Description
There are n soldiers standing in a line. Each soldier is assigned a unique rating value.
You have to form a team of 3 soldiers amongst them under the following rules:
- Choose 3 soldiers with index (
i, j, k) with rating (rating[i], rating[j], rating[k]).
- A team is valid if: (
rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).
Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).
Example 1:
Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1).
Example 2:
Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.
Example 3:
Input: rating = [1,2,3,4]
Output: 4
Constraints:
n == rating.length3 <= n <= 10001 <= rating[i] <= 105- All the integers in
rating are unique.
Solutions
Solution 1: Enumerate Middle Element
We can enumerate each element rating[i] in the array rating as the middle element, then count the number of elements l that are smaller than it on the left, and the number of elements r that are larger than it on the right. The number of combat units with this element as the middle element is l × r + (i - l) × (n - i - 1 - r). We can add this to the answer.
The time complexity is O(n2), and the space complexity is O(1). Where n is the length of the array rating.
PythonJavaC++GoTypeScript
class Solution:
def numTeams(self, rating: List[int]) -> int:
ans, n = 0, len(rating)
for i, b in enumerate(rating):
l = sum(a < b for a in rating[:i])
r = sum(c > b for c in rating[i + 1 :])
ans += l * r
ans += (i - l) * (n - i - 1 - r)
return ans(code-box)
class Solution {
public int numTeams(int[] rating) {
int n = rating.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
int l = 0, r = 0;
for (int j = 0; j < i; ++j) {
if (rating[j] < rating[i]) {
++l;
}
}
for (int j = i + 1; j < n; ++j) {
if (rating[j] > rating[i]) {
++r;
}
}
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}
}(code-box)
class Solution {
public:
int numTeams(vector<int>& rating) {
int n = rating.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
int l = 0, r = 0;
for (int j = 0; j < i; ++j) {
if (rating[j] < rating[i]) {
++l;
}
}
for (int j = i + 1; j < n; ++j) {
if (rating[j] > rating[i]) {
++r;
}
}
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}
};(code-box)
func numTeams(rating []int) (ans int) {
n := len(rating)
for i, b := range rating {
l, r := 0, 0
for _, a := range rating[:i] {
if a < b {
l++
}
}
for _, c := range rating[i+1:] {
if c < b {
r++
}
}
ans += l * r
ans += (i - l) * (n - i - 1 - r)
}
return
}(code-box)
function numTeams(rating: number[]): number {
let ans = 0;
const n = rating.length;
for (let i = 0; i < n; ++i) {
let l = 0;
let r = 0;
for (let j = 0; j < i; ++j) {
if (rating[j] < rating[i]) {
++l;
}
}
for (let j = i + 1; j < n; ++j) {
if (rating[j] > rating[i]) {
++r;
}
}
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}(code-box)
Solution 2: Binary Indexed Tree
We can use two binary indexed trees to maintain the number of elements l that are smaller than each element on the left in the array rating, and the number of elements r that are larger than it on the right. Then count the number of combat units with this element as the middle element as l × r + (i - l) × (n - i - 1 - r), and add this to the answer.
The time complexity is O(n × log n), and the space complexity is O(n). Where n is the length of the array rating.
PythonJavaC++GoTypeScript
class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] += v
x += x & -x
def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s
class Solution:
def numTeams(self, rating: List[int]) -> int:
nums = sorted(set(rating))
m = len(nums)
tree1 = BinaryIndexedTree(m)
tree2 = BinaryIndexedTree(m)
for v in rating:
x = bisect_left(nums, v) + 1
tree2.update(x, 1)
n = len(rating)
ans = 0
for i, v in enumerate(rating):
x = bisect_left(nums, v) + 1
tree1.update(x, 1)
tree2.update(x, -1)
l = tree1.query(x - 1)
r = n - i - 1 - tree2.query(x)
ans += l * r
ans += (i - l) * (n - i - 1 - r)
return ans(code-box)
class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
this.c = new int[n + 1];
}
public void update(int x, int v) {
while (x <= n) {
c[x] += v;
x += x & -x;
}
}
public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= x & -x;
}
return s;
}
}
class Solution {
public int numTeams(int[] rating) {
int n = rating.length;
int[] nums = rating.clone();
Arrays.sort(nums);
int m = 0;
for (int i = 0; i < n; ++i) {
if (i == 0 || nums[i] != nums[i - 1]) {
nums[m++] = nums[i];
}
}
BinaryIndexedTree tree1 = new BinaryIndexedTree(m);
BinaryIndexedTree tree2 = new BinaryIndexedTree(m);
for (int v : rating) {
int x = search(nums, v);
tree2.update(x, 1);
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int x = search(nums, rating[i]);
tree1.update(x, 1);
