Question

In the problem given here, i have to count total no. of swaps required while sorting an array using insertion sort.
here is my approach

#include <stdio.h>
int main()
{
    int t, N, swaps, temp, i, j;
    scanf("%d", &t);

    while(t--){
        scanf("%d", &N);
        
        int arr[N];
        swaps = 0;
    
        for(i=0; i<N; ++i){

            scanf("%d", &temp);

            j=i;
            while(j>0 && arr[j-1] > temp){
                arr[j] = arr[j-1];
                ++swaps;
                --j;
            }

            arr[j] = temp;
        }
        printf("%d\n", swaps);
    }

    return 0;
}

but, this soln is giving time limit exceeded.

How can i make it more fast?
and, what are the other better solutions of this problem?

Answer 1

This reminds me of a similar problem you may want to look at: http://www.spoj.pl/problems/YODANESS/

In your problem, you can’t afford the time to swap everything in case there are many swaps required. (imagine if the input was in reverse order 9,8,7,6.. then you would have to swap everything with everything basically.

I think in your case, each number must be swapped with all the numbers to the left of it that are smaller than it.

I suggest you use a range tree http://en.wikipedia.org/wiki/Range_tree

The great thing about a range tree is each node can know how many nodes are to its left and to its right. You could ask the tree “how many numbers are there greater than 10” very efficiently and that’s how many swaps you would have for a 9 say.

The trick is to build the range tree as you move from i=0 to i=N-1. At each point you can query the tree against the ith number before inserting the ith number into the range tree.

good luck!