Here is a solution of O(n) , taking 4 pointers 2 for each array
#include <cstdio> #include<iostream> using namespace std; #define N 10 int main(void) { int arr1[N] = {8,7,4,3,2,1,1,1,1,1}; int arr2[N] = {34,23,21,19,15,13,11,8,4,2}; int *p11,*p12,*p21,*p22; p11 = p12 = arr1; p21 = p22 = arr2; int f1; f1 = 0; for(int i=0;i<N;i++) { int ans=0; int a,b,c,d; a = *p11 + *p21; b = *p11 + *p22; c = *p21 + *p12; d = *(p11+1) + *(p21+1); //printf("a=%d b=%d c=%d d=%d\n",a,b,c,d); //debug if(f1==0) ans = a , p12++ , p22++ , f1=1; else if(b >= c && b >= d ) ans = b , p22++ ; else if(c >= b && c >= d ) ans = c , p12++ ; else ans = d , p11++ , p21++ ,printf("4 "); printf("%d\n",ans); } } Regards Jitendra Kushwaha Undergradute Student Computer Science & Eng. MNNIT, Allahabad -- 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.