Question

2179. Count Good Triplets in an Array

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Description

You are given two 0-indexed arrays nums1 and nums2 of length n, both of which are permutations of [0, 1, ..., n - 1].

A good triplet is a set of 3 distinct values which are present in increasing order by position both in nums1 and nums2. In other words, if we consider pos1v as the index of the value v in nums1 and pos2v as the index of the value v in nums2, then a good triplet will be a set (x, y, z) where 0 <= x, y, z <= n - 1, such that pos1x < pos1y < pos1z and pos2x < pos2y < pos2z.

Return _the total number of good triplets_.

 

Example 1:

Input: nums1 = [2,0,1,3], nums2 = [0,1,2,3]
Output: 1
Explanation: 
There are 4 triplets (x,y,z) such that pos1x < pos1y < pos1z. They are (2,0,1), (2,0,3), (2,1,3), and (0,1,3). 
Out of those triplets, only the triplet (0,1,3) satisfies pos2x < pos2y < pos2z. Hence, there is only 1 good triplet.

Example 2:

Input: nums1 = [4,0,1,3,2], nums2 = [4,1,0,2,3]
Output: 4
Explanation: The 4 good triplets are (4,0,3), (4,0,2), (4,1,3), and (4,1,2).

 

Constraints:

• n == nums1.length == nums2.length
• 3 <= n <= 105
• 0 <= nums1[i], nums2[i] <= n - 1
• nums1 and nums2 are permutations of [0, 1, ..., n - 1].

Solutions

Solution 1

class BinaryIndexedTree:
  def __init__(self, n):
    self.n = n
    self.c = [0] * (n + 1)

  @staticmethod
  def lowbit(x):
    return x & -x

  def update(self, x, delta):
    while x <= self.n:
      self.c[x] += delta
      x += BinaryIndexedTree.lowbit(x)

  def query(self, x):
    s = 0
    while x > 0:
      s += self.c[x]
      x -= BinaryIndexedTree.lowbit(x)
    return s


class Solution:
  def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
    pos = {v: i for i, v in enumerate(nums2, 1)}
    ans = 0
    n = len(nums1)
    tree = BinaryIndexedTree(n)
    for num in nums1:
      p = pos[num]
      left = tree.query(p)
      right = n - p - (tree.query(n) - tree.query(p))
      ans += left * right
      tree.update(p, 1)
    return ans

class Solution {
  public long goodTriplets(int[] nums1, int[] nums2) {
    int n = nums1.length;
    int[] pos = new int[n];
    BinaryIndexedTree tree = new BinaryIndexedTree(n);
    for (int i = 0; i < n; ++i) {
      pos[nums2[i]] = i + 1;
    }
    long ans = 0;
    for (int num : nums1) {
      int p = pos[num];
      long left = tree.query(p);
      long right = n - p - (tree.query(n) - tree.query(p));
      ans += left * right;
      tree.update(p, 1);
    }
    return ans;
  }
}

class BinaryIndexedTree {
  private int n;
  private int[] c;

  public BinaryIndexedTree(int n) {
    this.n = n;
    c = new int[n + 1];
  }

  public void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

  public int query(int x) {
    int s = 0;
    while (x > 0) {
      s += c[x];
      x -= lowbit(x);
    }
    return s;
  }

  public static int lowbit(int x) {
    return x & -x;
  }
}

class BinaryIndexedTree {
public:
  int n;
  vector<int> c;

  BinaryIndexedTree(int _n)
    : n(_n)
    , c(_n + 1) {}

  void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

  int query(int x) {
    int s = 0;
    while (x > 0) {
      s += c[x];
      x -= lowbit(x);
    }
    return s;
  }

  int lowbit(int x) {
    return x & -x;
  }
};

class Solution {
public:
  long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {
    int n = nums1.size();
    vector<int> pos(n);
    for (int i = 0; i < n; ++i) pos[nums2[i]] = i + 1;
    BinaryIndexedTree* tree = new BinaryIndexedTree(n);
    long long ans = 0;
    for (int& num : nums1) {
      int p = pos[num];
      int left = tree->query(p);
      int right = n - p - (tree->query(n) - tree->query(p));
      ans += 1ll * left * right;
      tree->update(p, 1);
    }
    return ans;
  }
};

type BinaryIndexedTree struct {
  n int
  c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
  c := make([]int, n+1)
  return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
  return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
  for x <= this.n {
    this.c[x] += delta
    x += this.lowbit(x)
  }
}

func (this *BinaryIndexedTree) query(x int) int {
  s := 0
  for x > 0 {
    s += this.c[x]
    x -= this.lowbit(x)
  }
  return s
}

func goodTriplets(nums1 []int, nums2 []int) int64 {
  n := len(nums1)
  pos := make([]int, n)
  for i, v := range nums2 {
    pos[v] = i + 1
  }
  tree := newBinaryIndexedTree(n)
  var ans int64
  for _, num := range nums1 {
    p := pos[num]
    left := tree.query(p)
    right := n - p - (tree.query(n) - tree.query(p))
    ans += int64(left) * int64(right)
    tree.update(p, 1)
  }
  return ans
}