tree2.update(x, -1);
int l = tree1.query(x - 1);
int r = n - i - 1 - tree2.query(x);
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}
private int search(int[] nums, int x) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
}
}(code-box)
class BinaryIndexedTree {
public:
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}
void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += x & -x;
}
}
int query(int x) {
int s = 0;
while (x) {
s += c[x];
x -= x & -x;
}
return s;
}
private:
int n;
vector<int> c;
};
class Solution {
public:
int numTeams(vector<int>& rating) {
vector<int> nums = rating;
sort(nums.begin(), nums.end());
nums.erase(unique(nums.begin(), nums.end()), nums.end());
int m = nums.size();
BinaryIndexedTree tree1(m);
BinaryIndexedTree tree2(m);
for (int& v : rating) {
int x = lower_bound(nums.begin(), nums.end(), v) - nums.begin() + 1;
tree2.update(x, 1);
}
int ans = 0;
int n = rating.size();
for (int i = 0; i < n; ++i) {
int x = lower_bound(nums.begin(), nums.end(), rating[i]) - nums.begin() + 1;
tree1.update(x, 1);
tree2.update(x, -1);
int l = tree1.query(x - 1);
int r = n - i - 1 - tree2.query(x);
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}
};(code-box)
type BinaryIndexedTree struct {
n int
c []int
}
func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}
func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += x & -x
}
}
func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= x & -x
}
return s
}
func numTeams(rating []int) (ans int) {
nums := make([]int, len(rating))
copy(nums, rating)
sort.Ints(nums)
m := 0
for i, x := range nums {
if i == 0 || x != nums[i-1] {
nums[m] = x
m++
}
}
nums = nums[:m]
tree1 := newBinaryIndexedTree(m)
tree2 := newBinaryIndexedTree(m)
for _, x := range rating {
tree2.update(sort.SearchInts(nums, x)+1, 1)
}
n := len(rating)
for i, v := range rating {
x := sort.SearchInts(nums, v) + 1
tree1.update(x, 1)
tree2.update(x, -1)
l := tree1.query(x - 1)
r := n - i - 1 - tree2.query(x)
ans += l * r
ans += (i - l) * (n - i - 1 - r)
}
return
}(code-box)
class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = new Array(n + 1).fill(0);
}
public update(x: number, v: number): void {
while (x <= this.n) {
this.c[x] += v;
x += x & -x;
}
}
public query(x: number): number {
let s = 0;
while (x > 0) {
s += this.c[x];
x -= x & -x;
}
return s;
}
}
function numTeams(rating: number[]): number {
let nums = [...rating];
nums.sort((a, b) => a - b);
const n = rating.length;
let m = 0;
for (let i = 0; i < n; ++i) {
if (i === 0 || nums[i] !== nums[i - 1]) {
nums[m++] = nums[i];
}
}
nums = nums.slice(0, m);
const search = (x: number): number => {
let l = 0;
let r = m;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
};
let ans = 0;
const tree1 = new BinaryIndexedTree(m);
const tree2 = new BinaryIndexedTree(m);
for (const x of rating) {
tree2.update(search(x), 1);
}
for (let i = 0; i < n; ++i) {
const x = search(rating[i]);
tree1.update(x, 1);
tree2.update(x, -1);
const l = tree1.query(x - 1);
const r = n - i - 1 - tree2.query(x);
ans += l * r;
ans += (i - l) * (n - i - 1 - r);
}
return ans;
}(code-box)
Solution 3: Recursion + Memoization
TypeScriptJavaScript
function numTeams(rating: number[]): number {
const n = rating.length;
const f: Record<Type, number[][]> = {
asc: Array.from({ length: n }, () => Array(3).fill(-1)),
desc: Array.from({ length: n }, () => Array(3).fill(-1)),
};
const fn = (i: number, available: number, type: Type) => {
if (!available) {
return 1;
}
if (f[type][i][available] !== -1) {
return f[type][i][available];
}
let ans = 0;
for (let j = i + 1; j < n; j++) {
if (rating[j] > rating[i]) {
if (type === 'asc') {
ans += fn(j, available - 1, 'asc');
}
} else {
if (type === 'desc') {
ans += fn(j, available - 1, 'desc');
}
}
}
f[type][i][available] = ans;
return ans;
};
let ans = 0;
for (let i = 0; i < n; i++) {
ans += fn(i, 2, 'asc') + fn(i, 2, 'desc');
}
return ans;
}
type Type = 'asc' | 'desc';(code-box)
/**
* @param {number[]} rating
* @return {number}
*/
var numTeams = function (rating) {
const n = rating.length;
const f = {
asc: Array.from({ length: n }, () => Array(3).fill(-1)),
desc: Array.from({ length: n }, () => Array(3).fill(-1)),
};
const fn = (i, available, type) => {
if (!available) {
return 1;
}
if (f[type][i][available] !== -1) {
return f[type][i][available];
}
let ans = 0;
for (let j = i + 1; j < n; j++) {
if (rating[j] > rating[i]) {
if (type === 'asc') {
ans += fn(j, available - 1, 'asc');
}
} else {
if (type === 'desc') {
ans += fn(j, available - 1, 'desc');
}
}
}
f[type][i][available] = ans;
return ans;
};
let ans = 0;
for (let i = 0; i < n; i++) {
ans += fn(i, 2, 'asc') + fn(i, 2, 'desc');
}
return ans;
};(code-box)
