@ross: I couldn't get reddy's solution. Please explain.
On Sun, Jun 5, 2011 at 10:50 PM, Deepak Jha deepak.127.0@gmail.comwrote:
the below solution should work given the input array is sorted ( I am
assuming ascending order)
void rearrangeArray(int[] a, int[] b){
int m = a.length;
int
@aakash Johari:
Let a and b be the 2 arrays.
At each stage of the process, if an element of A is greater than B,
then swap the largest element of A with the smallest element of B
and adjust pointers.
A : 2 4 15 12
B : 0.2 1 33 44
Now, 20, therefore swap 0 with 12..
Every step of the process,
the below solution should work given the input array is sorted ( I am
assuming ascending order)
void rearrangeArray(int[] a, int[] b){
int m = a.length;
int n = b.length;
int i = m - 1;
int j = 0;
while((i =0) (j = n-1)){
if(a[i] b[j]){
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
i--;
j++;
}
}
i think solution would be like this
eg:
A : 1 2 3 B: 0 1.5 4 5 9
Output:
A can contain any combination of nos 0,1,1.5
and B should contain 2 3 4 5 9 (in any order.)
this example is given by ROSS itself.
so sravanreddy solution is right , correct me if i'm wrong.
On Jun 3, 8:07 pm, bittu
Hi Rohit all,
Sorry that there was a small typo in the 'n' 'm' texts.
The example given by me is anyway the correct one.
Sravan Reddy's solution worked fine.
On Jun 4, 10:08 am, rohit rajuljain...@gmail.com wrote:
i think solution would be like this
eg:
A : 1 2 3 B: 0 1.5 4 5 9
Output:
A
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm 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
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++;
it can be done in O(m) . Use something like binary search .
code is here ...
#includestdio.h
void splitMN(int a[],int m , int b[], int n){
int al = 0 , bl = 0 ;
int ah = m-1 , bh = n-1 ;
int ai = (ah+al+1)/2;
int bi = (bh+bl+1)/2;
while(ai+bi!=m){
printf(Enter ai
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
--
You received this
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct me
if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross jagadish1...@gmail.com wrote:
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
@Ankit, The input should be 2 sorted arrays
Thanks,
Immanuel
On Sun, May 29, 2011 at 10:48 AM, ankit sambyal ankitsamb...@gmail.comwrote:
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct
me if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross
Oh I didn't read the question properly, Thanks for pointing out...
On Sat, May 28, 2011 at 10:28 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
@Ankit, The input should be 2 sorted arrays
Thanks,
Immanuel
On Sun, May 29, 2011 at 10:48 AM, ankit sambyal
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