【发布时间】:2020-06-05 21:28:08
【问题描述】:
目前想知道如何降低这个三元组计数函数的复杂性/时间?
这个方案可行,但是传递HackerRank太慢了
def count_triplets(arr, r):
left_map = {}
right_map = {}
for item in arr:
right_map[item] = arr.count(item) # store frequency of occurences of each item.
count = 0
for i in range(0, len(arr)):
mid_term = arr[i]
c1 = 0
c3 = 0
right_map[mid_term] -= 1
left_term = mid_term / r
if left_term in left_map and mid_term % r == 0:
c1 = left_map.get(left_term)
right_term = mid_term * r
if right_term in right_map:
c3 = right_map.get(right_term)
count += c1 * c3
if mid_term in left_map:
left_map[mid_term] += 1
else:
left_map[mid_term] = 1
return count
if __name__ == '__main__':
r = 3
arr = [1,3,9,9,9,27,81] # returns 9 as solution
# arr = [1,3,9,9,27,81] # returns 6 as solution
result = count_triplets(arr, r)
print("result = ", result)
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标签: python dictionary hashtable