| comments | difficulty | edit_url | rating | source | tags | |||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
true |
Hard |
2272 |
Biweekly Contest 72 Q4 |
|
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.length3 <= n <= 1050 <= nums1[i], nums2[i] <= n - 1nums1andnums2are permutations of[0, 1, ..., n - 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 ansclass 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
}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 ansclass 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;
}
};