Please try this solution. And tell if it fails at any case. If it works
fine, I will tell the logic.
> #include <stdio.h>
> #include <stdlib.h>
>
> void merge(int *a, int m, int *b, int n)
> {
> int i, j, k;
> int t;
>
> i = j = 0;
> k = -1;
>
> while ( i < m && a[i] < b[j] ) {
> i++;
> }
>
> while ( i < m && j < n ) {
> if ( k == -1 && b[j] < a[i] ) {
> t = a[i]; a[i] = b[j]; b[j] = t;
> k = j;
> j++;
> i++;
> } else if ( b[j] < b[k] ) {
> t = a[i]; a[i] = b[j]; b[j] = t;
> j++;
> i++;
> } else {
> t = a[i]; a[i] = b[k]; b[k] = t;
> i++;
> }
> }
> }
>
> int main()
> {
> int m, n;
> int *a, *b;
> int i;
>
> scanf ("%d%d", &m, &n);
>
> a = (int*) malloc(sizeof(int) * m);
> b = (int*) malloc(sizeof(int) * n);
>
> for ( i = 0; i < m; i++ )
> scanf ("%d", a + i);
>
> for ( i = 0; i < n; i++ )
> scanf ("%d", b + i);
>
> merge (a, m, b, n);
>
> printf ("After Merge Operation : \n");
>
> printf ("1st Array : ");
>
> for ( i = 0; i < m; i++ ) {
> printf ("%d ", a[i]);
> }
>
> printf ("\n2nd Array : " );
>
> for ( i = 0; i < n; i++ ) {
> printf ("%d ", b[i]);
> }
>
> return 0;
> }
>
>
> On Fri, Jun 3, 2011 at 8:07 AM, bittu <shashank7andr...@gmail.com> wrote:
> @sravanreddy...logical bugs if A is size of n & B is size m from your
> example assuming n<m so if you want smallest m elements in A then u
> only capacity of n elements & didn't allocate memory so these elements
> initialized by INT_MIN for m-n nodes so that array A can hold m
> smallest elements then what r u swapping u dude..isn't garbage
> value ?? you will get at 1st step only just run it ?? in you algo
> A_End=m-1(which 4th position in Array that DNE)..?? & also you have to
> free memory for m-n in array B as it contains n largest elements .
>
> take
> A= 1,2,3 n=3
> B= 0,1,4,5,9 m=5
>
> after allocating memory to Array A for m-n elements A will looks
> likes 1 2 3 INT_Max INT_Max
> now what you wants A should contains m smallest elements & B have n
> largest elements
> so O/P should be A=1,2,3,1,0 & B=INT_Max,INT_Max,4,5,9 now free
> memory used by 1st elements in array B so that A will represent M
> smallest elements & B will have n Largest elements
>
> so that above will work.
>
> Hope I am Correct let me know if any issue with explanation
>
> Thanks
> Shashank>>"The Best Way To Escape From The problem is To Solve It"
> CSE,BIT Mesra
>
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>
--
-Aakash Johari
(IIIT Allahabad)
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