How to find repeating sequence of Integers in an array of Integers?

Question

How to find repeating sequence of Integers in an array of Integers?

00 would be repeating, so would 123123, but 01234593623 would not be.

I have an idea to h

Answer 1

Try this:

string lookIn = "99123998877665544123"; 
// above has length of 20 (in positions 0 through 19)
int patternLength = 3;
// want to search each triple of letters 0-2, 1-3, 2-4 ... 17-19
//   however since there must be 3 chars after the 3-char pattern
//   we only want to search the triples up to 14-16 (20 - 3*2)
for (int i=0; i <= lookIn.Length - patternLength * 2; i++) {
   string lookingFor = lookIn.Substring(i, patternLength);
   // start looking at the pos after the pattern
   int iFoundPos = lookIn.IndexOf(lookingFor, i + patternLength);
   if (iFoundPos > -1) {
      string msg = "Found pattern '" + lookingFor 
                 + "' at position " + i 
                 + " recurs at position " + iFoundPos;
   }
}
// of course, you will want to validate that patternLength is less than
//   or equal to half the length of lookIn.Length, etc.

EDIT: improved and converted to javascript (from C# ... oops, sorry about that...)

function testfn() {
   var lookIn = "99123998877665544123"; 
   // above has length of 20 (in positions 0 through 19)
   var patternLength_Min = 2;
   var patternLength_Max = 5;
   if (patternLength_Max > (lookIn.length / 2) 
                || patternLength_Max < patternLength_Min
                || patternLength_Min < 1) {
      alert('Invalid lengths.')
   }
   var msg = "";
   for (var pLen = patternLength_Min; pLen <= patternLength_Max; pLen++)  {
      for (var i = 0; i <= lookIn.length - pLen * 2; i++) {
         var lookingFor = lookIn.substring(i, i + pLen);
         // start looking at the pos after the pattern
         var iFoundPos = lookIn.indexOf(lookingFor, i + pLen);
         if (iFoundPos > -1) {
            msg = msg + "Found '" + lookingFor 
                       + "' at pos=" + i 
                       + " recurs at pos=" + iFoundPos + "\n";
            ;
         }
      }
   }
   alert(msg);
}

The message box displays the following:

Found '99' at pos=0 recurs at pos=5
Found '12' at pos=2 recurs at pos=17
Found '23' at pos=3 recurs at pos=18
Found '123' at pos=2 recurs at pos=17

Answer 2

The answer posted by @AdrianLeonhard is perfectly working. But if I have a sequence of 0, 1, 2, 3, 4, 3, 5, 6, 4, 7, 8, 7, 8 many might be wondering on how to get all the repeated numbers from the array.

So, I wrote this simple logic which prints all the repeated numbers with their positions

    int[] arr = {0, 1, 2, 3, 4, 3, 5, 6, 4, 7, 8, 7, 8};

    for(int i=0; i<arr.length;i++){
        for(int j=i+1; j<arr.length;j++){
            if(arr[i] == arr[j]){
                System.out.println("Number: "+arr[i]+" found repeating at position: "+i+" , repeated at position "+j);
            }
        }
    }

Answer 3

You can always play with regular expressions to achieve a desired result. Use the regex backreference and combine it with the greedy quantifier:

    void printRepeating(String arrayOfInt)
    {
        String regex = "(\\d+)\\1";
        Pattern patt = Pattern.compile(regex);
        Matcher matcher = patt.matcher(arrayOfInt);           
        while (matcher.find())                              
        {               
            System.out.println("Repeated substring: " + matcher.group(1));
        } 
    }          

Answer 4

@MiljenMikic answer's is great, especially since the grammar isn't actually regular. :D

If you want to do it on an array in general, or want to understand it, this does pretty much exactly what the regex does:

public static void main(String[] args) {
    int[] arr = {0, 1, 2, 3, 2, 3}; // 2, 3 repeats at position 2.

    // for every position in the array:
    for (int startPos = 0; startPos < arr.length; startPos++) {
        // check if there is a repeating sequence here:

        // check every sequence length which is lower or equal to half the
        // remaining array length: (this is important, otherwise we'll go out of bounds)
        for (int sequenceLength = 1; sequenceLength <= (arr.length - startPos) / 2; sequenceLength++) {

            // check if the sequences of length sequenceLength which start
            // at startPos and (startPos + sequenceLength (the one
            // immediately following it)) are equal:
            boolean sequencesAreEqual = true;
            for (int i = 0; i < sequenceLength; i++) {
                if (arr[startPos + i] != arr[startPos + sequenceLength + i]) {
                    sequencesAreEqual = false;
                    break;
                }
            }
            if (sequencesAreEqual) {
                System.out.println("Found repeating sequence at pos " + startPos);
            }
        }
    }
}