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 <rahul.deshmukhpa...@gmail.com> 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 <manjunath.n...@gmail.com> wrote: > > > will this work in two sorted arrays of equal length.. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.