On Jul 10, 5:18 pm, Gene <gene.ress...@gmail.com> wrote:
> On Jul 9, 3:55 pm, Devendra Pratap Singh <dpsingh.ii...@gmail.com>
> wrote:
>
> > plz write a code to
>
> > Sort n integer in the range 1 to n^2 in O(n)
>
> Use radix sort with radix n. Three O(n) passes will always do the
> job. If you subtract 1 from each value so the range is 0 to n^2-1, two
> passes will be enough.
Nice lunchtime puzzle. Here is some code for your pleasure:
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE (10 * 1024)
struct node_s {
int next, val;
} nodes[MAX_SIZE];
// Radix-n sort descending. Assumes data are non-negative.
// Will take K passes if largest datum does not exceed n^K.
int sort_descending(struct node_s *nodes, int n)
{
int i, k, head, next, shifted, digit, more_p, n_passes, div;
int buckets[MAX_SIZE];
// Chain nodes into a single list.
head = 0;
for (i = 0; i < n; i++)
nodes[i].next = i + 1;
nodes[n - 1].next = -1;
// Make passes until all remainders are zero.
// Always 2 if max datum is less than n^2.
n_passes = 0;
div = 1;
do {
// Empty the buckets.
for (k = 0; k < n; k++)
buckets[k] = -1;
// Fill the buckets, noting whether we need more passes.
more_p = 0;
for (i = head; i != -1; i = next) {
next = nodes[i].next;
shifted = nodes[i].val / div;
digit = shifted % n;
nodes[i].next = buckets[digit];
buckets[digit] = i;
if (shifted >= div) more_p = 1;
}
// Concatenate the buckets.
head = -1;
for (k = 0; k < n; k++) {
for (i = buckets[k]; i != -1; i = next) {
next = nodes[i].next;
nodes[i].next = head;
head = i;
}
}
n_passes++;
div *= n;
} while (more_p);
printf("sort took %d passes\n", n_passes);
return head;
}
int main(void)
{
int i, n, head, last;
n = 500;
for (i = 0; i < n; i++)
nodes[i].val = rand() % (n * n);
head = sort_descending(nodes, n);
// Make sure we're sorted descending and print.
last = -1;
for (i = head; i != -1; i = nodes[i].next) {
printf("%d ", nodes[i].val);
if (last != -1 && nodes[i].val > last) {
printf("oops!\n");
return 1;
}
last = nodes[i].val;
}
printf("\n");
return 0;
}
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