Find the number of unordered pair in an array

问题

I ran into an interesting algorithm problem:

Given an array of integer, find the number of un-ordered pairs in that array, say given {1, 3, 2}, the answer is 1 because {3, 2} is un-ordered, and for array {3, 2, 1}, the answer is 3 because {3, 2}, {3, 1}, {2, 1}.

Obviously, this can be solved by brute force with O(n^2) running time, or permute all possible pairs then eliminate those invalid pairs.

My question is does any body have any better solution and how would you do it because it seems like a dynamic programming problem. A snippet of code would be helpful

回答1:

It is possible to solve this problem in O(n log n) time using a balanced binary search tree. Here is a pseudo-code of this algorithm:

tree = an empty balanced binary search tree
answer = 0
for each element in the array:
     answer += number of the elements in the tree greater then this element
     add this element to the tree

回答2:

You can use a modified version of merge sort to count the number of inversions. The trick is that while merging two sorted sub arrays you can come to know the elements which are out of place. If there are any elements in right subarray which need to go before the ones in left subarray, they are the inverted ones. I've written the code for this in python. You can check the explanation below it for better understanding. If you not able to understand merge sort I'd suggest you to revist merge sort after which this would be intuitive.

def merge_sort(l):
    if len(l) <= 1:
        return (0, l)
    else:
        mid = len(l) / 2
        count_left, ll = merge_sort(l[0:mid])
        count_right, lr = merge_sort(l[mid:])
        count_merge, merged = merge(ll, lr)
        total = count_left + count_right + count_merge
        return total, merged

def merge(left, right):
    li, ri = 0, 0
    merged = []        
    count = 0
    while li < len(left) and ri < len(right):
        if left[li] < right[ri]:
            merged.append(left[li])
            li += 1
        else:
            count += 1
            merged.append(right[ri])
            ri += 1

    if li < len(left):
        merged.extend(left[li:])
    elif ri < len(right):
        merged.extend(right[ri:])
    return count, merged

if __name__ == '__main__':
    # example 
    l = [6, 1 , 2, 3, 4, 5]
    print 'inverse pair count is %s'%merge_sort(l)[0]

Merge sort runs in n * log(n) time.
for the passed list l, merge_sort returns a tuple (in the form of (inversion_count, list)) of number of inversions and the sorted list
Merge step counts the number of inversions and stores it in the variable count.

回答3:

If you are just looking for the number of un-ordered pair and the array is sorted in ascending order. You can use this formula n * (n - 1) / 2. Suppose your array has n elements, for example 3 in your case. It will be 3 * 2 / 2 = 3. Assuming there are no duplicate elements.

回答4:

It was in one of my practice midterms and i think a nested for loop does the job pretty nice.

public static void main(String args[])
{
    int IA[] = {6,2,9,5,8,7};

    int cntr = 0;

    for(int i = 0; i <= IA.length-1;i++)
    {
        for(int j = i; j <= IA.length-1; j++)
        {
            if(IA[i]>IA[j])
            {

                System.out.print("("+IA[i]+","+ IA[j]+")"+";");
                cntr++;
            }

        }
    }

    System.out.println(cntr);

}

回答5:

You can use a modified merge-sort algorithm. Merging would look something like this.

merge(a, b):
    i = 0
    j = 0
    c = new int[a.length+b.length]
    inversions = 0
    for(k = 0 ; k < Math.min(a.length, b.length); k++)
        if(a[i] > b[j]):
            inversions++
            c[k] = b[j]
            j++
        else:
            c[k] = a[i]
            i++
    //dump the rest of the longer array in c
    return inversions

Merging is done in O(n) time. Time complexity of the whole merge sort is O(n log n)

来源：https://stackoverflow.com/questions/26702051/find-the-number-of-unordered-pair-in-an-array

标签

arrays

algorithm

dynamic-programming