If the ranges of the arrays are 1..n & 1..m, then we can solve it this
way....
if ((m+n)&1){
we can go with the method same as rahul patil's and in the condition
we can use count<=(m+n)/2+1, the median will be stored in res.
}
else{
we can go with the method same as rahul patil's and in the condition
we can use count<=(m+n)/2+1 and the median in this case will be the
average of elements at count (m+n)/2 & at count (m+n)/2+1.So, we will
have to store the last element in this case.
}
On Aug 11, 5:25 pm, rahul patil <[email protected]> wrote:
> is there any time complexity?
>
> the also can be like this
>
> char *res;
> char *ptr1 =arr1;
> char *ptr2 =arr2;
> int count =0, n= len(arr1) ,m=len(arr2);
> while(1){
> while(*ptr1 > *ptr2){
> ptr2++;
> count ++;
> if( count == (n+m)/2 ){
> res=ptr1;
> break out of outer while loop;
> }
> }
>
> while(*ptr1 < *ptr2){
> ptr1++;
> count ++;
> if( count == (n+m)/2 ){
> res=ptr2;
> break out of outer while loop;
> }
> }
>
> }
>
> On Aug 6, 7:20 pm, Manjunath Manohar <[email protected]> wrote:
>
> > will this work in two sorted arrays of equal length..
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