Return value of smallest index of repeated number in array

Question

I have an array of Integer from which I want to return the index value of the smallest number appearence.

Integer[] array = {8,2,10,7,2,10}; 

So, in this case if I look for number 2, i find that 2 is contained in array[1] and array[4]. I'm a little bit lost in managing an array.

What I have done so far is to check if the given number exists

public int exist( Integer number  ) {
        int index = 0;
        int position = 0;
        while (position <= array.length && number == array[position]){
            index = index + 1;
            if( index <= array.length ){
                index = position ;
            } else {
                index = -1;
            }
        }
        return index ;      
}

and I find from Internet this code which finds the first repeating element in an array of integers

class Main {
    // This function prints the first repeating element in arr[]
    static void printFirstRepeating(int arr[]) {
        // Initialize index of first repeating element
        int min = -1;

        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();

        // Traverse the input array from right to left
        for (int i=arr.length-1; i>=0; i--) {
            // If element is already in hash set, update min
            if (set.contains(arr[i]))
                min = i;
            else   // Else add element to hash set
                set.add(arr[i]);
        }
        // Print the result
        if (min != -1)
          System.out.println("The first repeating element is " + arr[min]);
        else
          System.out.println("There are no repeating elements");
    }

    // Driver method to test above method
    public static void main (String[] args) throws java.lang.Exception {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        printFirstRepeating(arr); // output 5
    }
}

I'm still unable to combine this code with the first to get what I really want.

Answer 1

In your coded example 10, 5, 3, 4, 3, 5, 6, the first repeated value is 3 at index 4, a repeat of the 3 at index 2. The other repeated value is 5 at index 5, a repeat of the 5 at index 1.

Illustrated:

10, 5, 3, 4, 3, 5, 6
             ↑        at index 4: 3 is first repeated value
       ↑              at index 2: 3 was first found here
                ↑     at index 5: 5 is second repeated value
    ↑                 at index 1: 5 was first found here

So, possible results, depending on what you really want:

• 4, index of first repeated value
• 2, index of value that is repeated first
• 1, first index of a repeated value

Here are solutions for all three:

private static void printIndexOfFirstRepeated(int ... values) {
    Set<Integer> set = new HashSet<>();
    for (int i = 0; i < values.length; i++)
        if (! set.add(values[i])) { // add() returns false if value already in set
            System.out.println("Index of first repeated: " + i);
            return;
        }
    System.out.println("No repeat found");
}
private static void printIndexOfRepeatedFirst(int ... values) {
    Map<Integer, Integer> mapValueToIndex = new HashMap<>();
    for (int i = 0; i < values.length; i++) {
        Integer prevIndex = mapValueToIndex.put(values[i], i); // put() returns old value, or null
        if (prevIndex != null) {
            System.out.println("Index of repeated first: " + prevIndex);
            return;
        }
    }
    System.out.println("No repeat found");
}
private static void printFirstIndexOfRepeated(int ... values) {
    Integer firstIndex = null;
    Map<Integer, Integer> mapValueToIndex = new HashMap<>();
    for (int i = 0; i < values.length; i++) {
        Integer prevIndex = mapValueToIndex.put(values[i], i); // put() returns old value, or null
        if (prevIndex != null && (firstIndex == null || prevIndex < firstIndex))
            firstIndex = prevIndex;
    }
    if (firstIndex != null)
        System.out.println("First index of repeated: " + firstIndex);
    else
        System.out.println("No repeat found");
}

TEST

printIndexOfFirstRepeated(10, 5, 3, 4, 3, 5, 6);
printIndexOfRepeatedFirst(10, 5, 3, 4, 3, 5, 6);
printFirstIndexOfRepeated(10, 5, 3, 4, 3, 5, 6);

OUTPUT

Index of first repeated: 4
Index of repeated first: 2
First index of repeated: 1

Answer 2

One approach to solving this problem is to use a Map instead of HashSet.

Make a Map<Integer,Integer> that maps a value from the array to the first occurrence of that number. Go through the array, and check the map if the value already exists.

• If the key is not in the map, insert the current index into the map
• If the key is in the map, return the corresponding value from the map.

The search code would look like this:

Map<Integer,Integer> firstAppearance = new HashMap<>();
for (int i = 0 ; i != arr.length ; i++) {
    if (firstAppearance.containsKey(arr[i])) {
        return firstAppearance.get(arr[i]);
    } else {
        firstAppearance.put(arr[i], i);
    }
}