referring the max frequency, and return the next one:

public static String get2ndMostFrequent(String str) {
    if (null == str || str.length() < 2) {
        return null;
    }
    return Arrays.stream(str.split("")) // Stream<String> chars
        .collect(Collectors.groupingBy(
            s -> s, LinkedHashMap::new, Collectors.summingInt(s -> 1)
        ))
        .entrySet()
        .stream()
        .collect(Collectors.toMap(
            Map.Entry::getValue,
            Map.Entry::getKey,
            (s1, s2) -> s1   s2,
            () -> new TreeMap<Integer, String>(Comparator.reverseOrder())
        ))
        .values()
        .stream()
        .skip(1)
        .findFirst()
        .orElse(null);
}

Tests:

for (String s : Arrays.asList("ababababd", "ababdababc", "dabbcacd", "acbdagh")) {
    System.out.println(s   " -> "   get2ndMostFrequent(s));
}

Output

ababababd -> d
ababdababc -> dc
dabbcacd -> null  // no 2nd frequency
acbdagh -> cbdgh