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2179. Count Good Triplets in an Array 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class FenwickTree {
 public:
  FenwickTree(int n) : sums(n + 1) {}

  void add(int i, int delta) {
    while (i < sums.size()) {
      sums[i] += delta;
      i += lowbit(i);
    }
  }

  int get(int i) const {
    int sum = 0;
    while (i > 0) {
      sum += sums[i];
      i -= lowbit(i);
    }
    return sum;
  }

 private:
  vector<int> sums;

  static inline int lowbit(int i) {
    return i & -i;
  }
};

class Solution {
 public:
  long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {
    const int n = nums1.size();
    long ans = 0;
    unordered_map<int, int> numToIndex;
    vector<int> A;
    // leftSmaller[i] := the number of A[j] < A[i], where 0 <= j < i
    vector<int> leftSmaller(n);
    // rightLarger[i] := the number of A[j] > A[i], where i < j < n
    vector<int> rightLarger(n);
    FenwickTree tree1(n);  // Calculates `leftSmaller`.
    FenwickTree tree2(n);  // Calculates `rightLarger`.

    for (int i = 0; i < n; ++i)
      numToIndex[nums1[i]] = i;

    // Remap each number in `nums2` to the according index in `nums1` as `A`.
    // Rephrase the problem as finding the number of increasing tripets in `A`.
    for (const int num : nums2)
      A.push_back(numToIndex[num]);

    for (int i = 0; i < n; ++i) {
      leftSmaller[i] = tree1.get(A[i]);
      tree1.add(A[i] + 1, 1);
    }

    for (int i = n - 1; i >= 0; --i) {
      rightLarger[i] = tree2.get(n) - tree2.get(A[i]);
      tree2.add(A[i] + 1, 1);
    }

    for (int i = 0; i < n; ++i)
      ans += static_cast<long>(leftSmaller[i]) * rightLarger[i];

    return ans;
  }
};
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class FenwickTree {
  public FenwickTree(int n) {
    sums = new int[n + 1];
  }

  public void add(int i, int delta) {
    while (i < sums.length) {
      sums[i] += delta;
      i += lowbit(i);
    }
  }

  public int get(int i) {
    int sum = 0;
    while (i > 0) {
      sum += sums[i];
      i -= lowbit(i);
    }
    return sum;
  }

  private int[] sums;

  private static int lowbit(int i) {
    return i & -i;
  }
}

class Solution {
  public long goodTriplets(int[] nums1, int[] nums2) {
    final int n = nums1.length;
    long ans = 0;
    Map<Integer, Integer> numToIndex = new HashMap<>();
    int[] A = new int[n];
    // leftSmaller[i] := the number of A[j] < A[i], where 0 <= j < i
    int[] leftSmaller = new int[n];
    // rightLarger[i] := the number of A[j] > A[i], where i < j < n
    int[] rightLarger = new int[n];
    FenwickTree tree1 = new FenwickTree(n); // Calculates `leftSmaller`.
    FenwickTree tree2 = new FenwickTree(n); // Calculates `rightLarger`.

    for (int i = 0; i < n; ++i)
      numToIndex.put(nums1[i], i);

    // Remap each number in `nums2` to the according index in `nums1` as `A`.
    // Rephrase the problem as finding the number of increasing tripets in `A`.
    for (int i = 0; i < n; ++i)
      A[i] = numToIndex.get(nums2[i]);

    for (int i = 0; i < n; ++i) {
      leftSmaller[i] = tree1.get(A[i]);
      tree1.add(A[i] + 1, 1);
    }

    for (int i = n - 1; i >= 0; --i) {
      rightLarger[i] = tree2.get(n) - tree2.get(A[i]);
      tree2.add(A[i] + 1, 1);
    }

    for (int i = 0; i < n; ++i)
      ans += (long) leftSmaller[i] * rightLarger[i];

    return ans;
  }
}
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class FenwickTree:
  def __init__(self, n: int):
    self.sums = [0] * (n + 1)

  def add(self, i: int, delta: int) -> None:
    while i < len(self.sums):
      self.sums[i] += delta
      i += FenwickTree.lowbit(i)

  def get(self, i: int) -> int:
    summ = 0
    while i > 0:
      summ += self.sums[i]
      i -= FenwickTree.lowbit(i)
    return summ

  @staticmethod
  def lowbit(i: int) -> int:
    return i & -i


class Solution:
  def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
    n = len(nums1)
    numToIndex = {num: i for i, num in enumerate(nums1)}
    # Remap each number in `nums2` to the according index in `nums1` as `A`.
    # Rephrase the problem as finding the number of increasing tripets in `A`.
    A = [numToIndex[num] for num in nums2]
    # leftSmaller[i] := the number of A[j] < A[i], where 0 <= j < i
    leftSmaller = [0] * n
    # rightLarger[i] := the number of A[j] > A[i], where i < j < n
    rightLarger = [0] * n
    tree1 = FenwickTree(n)  # Calculates `leftSmaller`.
    tree2 = FenwickTree(n)  # Calculates `rightLarger`.

    for i, a in enumerate(A):
      leftSmaller[i] = tree1.get(a)
      tree1.add(a + 1, 1)

    for i, a in reversed(list(enumerate(A))):
      rightLarger[i] = tree2.get(n) - tree2.get(a)
      tree2.add(a + 1, 1)

    return sum(a * b for a, b in zip(leftSmaller, rightLarger))