A. Count Equal and Divisible Pairs in an Array (opens new window)

Brute Force, check every pair of indexes (i, j).

DETAILS

def countPairs(self, A: List[int], k: int) -> int:
        ans, n = 0, len(A)
        for i in range(n - 1):
            for j in range(i + 1, n):
                ans += int(A[i] == A[j] and not (i * j) % k):
        return ans

B. Find Three Consecutive Integers That Sum to a Given Number (opens new window)

Check if num is divisible by 3.

DETAILS

def sumOfThree(self, num: int) -> List[int]:
        return [] if num % 3 else [num // 3 - 1, num //3, num // 3 + 1]

C. Maximum Split of Positive Even Integers (opens new window)

Greedy. Start from the smallest even number 2.

DETAILS

def maximumEvenSplit(self, A: int) -> List[int]:
      if A % 2: return []
      ans, cur, A = [2], 4, A - 2
      while A >= cur:
          ans.append(cur)
          A -= cur
          cur += 2
      ans[-1] += A
      return ans

D. Count Good Triplets in an Array (opens new window)

Please refer to This LC Discussion (opens new window) for my solutions in English.

思路是：只看triplet里中间的那个数： 对于A中每一个数字a，我们需要找到：以a为中心的triplet有多少？这样，把所有数的triplets加一起即可。 所以，对于a我们需要找到：

• 有多少个数在A, B里位置均在a的左边，比如有x个数。
• 有多少个数在A, B里的位置均在a的右边，比如有y个数。
• 以a为中心的triple共有x*y种。

遍历过程需要logN时间内的插入/二分查找，数据结构用BST或者BIT都可以。我用的是Python有现成的BST容器SortedList。

DETAILS

from sortedcontainers import SortedList
class Solution:
    def goodTriplets(self, A: List[int], B: List[int]) -> int:
        # Index of a (from A) in B.
        pos = [0] * len(A)               
        for idx, b in enumerate(B):
            pos[b] = idx
        
        # Build pre_a[i]: number of elements on a[i]'s left in both A and B.
        # pos_in_b: sorted indexes (in B) of all the visited elements in A.
        pos_in_b, pre_a = SortedList([pos[A[0]]]), [0]      
        for a in A[1:]:       
            pos_in_b.add(pos[a])
            pre_a.append(pos_in_b.bisect_left(pos[a]))
    
        # Build suf_a[i]: number of elements on a[i]'s right in both A and B.
        pos_in_b, suf_a = SortedList([pos[A[-1]]]), [0]
        for a in reversed(A[:len(A)-1]):
            idx = pos_in_b.bisect(pos[a])
            suf_a.append(len(pos_in_b) - idx)
            pos_in_b.add(pos[a])
        suf_a.reverse()
        
        # Sum up all unique triplets centered on A[i].
        ans = 0
        for x, y in zip(pre_a, suf_a):
            ans += x * y
        return ans