数组的最小乘积子集的Javascript程序

给定一个数组 a，我们必须找到数组中存在的元素子集的最小乘积。最小乘积也可以是单个元素。

例子：

Input : a[] = { -1, -1, -2, 4, 3 }
Output : -24
Explanation : Minimum product will be ( -2 * -1 * -1 * 4 * 3 ) = -24

Input : a[] = { -1, 0 }
Output : -1
Explanation : -1(single element) is minimum product possible
 
Input : a[] = { 0, 0, 0 }
Output : 0

一个简单的解决方案是生成所有子集，找到每个子集的乘积并返回最小乘积。
更好的解决方案是使用以下事实。

1. 如果有偶数个负数且没有零，则结果是除最大值负数之外的所有负数的乘积。
2. 如果有奇数个负数并且没有零，则结果只是所有的乘积。
3. 如果有零和正数，没有负数，则结果为 0。例外情况是当没有负数且所有其他元素为正数时，我们的结果应该是第一个最小正数。

Javascript

  
// Javascript program to find maximum 
// product of a subset.
function minProductSubset(a, n)
{
    if (n == 1)
        return a[0];
  
    // Find count of negative numbers,
    // count of zeros, maximum valued
    // negative number, minimum valued
    // positive number and product of
    // non-zero numbers
    let negmax = Number.MIN_VALUE;
    let posmin = Number.MIN_VALUE;
    let count_neg = 0, count_zero = 0;
    let product = 1;
  
    for(let i = 0; i < n; i++)
    {
          
        // If number is zero, count it
        // but dont multiply
        if (a[i] == 0)
        {
            count_zero++;
            continue;
        }
  
        // Count the negative numbers
        // and find the max negative number
        if (a[i] < 0) 
        {
            count_neg++;
            negmax = Math.max(negmax, a[i]);
        }
  
        // Find the minimum positive number
        if (a[i] > 0 && a[i] < posmin) 
        {
            posmin = a[i];
        }
  
        product *= a[i];
    }
  
    // If there are all zeroes
    // or zero is present but no
    // negative number is present
    if (count_zero == n || (count_neg == 0 && 
        count_zero > 0))
        return 0;
  
    // If there are all positive
    if (count_neg == 0)
        return posmin;
  
    // If there are even number except
    // zero of negative numbers
    if (count_neg % 2 == 0 && count_neg != 0)
    {
  
        // Otherwise result is product of
        // all non-zeros divided by maximum
        // valued negative.
        product = parseInt(product / negmax, 10);
    }
    return product;
}
  
// Driver code
let a = [ -1, -1, -2, 4, 3 ];
let n = 5;
  
document.write(minProductSubset(a, n));
  
// This code is contributed by rameshtravel07
  

输出：

-24

时间复杂度： O(n)
辅助空间： O(1)

有关更多详细信息，请参阅有关数组的最小产品子集的完整文章！