On May 3, 9:43 am, Jitendra Kushwaha <jitendra.th...@gmail.com> wrote:
> @divya
>
> I try to simulate what you said for the given array
>
> index : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
> array1 : 8, 7, 4 ,3 , 2, 1, 1, 1, 1, 1
> ^ ^
> p11 p12
>
> *p11 = 8 and *p12 = 3
>
> index : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
> array2 : 34, 23, 21, 19, 15, 13, 11, 8, 4, 2
> ^ ^
> p21 p22
>
> *p21 = 34 and *p22 = 23
>
> a = 8 + 34 = 41 //arr1[0] + arr2[0]
> b = 8 +23 = 31 //arr1[0] + arr2[1]
> c = 34 + 3 = 37 //arr1[4] + arr2[0] greatest !!!
> d = 7 +23 = 30 //arr1[1] + arr2[1]
>
> arr1[0] + arr2[2] = 29 which is less than c..
>
> here is my output
>
> arr1[0] + arr2[0] = 42
> arr1[1] + arr2[0] = 41
> arr1[2] + arr2[0] = 38
> arr1[3] + arr2[0] = 37
> arr1[4] + arr2[0] = 36
> arr1[5] + arr2[0] = 35
> arr1[6] + arr2[0] = 35
> arr1[7] + arr2[0] = 35
> arr1[8] + arr2[0] = 35
> arr1[9] + arr2[0] = 35
>
> i have attached the code try with replacing arr1 and arr2 with
> following.and the case u wanted to point out will be taken throught in
> following test case..(arr1[0] + arr2[1] = 9+27 = 36 will be taken before
> arr1[6] + arr2[0] = 1+34 = 35 )
>
> int arr1[N] = {9,7,4,3,2,1,1,1,1,1};
> int arr2[N] = {34,27,21,19,15,13,11,8,4,2};
>
> Regards
> Jitendra Kushwaha
>
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I propose this solution (it's C89):
#include <stdio.h>
//int a[] = { 8, 7, 4, 3, 2, 1, 1, 1, 1, 1 };
//int b[] = { 34, 23, 21, 19, 15, 13, 11, 8, 4, 2 };
int a[] = { 6, 5, 4, 3, 2, 1 };
int b[] = { 9, 8, 6, 5, 3, 2 };
void find_pairs(int *a, int *b, int n)
{
int iamax[100], ibmax[100], i, ia, ib;
for (i = 0; i < n; i++)
iamax[i] = ibmax[i] = -1;
ia = ib = 0;
for (i = 0; i < n; i++) {
printf("(%d,%d)\n", a[ia], b[ib]);
if (a[ia + 1] + b[ibmax[ia + 1] + 1] > b[ib + 1] + a[iamax[ib +
1] +
1])
ib = ++ibmax[++ia];
else
ia = ++iamax[++ib];
}
}
int main(void)
{
find_pairs(a, b, sizeof a / sizeof a[0]);
return 0;
}
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