Find max non-repeatable value in the array with a

问题:

How we could find max non-repeatable value in the array with a time ~O(N)

I was trying to do it during huge time, but I found only ~O(2N):

int solution(int[] A) {

    List<Integer> arr = new ArrayList<Integer>();
    Set<Integer> set = new HashSet<Integer>();

    int max = 0;

    for(int i = 0; i < A.length; i++) {

        if(!set.add(A[i])) arr.add(A[i]);

    }

    for(int i = 0; i < A.length; i++) {
        if (max < A[i] && !arr.contains(A[i])) max = A[i];
    }

    return max;

}

Could we do it a little bit faster?!

回答1:

    int A[] = {5, 5, 3, 2, 3, 1};
    Map<Integer, Integer> map = new HashMap<>();

    for(int i : A) {
        Integer count = map.get(i);
        // according to the comments
        // map.merge(i, 1, Integer::sum)
        map.put(i, count == null ? 1 : count + 1);
    }

    int max = 0;
    for (Map.Entry<Integer, Integer> e : map.entrySet()) {
        if (e.getValue() == 1 && max < e.getKey()) {
            max = e.getKey();
        }
    }

    System.out.println(max);

Here you map each number to a number of times it present in the array.

In this case complexity is O(n).

And just out of integer possible the fastest implementation possible

// this code for academic purpose only
// it can work only with integers less than
// 2^nextPowerOfTwo(array.lenght) as
// hash collision doesn't resolved
public static int nextPowerOfTwo(int value) {
    value--;
    value |= value >> 1;
    value |= value >> 2;
    value |= value >> 4;
    value |= value >> 8;
    value |= value >> 16;

    return ++value;
}

public static int findMaxUnique(int[] array) {
    final int hashSize = nextPowerOfTwo(array.length);
    final int[] hashArray = new int[hashSize];

    for (int n : array) {
        int hash = n ^ (n >>> 16);
        hash &= hashSize - 1;
        hashArray[hash]++;
    }

    int max = 0;
    for (int n : array) {
        int hash = n ^ (n >>> 16);
        hash &= hashSize - 1;
        if (hashArray[hash] == 1 && max < n) {
            max = n;
        }
    }

    return max;
}

public static void main(String[] args) {
    int[] array = {5, 4, 5, 3, 1, 5, 4, 0};
    System.out.println(findMaxUnique(array));
}

回答2:

The idea is to count the number of occurrence, and store it in a treeMap. since treeMap is sorted, we can find the last (biggest) number that is not repeated (value of 1)

//tree map is sorted... so the idea is to find the last item with value = 1
TreeMap<Integer,Integer> occurrence = new TreeMap<>();

//tabulate the number of occurrence of each key 
//O(2N)
Stream.of(array).forEach( key ->
    occurance.compute(key, (k, v) -> v == null ? 1 : v + 1));

//now that we have the treeMap with the number of occurrence, 
//we can find the last non-repeated value.


while(!occurance.isEmpty()){
  //iterate from the back up
  Map.Entry<Integer,Integer> le = occurance.pollLastEntry(); //or var if JAVA10
  if (le.getValue==1) return le.getKey(); 

}

//no unique numbers found
return Integer.MIN_VALUE;

回答3:

You could probably sort your array in descending order using the Radix sort and then loop over it and check wheter currentIndex == currentIndex + 1.

This will give you a maximum complexity of O(n*x) which is basically the same as O(n) if I'm not mistaking and still better than the actual O(n^2) you're using