I'm sorry for deleting the original question, here it is: We have a bag or an array of n integers, we need to find the product of each of the (n-1) subsets. e.g:
S = {1, 0, 3, 6}
ps[1] = 0*3*6 = 0;
ps[2] = 1*3*6 = 18; etc.
After discussions, we need to take care of the three cases and they are illustrated in the following:
1. S is a set (contains one zero element)
for i=1 to n
if s[i]=0
sp[i] = s[1] * s[2] * ...* s[i-1] * s[i+1] *.....*s[n]
else
sp[i] = 0;
2. S is a bag (contains more than one zero element)
for i=1 to n
sp[i] = 0;
3. S is a set (contains no zero elements)
product = 1
for i=1 to n
product *= s[i];
for i=1 to n
sp[i] = product / s[i];
Thanks.
If the set is very large, it may be convenient to:
If the set contains zero (i.e. P=0, x=0), you must deal with it as a special case.
EDIT. Here is a solution in Scheme, taking into account andand's answer. I'm a complete beginner - can someone help me improve the following code (make it more efficient, more readable, more lisp-ish)? (Feel free to edit my answer.)
#!/usr/bin/env guile !#
(use-modules (ice-9 pretty-print))
(define (count-zeros l)
(cond ((null? l) 0)
((= 0 (car l)) (+ 1 (count-zeros (cdr l))))
(else (count-zeros (cdr l)))))
(define (non-zero-product l)
(define (non-zero-product-loop l product)
(cond ((null? l) product)
((= 0 (car l)) (non-zero-product-loop (cdr l) product))
(else (non-zero-product-loop (cdr l) (* (car l) product)))))
(non-zero-product-loop l 1))
(define (n-1-products l)
(let ((nzeros (count-zeros l)))
(cond ((> nzeros 1)
(map (lambda (x) 0) l))
((= 1 nzeros)
(map (lambda (x) (if (= 0 x) (non-zero-product l) 0)) l))
(else
(map (lambda (x) (/ (non-zero-product l) x)) l)))))
(pretty-print (n-1-products '(1 2 3 4 5)))
(pretty-print (n-1-products '(0 1 2 3 4)))
(pretty-print (n-1-products '(0 1 2 3 0)))
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