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Okk As programmer we love get involved in logic building but that is not the case some time we become blank over some type of puzzle as below mentioned. Let me declare that this is not any kind of homework or job stuff it simply a logic and performance practice puzzle.Okk the puzzle of given an Strings` with comma separated words like 

String S= peas,sugar,rice,soup

Now crux is to find out length of longest chain of the words like last character of word should be the first character of next word and so on to create a longest possible chain and finally to calculate the length of that chain.

Now I had tried to figure out some sort of solution like

  1. split the string with comma
  2. add them in list
  3. sort that list etc

but now how to develop further logic As I m little poor over logic development,Help is appreciated and if above half logic is not proper as it should be than what must the simple sort and perfect way to get the length of the longest chain of words.

Summary
input: String S= peas,sugar,rice,soup.
output: 4 length of words (peas->sugar->rice->soup) or (soup->peas->sugar->rice) etc

Bhargav Modi
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4 Answers4

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Once you have list (or array) you can iterate over the array checking your condition (equality of last letter of n-th words with the first letter of first word) and increase counter each time. Once the condition is false just escape the loop. Your counter will hold value you need.

AlexR
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1

okk friends here the logic and core part which I had made and my puzzle got solved

import java.util.Map;
import java.util.Stack;

public class CandidateCode 
{ 
public static int chainLength=0;
public static void main(String[] args) {
    String s= "peas,sugar,rice,soup";
    int chainLengthfinal=wordChain(s);
    System.out.println("final length:"+chainLengthfinal);
}
public static int wordChain(String input1)
{
    List<String> stringList = new ArrayList<String>();
    stringList= Arrays.asList(input1.split(","));
    boolean ischain = new CandidateCode().hasChain(stringList);
    if (ischain) {
     return chainLength;    
    }
    return 0;

}
Map<Character, List<String>> startsWith = new HashMap<Character, List<String>>();
Map<Character, List<String>> endsWith = new HashMap<Character, List<String>>();

private Character getFirstChar(String str) {
    return str.charAt(0);
}

private Character getLastChar(String str) {
    return str.charAt(str.length() - 1);
}

boolean hasChain(List<String> stringList) {
    for (String str : stringList) {
        Character start = getFirstChar(str);
        Character end = getLastChar(str);
        List<String> startsWithList;
        List<String> endsWithList;

        if (startsWith.containsKey(start)) {
            startsWithList = startsWith.get(start);
        } else {
            startsWithList = new ArrayList<String>();
            startsWith.put(start, startsWithList);
        }

        if (endsWith.containsKey(end)) {
            endsWithList = endsWith.get(end);
        } else {
            endsWithList = new ArrayList<String>();
            endsWith.put(end, endsWithList);
        }
        startsWithList.add(str);
        endsWithList.add(str);
    }

    Stack<String> stringStack = new Stack<String>();
    for (String str : stringList) {
        if (hasChain(stringList.size(), str, stringStack)) {
            System.out.println(stringStack);
            System.out.println("size "+stringStack.size());
            chainLength= stringStack.size();
            return true;
        }
    }

    return false;
}

private boolean hasChain(int size, String startString, Stack<String> stringStack) {
    if (size == stringStack.size()) return true;
    Character last = getLastChar(startString);
    if (startsWith.containsKey(last)) {
        List<String> stringList = startsWith.get(last);
        for (int i = 0; i < stringList.size(); i++) {
            String candidate = stringList.remove(i--);
            stringStack.push(candidate);
            if (hasChain(size, candidate, stringStack)) {
                return true;
            }
            stringStack.pop();
            stringList.add(++i, candidate);
        }
    }

    return false;
 }
}

output of the above program will be
[soup, peas, sugar, rice]

size 4.

final length:4.

Bhargav Modi
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0

initialize a " " string named last(String last=" ")

get the first string by splitting with comma

substring the last char of the string and store it to last

boolean brokenchain=false;
length=0;
while(more string to split with comma)&&(!brokenchain){
     split string with comma
     substring to get first char
     if(first char!=last){
        brokenchain=true;
      }else{
         length++;
         get last char of this string with substring and store it to last
      }   
}

if you have for input a sequence of legth 5 and the it brokes and there is a sequence of length 6 following which you want to count and print as output, you have to store the count variable in a map, for example, as a key associated with the sequence as far. then you continue the loop(you have to make the brokenchain=false again) until the input string sequence ends. then you get the bigger key from your map and print it with his associated value(the biggest sequence)

Kwnstantinos Papadas
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0

I think you need to find the largest and smallest number.

  1. split the string with comma

  2. add them as list_item

  3. compare list_item1 and list_item2, the largest value becomes list_item_X

  4. compare list_item3 and list_item4, the largest value becomes list_item_Y Now compare list_item1 and list_item_X, the largest value becomes

  5. So the largest value is list_item_Z, here is implimentation through code.

$s = 'peas,sugar,rice,soup';
$list_items = explode(',', $s);
$lengths = array_map('strlen', $list_items);
echo "The shortest is " . min($lengths) . ". The longest is " . max($lengths);

Omer Farooq
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