Suppose I have an array
a[3]={1,3,8}
I want the output to be an array containing the numbers obtained by adding numbers from array a, as well as the elements of array a. i.e.,
b[0]=1
b[1]=3
b[2]=8
b[3]=4 //(1+3)
b[4]=9 //(1+8)
b[5]=11 //(3+8)
b[6]=12 //(1+3+8)
How do I do this?
So you want to generate all possible subsets of a set of numbers and list their sum.
First, you want to enumerate all the subsets. Since there are 2 ^ N subsets in a set containing N elements, you can simply do this by iterating through the natural numbers from 0 to 1 << (sizeof(arr) / sizeof(arr[0])) and treating the binary representation of the number in the following manner: if a particular bit is set at position k, then the kth element is in the currently generated subset, else it isn't.
Then, you should add all the selected elements together and store the result in the next slot of another array (sized 2 ^ N, obviously).
#define COUNT(a) (sizeof(a) / sizeof(a[0]))
#include <limits.h> // for CHAR_BIT
unsigned a[3] = { 1, 3, 8 };
unsigned sums[1 << COUNT(a)] = { 0 };
for (unsigned i = 0; i < COUNT(sums); i++) {
for (unsigned j = 0; j < sizeof(i) * CHAR_BIT; j++) {
if ((i >> j) & 1) {
sums[i] += a[j];
}
}
}
for (unsigned i = 0; i < COUNT(sums); i++) {
printf("%d\n", sums[i]);
}
I would do it like this:
b[0] = 0; //This is not listed in the question, but should be included in the result IMHO.
for(long i = 0; i < elementsA; i++) {
long preexistingElements = 1 << i;
long curElement = a[i];
for(long j = 0; j < preexistingElements; j++) {
b[j + preexistingElements] = b[j] + curElement;
}
}
This algorithm is linear in the size of the result array as each of its elements is computed with constant cost.
If you really must exclude the zero and want to malloc the result array, turn the algorithm around so that b is filled from the back with the zero element as the last element.
