从具有重复元素的数组中随机找到一个组合，它的总和等于 n

Question

如何从具有重复元素的array中随机找到一个组合，并且它的和等于n。

示例

• array是[1, 2, 2, 3]，n是3
• 答案是1+2、1+2、3
• 如果randomSubsetSum(array, n)是解，则randomSubsetSum([1,2,2,3], 3)将返回1+2、1+2、3中的一个。注：1+2的出现频率是3的两倍
• 真实世界的场景：从题库中随机选择问题进行考试

我发现了一些类似的问题和解决方案：

1. Q：找到所有可能的数字组合，以达到给定的和

   A： 溶液A和溶液B

2. Q： 具有k部分的秩和非秩整数分区

   A： 解决方案C

缺陷

CCD_ 17和CCD_。solution C不允许重复元素。

我的Java解决方案

public List<Integer> randomSubsetSum(List<Integer> list, Integer n) {
list.removeIf(e -> e > n);
int maxSum = list.stream().reduce(0, Integer::sum);
if (maxSum < n) {
throw new RuntimeException("maxSum of list lower than n!");
}
if (maxSum == n) {
return list;
}
final SecureRandom random = new SecureRandom();
// maybe helpful, not important
final Map<Integer, List<Integer>> map = list.stream().collect(Collectors.groupingBy(Function.identity()));
final List<Integer> keys = new ArrayList<>(map.keySet());
final List<Integer> answers = new ArrayList<>();
int sum = 0;
while (true) {
int keyIndex = random.nextInt(keys.size());
Integer key = keys.get(keyIndex);
sum += key;
// sum equal n
if (sum == n) {
List<Integer> elements = map.get(key);
answers.add(elements.get(random.nextInt(elements.size())));
break;
}
// sum below n
if (sum < n) {
List<Integer> elements = map.get(key);
answers.add(elements.remove(random.nextInt(elements.size())));
if (elements.isEmpty()) {
map.remove(key);
keys.remove(keyIndex);
}
continue;
}
// sum over n: exists (below  = n - sum + key) in keys
int below = n - sum + key;
if (CollectionUtils.isNotEmpty(map.get(below))) {
List<Integer> elements = map.get(below);
answers.add(elements.get(random.nextInt(elements.size())));
break;
}
// sum over n: exists (over  = sum - n) in answers
int over = sum - n;
int answerIndex =
IntStream.range(0, answers.size())
.filter(index -> answers.get(index) == over)
.findFirst().orElse(-1);
if (answerIndex != -1) {
List<Integer> elements = map.get(key);
answers.set(answerIndex, elements.get(random.nextInt(elements.size())));
break;
}
// Point A. BUG: may occur infinite loop
// sum over n: rollback sum
sum -= key;
// sum over n: remove min element in answer
Integer minIndex =
IntStream.range(0, answers.size())
.boxed()
.min(Comparator.comparing(answers::get))
// never occurred
.orElseThrow(RuntimeException::new);
Integer element = answers.remove((int) minIndex);
sum -= element;
if (keys.contains(element)) {
map.get(element).add(element);
} else {
keys.add(element);
map.put(element, new ArrayList<>(Collections.singleton(element)));
}
}
return answers;
}

在Point A，可能会发生无限循环(例如randomSubsetSum([3,4,8],13)(或使用大量时间。如何修复这个错误，或者有其他解决方案吗？

Answer 1

这是一个从解决方案a.中稍作调整的解决方案

from random import random
def random_subset_sum(array, target):
sign = 1
array = sorted(array)
if target < 0:
array = reversed(array)
sign = -1
# Checkpoint A
last_index = {0: [[-1,1]]}
for i in range(len(array)):
for s in list(last_index.keys()):
new_s = s + array[i]
total = 0
for index, count in last_index[s]:
total += count
if 0 < (new_s - target) * sign:
pass # Cannot lead to target
elif new_s in last_index:
last_index[new_s].append([i,total])
else:
last_index[new_s] = [[i, total]]
# Checkpoint B
answer_indexes = []
last_choice = len(array)
while -1 < last_choice:
choice = None
total = 0
for i, count in last_index[target]:
if last_choice <= i:
break
total += count
if random() <= count / total:
choice = i
target -= array[choice]
last_choice = choice
if -1 < choice:
answer_indexes.append(choice)
return [array[i] for i in reversed(answer_indexes)]