Solution 2

class Node:
  def __init__(self):
    self.l = 0
    self.r = 0
    self.v = 0


class SegmentTree:
  def __init__(self, n):
    self.tr = [Node() for _ in range(4 * n)]
    self.build(1, 1, n)

  def build(self, u, l, r):
    self.tr[u].l = l
    self.tr[u].r = r
    if l == r:
      return
    mid = (l + r) >> 1
    self.build(u << 1, l, mid)
    self.build(u << 1 | 1, mid + 1, r)

  def modify(self, u, x, v):
    if self.tr[u].l == x and self.tr[u].r == x:
      self.tr[u].v += v
      return
    mid = (self.tr[u].l + self.tr[u].r) >> 1
    if x <= mid:
      self.modify(u << 1, x, v)
    else:
      self.modify(u << 1 | 1, x, v)
    self.pushup(u)

  def pushup(self, u):
    self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v

  def query(self, u, l, r):
    if self.tr[u].l >= l and self.tr[u].r <= r:
      return self.tr[u].v
    mid = (self.tr[u].l + self.tr[u].r) >> 1
    v = 0
    if l <= mid:
      v += self.query(u << 1, l, r)
    if r > mid:
      v += self.query(u << 1 | 1, l, r)
    return v


class Solution:
  def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
    pos = {v: i for i, v in enumerate(nums2, 1)}
    ans = 0
    n = len(nums1)
    tree = SegmentTree(n)
    for num in nums1:
      p = pos[num]
      left = tree.query(1, 1, p)
      right = n - p - (tree.query(1, 1, n) - tree.query(1, 1, p))
      ans += left * right
      tree.modify(1, p, 1)
    return ans

class Solution {
  public long goodTriplets(int[] nums1, int[] nums2) {
    int n = nums1.length;
    int[] pos = new int[n];
    SegmentTree tree = new SegmentTree(n);
    for (int i = 0; i < n; ++i) {
      pos[nums2[i]] = i + 1;
    }
    long ans = 0;
    for (int num : nums1) {
      int p = pos[num];
      long left = tree.query(1, 1, p);
      long right = n - p - (tree.query(1, 1, n) - tree.query(1, 1, p));
      ans += left * right;
      tree.modify(1, p, 1);
    }
    return ans;
  }
}

class Node {
  int l;
  int r;
  int v;
}

class SegmentTree {
  private Node[] tr;

  public SegmentTree(int n) {
    tr = new Node[4 * n];
    for (int i = 0; i < tr.length; ++i) {
      tr[i] = new Node();
    }
    build(1, 1, n);
  }

  public void build(int u, int l, int r) {
    tr[u].l = l;
    tr[u].r = r;
    if (l == r) {
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
  }

  public void modify(int u, int x, int v) {
    if (tr[u].l == x && tr[u].r == x) {
      tr[u].v += v;
      return;
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (x <= mid) {
      modify(u << 1, x, v);
    } else {
      modify(u << 1 | 1, x, v);
    }
    pushup(u);
  }

  public void pushup(int u) {
    tr[u].v = tr[u << 1].v + tr[u << 1 | 1].v;
  }

  public int query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) {
      return tr[u].v;
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    int v = 0;
    if (l <= mid) {
      v += query(u << 1, l, r);
    }
    if (r > mid) {
      v += query(u << 1 | 1, l, r);
    }
    return v;
  }
}

class Node {
public:
  int l;
  int r;
  int v;
};

class SegmentTree {
public:
  vector<Node*> tr;

  SegmentTree(int n) {
    tr.resize(4 * n);
    for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
    build(1, 1, n);
  }

  void build(int u, int l, int r) {
    tr[u]->l = l;
    tr[u]->r = r;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
  }

  void modify(int u, int x, int v) {
    if (tr[u]->l == x && tr[u]->r == x) {
      tr[u]->v += v;
      return;
    }
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    if (x <= mid)
      modify(u << 1, x, v);
    else
      modify(u << 1 | 1, x, v);
    pushup(u);
  }

  void pushup(int u) {
    tr[u]->v = tr[u << 1]->v + tr[u << 1 | 1]->v;
  }

  int query(int u, int l, int r) {
    if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    int v = 0;
    if (l <= mid) v += query(u << 1, l, r);
    if (r > mid) v += query(u << 1 | 1, l, r);
    return v;
  }
};

class Solution {
public:
  long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {
    int n = nums1.size();
    vector<int> pos(n);
    for (int i = 0; i < n; ++i) pos[nums2[i]] = i + 1;
    SegmentTree* tree = new SegmentTree(n);
    long long ans = 0;
    for (int& num : nums1) {
      int p = pos[num];
      int left = tree->query(1, 1, p);
      int right = n - p - (tree->query(1, 1, n) - tree->query(1, 1, p));
      ans += 1ll * left * right;
      tree->modify(1, p, 1);
    }
    return ans;
  }
};