CareerCup

I think you should build trie from these strings, so when you go from root to nodes, all strings that would be under current node - will be answers. For better performance you can store all references on all strings each node belongs to. so complexity wiil be O(n), where n - is lenght of input.

You should make use of n-ary tree. store characters of the string as node of the n-ary tree. Now make use of DFS as :
say the input is aa -- now to give the suggestions take 2nd 'a' as main node and apply the DFS on it. Print the result of the DFS .

Please apply Weighted Quick Union Find Algorithm. It will help to distribute the character on the wight. We can decide the way of representation of the string representation and can take the sub tree for the node for which the string matches.

/*This code Needs some modification anybody who got the logic can try it.. and it does work for "a" and "aa", but doesn't work for input "ac" */

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class StringChallenge {
    public static void main(String[] args) {
	// TODO Auto-generated method stub

	String[] input = { "aabb", "aafd", "acff", "aacg", "acfg" };
	List<String> lst = new ArrayList<String>();

	for (String str : input) {
	    lst.add(str);
	}

	System.out.println("enter your option: ");
	Scanner in = new Scanner(System.in);
	String expt = in.nextLine();
	char[] inp = expt.toCharArray();
	
	for (int i = 0; i < inp.length; i++) {
	    while (lst.get(i).contains(expt)) {
		System.out.println(lst.get(i) + " ");
		lst.remove(i);
		continue;
	    }
	}
    }
}

import java.util.Scanner;

public class CheckSubString {

public static void main(String[] args) {

String[] str={"aabd","adbd","bfdbd","abcd","absbd"};
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();

for(int i=0;i<str.length;i++){
System.out.println(str[i].substring(0,input.length()));
if(str[i].substring(0,input.length()).equals(input))
System.out.println(str[i]);
else
continue;
}

}

}

package com.example.itext;

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;




/**
* Hello world!
*
*/
public class App implements IExample
{
public static void main( String[] args )
{
String[] strArr = {"aBB","aaCC","aBc","aaaVBF","bbaa","bAAA","aaaHHHH","d"};
List<String> strArrayList = new ArrayList<String>();
for(String str:strArr){
strArrayList.add(str);
}
Collections.sort(strArrayList);
String inputString = "d";
StringBuilder strb = new StringBuilder("");
for(String listElm: strArrayList){
if(inputString.charAt(0)== listElm.charAt(0)){
if(listElm.startsWith(inputString))
strb.append(listElm).append(" ");
}else{
continue;
}

}

System.out.println("strb: " + strb);

}

}

public static void findStrings(String[] sa, String str)
	{
		if(sa==null)
			return;
		else
		{
			for(int i=0;i<sa.length;i++)
			{
				if(sa[i].startsWith(str))
					System.out.println(sa[i]);
			}
		}

}

/*here is the code in C*/

#include<stdio.h>
#include<string.h>
int main()
{
int i;
char strin[][10] = {"abc","aab","aacf","gffg","saas","dddd","lkll","dfgfd","dfgdete","aafgdgr"};
char str[5];
printf("Enter the string to be searched\n");
scanf("%s",str);
for(i = 0;i<10;i++){
if(strncmp(strin[i],str,strlen(str)) == 0)
printf("%s\n",strin[i]);
}
return 0;
}