CareerCup

static void printWordsWithAdjacentDupChars(){
		
		String[] wl = {"app","By","Hello","Cat","Summer","way"};
		
		ArrayList<String> al = new ArrayList<String>();
		
		for (String word: wl){
			
			for (int i=0; i < word.length()-1;i++){
				
				if (word.charAt(i)== word.charAt(i+1)){
					al.add(word);
					break;
				}
				
			}
			
		}
		
		System.out.print(al);
		
	}

public class FindConcecutiveCharacter {

	public static String[] findConcecutive(String[] arr) {

		String temp = null;
		
		int len=arr.length;
		
		String[] finalArray = new String[len];
		for(int i=0;i<len;i++){
			char[] temparr=arr[i].toCharArray();
			for(int j=0;j<temparr.length-1;j++){
				if(temparr[j]==temparr[j+1]){
					finalArray[i]=arr[i];
				}
					
			}
		}
		
		return finalArray;
	}
	
	public static void main(String[] args) {
		String [] array={"hello","robot","summer","elephant"};
		String [] array1=findConcecutive(array);
		for(int i=0;i<array1.length;i++){
			System.out.println(array1[i]);
		}
	}
}

public class AmdocsInterviewQuestionDeveloperProgramEngineers {

    public static void main(String[] args) {
        String[] wordArray = {"hello", "robot", "summer", "elephant"};
        System.out.println(Arrays.toString(repeatChars(wordArray)));
    }

    private static String[] repeatChars(String[] wordArray) {
        List<String> result = new LinkedList<String>();
        //
        for (String word : wordArray) {
            for (int i = 1; i < word.length(); i++) {
                if (word.charAt(i - 1) == word.charAt(i)) {
                    result.add(word);
                    break;
                }
            }
        }
        //
        return result.toArray(new String[result.size()]);
    }
}

Usage:

python consecutivechars.py hello,robot,summer,elephant
hello
summer

Script:

#!/usr/bin/env python


import sys

def has_consecutive(word, prev_ch):
    if not word:
        return False
    elif word[0] == prev_ch:
        return True
    else:
        return has_consecutive(word[1:], word[0])

if __name__ == '__main__':
    if len(sys.argv) < 2:
        sys.exit(2)
    words = [word.strip() for word in sys.argv[1].split(',') if word.strip()]
    print '\n'.join([word for word in words if has_consecutive(word, '')])

Could be optimized in C easily for efficiency, to have an array of pointers of all word beginnings (NULL truncate each, in place), and recurse on &word[1] and word[0] in has_consecutive instead of using the list operations in Python.

/*
        Problem: Return an array of strings that have repeated characters together
     */
    public static String[] findConsecutiveCharacters(String[] arr)
    {
        ArrayList<String> arrayList = new ArrayList<>();
        for (int i = 0; i < arr.length; i++)
        {
            String s = arr[i];
            for (int j = 0; j < s.length() - 1; j++) {
                if (s.charAt(j + 1) == s.charAt(j)) {
                     arrayList.add(arr[i]);
                }
            }
        }
        return arrayList.toArray(new String[arrayList.size()]);
    }

Similar solution to another poster here but I used ArrayList append so there aren't so many nulls in the final result array. This solution has a similar algorithm to what appears in Cracking the Coding Interview on Problem 1.5 Encode String.

public ArrayList<String> getRepeat(ArrayList<String> str){
int j;
ArrayList<String> atr = new ArrayList<String>();
for(String i : str){
for(j=1;j<i.length();j++){
if(i.charAt(j)==i.charAt(j-1))
{
atr.add(i);
}
}
}

return atr;
}

def repeatChar(words):
    duplicate_string = []
    for string in words:
        strings = list(string)
        for i in range(0,len(string) - 1):
            if strings[i] == strings[i + 1]:
                duplicate_string.append(string)
    return duplicate_string