Simplest (though not most efficient) way will be to sort the array at take the subarray containing the last 4 elements.
You can use Arrays.sort() to sort and Arrays.copyOfRange() to take the subarray.
int[] arr = new int[] {1232, -1221, 0, 345, 78, 99};
Arrays.sort(arr);
int[] top4 = Arrays.copyOfRange(arr, arr.length-4,arr.length);
System.out.println(Arrays.toString(top4));
For more efficient solution, one can maintain a min-heap of top K elements or use selection algorithm to find the top 4th element. The two approaches are described in this thread.
Though the selection algorithm offers O(n) solution, the min-heap solution (which is O(nlogK)) should have better constants, and especially for small k is likely to be faster.
P.S. (EDIT):
For 4 elements, you might find that invoking a loop 4 times, and finding a max in each of them (and changing the old max to -infinity in each iteration) will be more efficient then the more "complex" approaches, since it requires sequential reads and have fairly small constants. This is of course not true for larger k, and decays into O(n^2) for k->n
EDIT2: benchmarking:
for the fun of it, I ran a benchmark on the attached code. The results are:
[naive, sort, heap] = [9032, 214902, 7531]
We can see that the naive and heap are much better then the sort based approach, and the naive is slightly slower then the heap based. I did a wilcoxon test to check if the difference between naive and heap is statistically significant, and I got a P_Value of 3.4573e-17. This means that the probability of the two approaches are "identical" is 3.4573e-17 (extremely small). From this we can conclude - heap based solution gives better performance then naive and sorting solution (and we empirically proved it!).
Attachment: The code I used:
public static int[] findTopKNaive(int[] arr, int k) {
int[] res = new int[k];
for (int j = 0; j < k; j++) {
int max=Integer.MIN_VALUE, maxIdx = -1;
for (int i = 0; i < arr.length; i++) {
if (max < arr[i]) {
max = arr[i];
maxIdx = i;
}
}
arr[maxIdx] = Integer.MIN_VALUE;
res[k-1-j] = max;
}
return res;
}
public static int[] findTopKSort(int[] arr, int k) {
Arrays.sort(arr);
return Arrays.copyOfRange(arr, arr.length-k,arr.length);
}
public static int[] findTopKHeap(int[] arr, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
for (int x : arr) {
if (pq.size() < k) pq.add(x);
else if (pq.peek() < x) {
pq.poll();
pq.add(x);
}
}
int[] res = new int[k];
for (int i =0; i < k; i++) res[i] = pq.poll();
return res;
}
public static int[] createRandomArray(int n, Random r) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.nextInt();
return arr;
}
public static void main(String... args) throws Exception {
Random r = new Random(1);
int k = 4;
int repeats = 200;
int n = 5000000;
long[][] results = new long[3][repeats];
for (int i = 0; i < repeats; i++) {
int[] arr = createRandomArray(n, r);
int[] myCopy;
myCopy = Arrays.copyOf(arr, n);
long start = System.currentTimeMillis();
findTopKNaive(myCopy, k);
results[0][i] = System.currentTimeMillis() - start;
myCopy = Arrays.copyOf(arr, n);
start = System.currentTimeMillis();
findTopKSort(myCopy, k);
results[1][i] = System.currentTimeMillis() - start;
myCopy = Arrays.copyOf(arr, n);
start = System.currentTimeMillis();
findTopKHeap(myCopy, k);
results[2][i] = System.currentTimeMillis() - start;
}
long[] sums = new long[3];
for (int i = 0; i < repeats; i++)
for (int j = 0; j < 3; j++)
sums[j] += results[j][i];
System.out.println(Arrays.toString(sums));
System.out.println("results for statistic test:");
for (int i = 0; i < repeats; i++) {
System.out.println(results[0][i] + " " + results[2][i]);
}
}
