Get indices of n maximums in java array

Question

I have an array of size 1000. How can I find the indices (indexes) of the five maximum elements?

An example with setup code and my attempt are displayed below:

Random rand = new Random();
int[] myArray = new int[1000];
int[] maxIndices = new int[5];
int[] maxValues = new int[5];

for (int i = 0; i < myArray.length; i++) {
  myArray[i] = rand.nextInt();
}

for (int i = 0; i < 5; i++) {
  maxIndices[i] = i;
  maxValues[i] = myArray[i];
}

for (int i = 0; i < maxIndices.length; i++) {
  for (int j = 0; j < myArray.length; j++) {
    if (myArray[j] > maxValues[i]) {
      maxIndices[i] = j;
      maxValues[i] = myArray[j];
    }
  }
}

for (int i = 0; i < maxIndices.length; i++) {
  System.out.println("Index: " + maxIndices[i]);
}

I know the problem is that it is constantly assigning the highest maximum value to all the maximum elements. I am unsure how to remedy this because I have to preserve the values and the indices of myArray.

I don't think sorting is an option because I need to preserve the indices. In fact, it is the indices that I need specifically.

Answer 1

Sorting is an option, at the expense of extra memory. Consider the following algorithm.

1. Allocate additional array and copy into - O(n)
2. Sort additional array - O(n lg n)
3. Lop off the top k elements (in this case 5) - O(n), since k could be up to n
4. Iterate over the original array - O(n)
    4.a search the top k elements for to see if they contain the current element - O(lg n)

So it step 4 is (n * lg n), just like the sort. The entire algorithm is n lg n, and is very simple to code.

Here's a quick and dirty example. There may be bugs in it, and obviously null checking and the like come into play.

import java.util.Arrays;

class ArrayTest {

    public static void main(String[] args) {
        int[] arr = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10};
        int[] indexes = indexesOfTopElements(arr,3);
        for(int i = 0; i < indexes.length; i++) {
            int index = indexes[i];
            System.out.println(index + " " + arr[index]);
        }
    }

    static int[] indexesOfTopElements(int[] orig, int nummax) {
        int[] copy = Arrays.copyOf(orig,orig.length);
        Arrays.sort(copy);
        int[] honey = Arrays.copyOfRange(copy,copy.length - nummax, copy.length);
        int[] result = new int[nummax];
        int resultPos = 0;
        for(int i = 0; i < orig.length; i++) {
            int onTrial = orig[i];
            int index = Arrays.binarySearch(honey,onTrial);
            if(index < 0) continue;
            result[resultPos++] = i;
        }
        return result;
    }

}

There are other things you can do to reduce the overhead of this operation. For example instead of sorting, you could opt to use a queue that just tracks the largest 5. Being ints they values would probably have to be boxed to be added to a collection (unless you rolled your own) which adds to overhead significantly.

Answer 2

Sorry to answer this old question but I am missing an implementation which has all following properties:

• Easy to read
• Performant
• Handling of multiple same values

Therefore I implemented it:

    private int[] getBestKIndices(float[] array, int num) {
        //create sort able array with index and value pair
        IndexValuePair[] pairs = new IndexValuePair[array.length];
        for (int i = 0; i < array.length; i++) {
            pairs[i] = new IndexValuePair(i, array[i]);
        }

        //sort
        Arrays.sort(pairs, new Comparator<IndexValuePair>() {
            public int compare(IndexValuePair o1, IndexValuePair o2) {
                return Float.compare(o2.value, o1.value);
            }
        });

        //extract the indices
        int[] result = new int[num];
        for (int i = 0; i < num; i++) {
            result[i] = pairs[i].index;
        }
        return result;
    }

    private class IndexValuePair {
        private int index;
        private float value;

        public IndexValuePair(int index, float value) {
            this.index = index;
            this.value = value;
        }
    }