CareerCup

static String findLargestRepeatedSubString(String s){
		if(s == null)
			throw new IllegalArgumentException("Provided String is null");
		if(s.length() == 1){
			return null;
		}
		String result = null;
		for(int start = 0; start < s.length(); start++){
			for(int end = start+1; end <= s.length(); end++) {
				String subString = s.substring(start, end);
				if(s.substring(end).contains(subString)){
					if(result == null)
						result = s.substring(start, end);
					else if(subString.length() > result.length()){
						result = subString;
					}
				}
			}
		}
		return result;
	}

static String findLargestRepeatedSubString(String s){
		if(s == null)
			throw new IllegalArgumentException("Provided String is null");
		if(s.length() == 1){
			return null;
		}
		String result = null;
		for(int start = 0; start < s.length(); start++){
			for(int end = start+1; end <= s.length(); end++) {
				String subString = s.substring(start, end);
				if(s.substring(end).contains(subString)){
					if(result == null)
						result = s.substring(start, end);
					else if(subString.length() > result.length()){
						result = subString;
					}
				}
			}
		}
		return result;
	}

string FindLongestRepSubstring(string s)
{
        if(s == null)
	   return throw new ArgumentException("s");
	HashTable<string, int> h = new HashTable<string, h>();
	string longRepStr = null;	
	for(int i=0;i<s.Length; i++)
	{
		for(int = i+1; j<s.Length; j++)
			{
			    string nextSubStr = s.substring(i, j);
			    if(h.ContainsKey(nextSubStr))
				{
				    h[nextSubStr].Value++;
				    if(longRepStr == null)
				    {
					longRepStr = nextSubStr;
				    }
				    else if(longRepStr.Length < nextSubStr.Length) //if(h[longRepStr].Value < h[nextSubStr].Value)
				    {					
					longRepStr = nextSubStr;
				    }
				}
			    else
				{
				   h[nextSubStr].Value = 1;
				}
			   }
       }
       return longRepStr;
}

Above codes time Complexity is n^3. We can do this n^2 using dynamic programming.

{
// Time - O(n^2), Space - O(n^2)

static String findLongestRepeatedSubString(String s) {
if(s==null || s.trim().length()==0)
return "";
int len = s.length();
int[][] matrix = new int[len][len];
int maxLen = 0;
int maxIndex = 0;
for(int i=0; i< len; i++) {
for(int j=i+1; j < len; j++) {
if(s.charAt(i)==s.charAt(j)) {
matrix[i][j] = ((i-1<0 || j-1<0)?0:matrix[i-1][j-1]) +1;
if(matrix[i][j]>maxLen) {
maxLen = matrix[i][j];
maxIndex = j;
}
}
}
}

return s.substring(maxIndex-maxLen+1,maxIndex+1);
}
}

use a suffix tree. Find the deepest node having more than one child.