Arrays.sort()会不会增加时间复杂度和空间时间复杂度？

Question

There is an array related problem, the requirement is that time complexity is O(n) and space complexity is O(1).

If I use Arrays.sort(arr) , and use a for loop to one pass loop, for example:

public static int hello(int[]A){
  Arrays.sort(A);
  for(int i=0;i<A.length;i++){
     ....................
  }
  return ....;

}

So the loop will cost O(n) time. My question is: will Arrays.sort() cost more time? If I use Arrays.sort() , will this time complexity still be O(n)? And will Arrays.sort() cost more space?

Answer 1

I am assuming you are talking about Java here.

So the loop will cost O(n) time, my question is that will Arrays.sort() cost more time?

Yes , Arrays.sort(int[]) in all Java standard library implementations that I know, is an example of a comparison-based sort and thus must have worst-case complexity Ω(n log n) . In particular, Oracle Java 7 uses a dual-pivot quicksort variant for the integer overloads, which actually has an Ω(n ² ) worst case.

and will Arrays.sort() cost more space?

In all likelihood it will use ω(1) space (which means another yes , the space usage is not O(1)). While it's not impossible to implement a comparison-based sort with only constant extra space, it's highly impractical.

That said, under certain conditions it is possible to sort specific types of data in linear time, see for example:

• http://en.wikipedia.org/wiki/Counting_sort
• http://en.wikipedia.org/wiki/Pigeonhole_sort
• http://en.wikipedia.org/wiki/Radix_sort

With a constant range of input integers (for example if abs(A[i]) <= C for some constant C), then counting sort and radix sort use indeed only O(n) time and O(1) space, so that might be useful.

Answer 2

According to the java jvm 8 javadocs of Arrays.sort() method:

The sorting algorithm is a Dual-Pivot Quicksort by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm offers O(n log(n)) performance on many data sets that cause other quicksorts to degrade to quadratic performance, and is typically faster than traditional (one-pivot) Quicksort implementations.

So it will increase your time complexity from O(n) to O(n log(n))

Answer 3

It is more than O(n) time and requires more than O(1) space.

Arrays.sort() utilizes a modified Timsort in 1.7 which is a relatively recently developed sorting algorithm and it offers sorting with complexity x where O(n)< x < O(nlgn) and space of O(n/2)

Answer 4

最近 JDK 中的 Arrays.sort(int[] a) 是用双枢轴快速排序算法实现的，该算法具有 O(n log n) 平均复杂度并且就地执行（例如不需要额外的空间）。

Answer 5

Since you're talking about it in Java Language, the time complexity will surely increase from O(n) to O(nlogn). That's because in Java 8, Arrays.sort() is implemented in Dual-pivot quicksort algorithm, not single pivot . So it requires extra time. And space complexity of O(1) is not possible , since it requires more space, I guess O(n/2).

Answer 6

Arrays.sort(int[])

As well as Arrays.sort(long[]), Arrays.sort(float[]) and Arrays.sort(double[])

Time complexity

Time complexity of Arrays.sort(int[]) depends on the version of Java.

O(n ² ) prior to Java 14

A pretty ordinary quicksort was used with time complexity ranging from O(n) (when the array is already sorted and we are only checking that it is) to O(n ² ) for certain inputs that cause extremely uneven distribution of elements into parts with an average complexity of O(n log(n)) . You can find a detailed analysis here .

O(n log(n)) starting from Java 14

In Java 14 the implementation was improved to guarantee the worst-case time complexity of O(n log(n)) . The function was changed to resort to heapsort if recursion becomes too deep:

if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
  heapSort(a, low, high);
  return;
}

which prevents the method from degrading to quadratic time complexity.

Glimpse into the future

There is an initiative to switch to radix sort for almost random big enough arrays thus reducing the time complexity to O(n) in the worst-case .

O(n) space

In all versions, the algorithm has space complexity ranging from O(1) (when the array is already sorted and we only to check that it is) to O(n) (when the array is highly structured (there is a small number of sorted subarrays inside the original array and we merge those subarrays)).

Here's where allocation happens in the worst case:

/*
 * Merge runs of highly structured array.
 */
if (count > 1) {
  int[] b; int offset = low;

  if (sorter == null || (b = (int[]) sorter.b) == null) {
    b = new int[size];
  } else {
    offset = sorter.offset;
  }
  mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
}
return true;

DualPivotQuicksort.java

While the question asks specifically about Arrays.sort(int[]) method I still decided to include answers for other data types since this is the first result when you look for Arrays.sort() time and space complexity in Google and it is not easy to find correct answers to this simple question in other places.

Arrays.sort(short[])

As well as Arrays.sort(char[]) and Arrays.sort(byte[])

O(n) time, O(1) space

Although the documentation says:

The sorting algorithm is a Dual-Pivot Quicksort by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm offers O(n log(n)) performance on all data sets, and is typically faster than traditional (one-pivot) Quicksort implementations.

This is not true at least starting from Java 7. Actually, an in-place counting sort used for big enough arrays, which has linear time complexity and constant space complexity :

private static void countingSort(short[] a, int low, int high) {
    int[] count = new int[NUM_SHORT_VALUES];

    /*
     * Compute a histogram with the number of each values.
     */
    for (int i = high; i > low; ++count[a[--i] & 0xFFFF]);

    /*
     * Place values on their final positions.
     */
    if (high - low > NUM_SHORT_VALUES) {
        for (int i = MAX_SHORT_INDEX; --i > Short.MAX_VALUE; ) {
            int value = i & 0xFFFF;

            for (low = high - count[value]; high > low;
                a[--high] = (short) value
            );
        }
    } else {
        for (int i = MAX_SHORT_INDEX; high > low; ) {
            while (count[--i & 0xFFFF] == 0);

            int value = i & 0xFFFF;
            int c = count[value];

            do {
                a[--high] = (short) value;
            } while (--c > 0);
        }
    }
}

Counting sort implementation

Arrays.sort(Object[])

Unlike other methods, this one is well-documented and the documentation here corresponds to reality.

O(n log(n)) time

Starting from Java 7

This implementation is a stable, adaptive, iterative mergesort that requires far fewer than n lg(n) comparisons when the input array is partially sorted, while offering the performance of a traditional mergesort when the input array is randomly ordered. If the input array is nearly sorted, the implementation requires approximately n comparisons.

https://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#sort(java.lang.Object[])

Before Java 7

The sorting algorithm is a modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist). This algorithm offers guaranteed n*log(n) performance.

https://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort(java.lang.Object[])

O(n) space

Starting from Java 7

Temporary storage requirements vary from a small constant for nearly sorted input arrays to n/2 object references for randomly ordered input arrays.

https://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#sort(java.lang.Object[])

Before Java 7

The algorithm used by java.util.Arrays.sort and (indirectly) by java.util.Collections.sort to sort object references is a "modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist)." It is a reasonably fast stable sort that guarantees O(n log n) performance and requires O(n) extra space.

https://bugs.openjdk.org/browse/JDK-6804124

Answer 7

import java.util.Arrays;
public class MyClass {
    
    
    static void hello(int ac[]){
        
    }
    
    public static void main(String args[]) {
  
      int ac[] ={1,4,2,3,5};
    
       int i=0;
       int temp=0;
       
       while(i!=5-1){
            if( ac[i]>ac[i+1]){
               temp= ac[i];
               ac[i]=ac[i+1];
               ac[i+1]=temp;
               i= -1;
           }
           
          i++;
       }
       
      System.out.println(Arrays.toString(ac));



    }
}