Skip to content

1395. Count Number of Teams

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.length
  • 3 <= n <= 1000
  • 1 <= 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 $l \times r + (i - l) \times (n - i - 1 - r)$. We can add this to the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(1)$. Where $n$ is the length of the array $rating$.

1
2
3
4
5
6
7
8
9
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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;
}

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 $l \times r + (i - l) \times (n - i - 1 - r)$, and add this to the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $rating$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
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;
}

Solution 3: Recursion + Memoization

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
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';

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
/**
 * @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;
};

Comments