Puzzle.. solving product of values in array X

问题

Can you please help me solving this one?

You have an unordered array X of n integers. Find the array M containing n elements where Mi is the product of all integers in X except for Xi. You may not use division. You can use extra memory. (Hint: There are solutions faster than O(n^2).)

The basic ones - O(n^2) and one using division is easy. But I just can't get another solution that is faster than O(n^2).

回答1:

Let left[i] be the product of all elements in X from 1..i. Let right[i] be the product of all elements in X from i..N. You can compute both in O(n) without division in the following way: left[i] = left[i - 1] * X[i] and right[i] = right[i + 1] * X[i];

Now we will compute M: M[i] = left[i - 1] * right[i + 1]

Note: left and right are arrays.

Hope it is clear:)

回答2:

Here's a solution in Python. I did the easy way with division to compare against the hard way without. Do I get the job?

L = [2, 1, 3, 5, 4]

prod = 1
for i in L: prod *= i
easy = map(lambda x: prod/x, L)
print easy

hard = [1]*len(L)
hmm = 1
for i in range(len(L) - 1):
    hmm *= L[i]
    hard[i + 1] *= hmm
huh = 1
for i in range(len(L) - 1, 0, -1):
    huh *= L[i]
    hard[i - 1] *= huh
print hard

回答3:

O(n) - http://nbl.cewit.stonybrook.edu:60128/mediawiki/index.php/TADM2E_3.28

two passes -

 int main (int argc, char **argv) {
    int array[] = {2, 5, 3, 4};
    int fwdprod[] = {1, 1, 1, 1};
    int backprod[] = {1, 1, 1, 1};
    int mi[] = {1, 1, 1, 1};
    int i, n = 4;
    for (i=1; i<=n-1; i++) {
        fwdprod[i]=fwdprod[i-1]*array[i-1];
    }
    for (i=n-2; i>=0; i--) {
        backprod[i] = backprod[i+1]*array[i+1];
    }
    for (i=0;i<=n-1;i++) {
        mi[i]=fwdprod[i]*backprod[i]; 
    }
    return 0;
}

回答4:

Old but very cool, I've been asked this at an interview myself and seen several solutions since but this is my favorite as taken from http://www.polygenelubricants.com/2010/04/on-all-other-products-no-division.html

static int[] products(int... nums) {
       final int N = nums.length;
       int[] prods = new int[N];
       java.util.Arrays.fill(prods, 1);
       for (int // pi----> * <----pj
          i = 0, pi = 1    ,  j = N-1, pj = 1  ;
         (i < N)           &          (j >= 0) ;
          pi *= nums[i++]  ,  pj *= nums[j--]  )
       {
          prods[i] *= pi   ;  prods[j] *= pj   ;
          System.out.println("pi up to this point is " + pi + "\n");
          System.out.println("pj up to this point is " + pj + "\n");
          System.out.println("prods[i]:" + prods[i] + "pros[j]:" +  prods[j] + "\n");
       }
       return prods;
    }

Here's what's going on, if you write out prods[i] for all the iterations, you'll see the following being calculated

prods[0], prods[n-1]
prods[1], prods[n-2]
prods[2], prods[n-3]
prods[3], prods[n-4]
.
.
.
prods[n-3], prods[2]
prods[n-2], prods[1]
prods[n-1], prods[0]

so each prods[i] get hit twice, one from the going from head to tail and once from tail to head, and both of these iterations are accumulating the product as they traverse towards the center so it's easy to see we'll get exactly what we need, we just need to be careful and see that it misses the element itself and that's where it gets tricky. the key lies in the

pi *= nums[i++], pj *= nums[j--]

in the for loop conditional itself and not in the body which do not happen until the end of the iteration. so for prods[0], it starts at 1*1 and then pi gets set to 120 after, so prods[0] misses the first elements prods[1], it's 1 * 120 = 120 and then pi gets set to 120*60 after so on and so on

回答5:

O(nlogn) approach:

int multiply(int arr[], int start, int end) {
    int mid;
    if (start > end) {
        return 1;
    }
    if (start == end) {
        return arr[start];
    }
    mid = (start+end)/2;
    return (multiply(arr, start, mid)*multiply(arr, mid+1, end));
}

int compute_mi(int arr[], int i, int n) {
    if ((i >= n) || (i < 0)) {
        return 0;
    }

    return (multiply(arr, 0, i-1)*multiply(arr, i+1, n-1));
}

来源：https://stackoverflow.com/questions/4057712/puzzle-solving-product-of-values-in-array-x

标签

algorithm

puzzle