Question

2475. Number of Unequal Triplets in Array

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Description

You are given a 0-indexed array of positive integers nums. Find the number of triplets (i, j, k) that meet the following conditions:

• 0 <= i < j < k < nums.length
• nums[i], nums[j], and nums[k] are pairwise distinct.
  • In other words, nums[i] != nums[j], nums[i] != nums[k], and nums[j] != nums[k].

Return _the number of triplets that meet the conditions._

 

Example 1:

Input: nums = [4,4,2,4,3]
Output: 3
Explanation: The following triplets meet the conditions:
- (0, 2, 4) because 4 != 2 != 3
- (1, 2, 4) because 4 != 2 != 3
- (2, 3, 4) because 2 != 4 != 3
Since there are 3 triplets, we return 3.
Note that (2, 0, 4) is not a valid triplet because 2 > 0.

Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: No triplets meet the conditions so we return 0.

 

Constraints:

• 3 <= nums.length <= 100
• 1 <= nums[i] <= 1000

Solutions

Solution 1: Brute Force Enumeration

We can directly enumerate all triples $(i, j, k)$ and count all the ones that meet the conditions.

The time complexity is $O(n^3)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

class Solution:
  def unequalTriplets(self, nums: List[int]) -> int:
    n = len(nums)
    ans = 0
    for i in range(n):
      for j in range(i + 1, n):
        for k in range(j + 1, n):
          ans += (
            nums[i] != nums[j] and nums[j] != nums[k] and nums[i] != nums[k]
          )
    return ans

class Solution {
  public int unequalTriplets(int[] nums) {
    int n = nums.length;
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        for (int k = j + 1; k < n; ++k) {
          if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) {
            ++ans;
          }
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int unequalTriplets(vector<int>& nums) {
    int n = nums.size();
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        for (int k = j + 1; k < n; ++k) {
          if (nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k]) {
            ++ans;
          }
        }
      }
    }
    return ans;
  }
};

func unequalTriplets(nums []int) (ans int) {
  n := len(nums)
  for i := 0; i < n; i++ {
    for j := i + 1; j < n; j++ {
      for k := j + 1; k < n; k++ {
        if nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k] {
          ans++
        }
      }
    }
  }
  return
}

function unequalTriplets(nums: number[]): number {
  const n = nums.length;
  let ans = 0;
  for (let i = 0; i < n - 2; i++) {
    for (let j = i + 1; j < n - 1; j++) {
      for (let k = j + 1; k < n; k++) {
        if (nums[i] !== nums[j] && nums[j] !== nums[k] && nums[i] !== nums[k]) {
          ans++;
        }
      }
    }
  }
  return ans;
}

impl Solution {
  pub fn unequal_triplets(nums: Vec<i32>) -> i32 {
    let n = nums.len();
    let mut ans = 0;
    for i in 0..n - 2 {
      for j in i + 1..n - 1 {
        for k in j + 1..n {
          if nums[i] != nums[j] && nums[j] != nums[k] && nums[i] != nums[k] {
            ans += 1;
          }
        }
      }
    }
    ans
  }
}

Solution 2: Sorting + Enumeration of Middle Elements + Binary Search

We can also sort the array $nums$ first.

Then traverse $nums$, enumerate the middle element $nums[j]$, and use binary search to find the nearest index $i$ on the left side of $nums[j]$ such that $nums[i] < nums[j]$; find the nearest index $k$ on the right side of $nums[j]$ such that $nums[k] > nums[j]$. Then the number of triples with $nums[j]$ as the middle element and meeting the conditions is $(i + 1) \times (n - k)$, which is added to the answer.

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

class Solution:
  def unequalTriplets(self, nums: List[int]) -> int:
    nums.sort()
    ans, n = 0, len(nums)
    for j in range(1, n - 1):
      i = bisect_left(nums, nums[j], hi=j) - 1
      k = bisect_right(nums, nums[j], lo=j + 1)
      ans += (i >= 0 and k < n) * (i + 1) * (n - k)
    return ans

class Solution {
  public int unequalTriplets(int[] nums) {
    Arrays.sort(nums);
    int ans = 0, n = nums.length;
    for (int j = 1; j < n - 1; ++j) {
      int i = search(nums, nums[j], 0, j) - 1;
      int k = search(nums, nums[j] + 1, j + 1, n);
      if (i >= 0 && k < n) {
        ans += (i + 1) * (n - k);
      }
    }
    return ans;
  }

  private int search(int[] nums, int x, int left, int right) {
    while (left < right) {
      int mid = (left + right) >> 1;
      if (nums[mid] >= x) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return left;
  }
}

class Solution {
public:
  int unequalTriplets(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int ans = 0, n = nums.size();
    for (int j = 1; j < n - 1; ++j) {
      int i = lower_bound(nums.begin(), nums.begin() + j, nums[j]) - nums.begin() - 1;
      int k = upper_bound(nums.begin() + j + 1, nums.end(), nums[j]) - nums.begin();
      if (i >= 0 && k < n) {
        ans += (i + 1) * (n - k);
      }
    }
    return ans;
  }
};

func unequalTriplets(nums []int) (ans int) {
  sort.Ints(nums)
  n := len(nums)
  for j := 1; j < n-1; j++ {
    i := sort.Search(j, func(h int) bool { return nums[h] >= nums[j] }) - 1
    k := sort.Search(n, func(h int) bool { return h > j && nums[h] > nums[j] })
    if i >= 0 && k < n {
      ans += (i + 1) * (n - k)
    }
  }
  return
}

function unequalTriplets(nums: number[]): number {
  const n = nums.length;
  const cnt = new Map<number, number>();
  for (const num of nums) {
    cnt.set(num, (cnt.get(num) ?? 0) + 1);
  }
  let ans = 0;
  let a = 0;
  for (const b of cnt.values()) {
    const c = n - a - b;
    ans += a * b * c;
    a += b;
  }
  return ans;
}

use std::collections::HashMap;
impl Solution {
  pub fn unequal_triplets(nums: Vec<i32>) -> i32 {
    let mut cnt = HashMap::new();
    for num in nums.iter() {
      *cnt.entry(num).or_insert(0) += 1;
    }
    let n = nums.len();
    let mut ans = 0;
    let mut a = 0;
    for v in cnt.values() {
      let b = n - a - v;
      ans += v * a * b;
      a += v;
    }
    ans as i32
  }
}

Solution 3: Hash Table

We can also use a hash table $cnt$ to count the number of each element in the array $nums$.

Then traverse the hash table $cnt$, enumerate the number of middle elements $b$, and denote the number of elements on the left as $a$. Then the number of elements on the right is $c = n - a - b$. At this time, the number of triples that meet the conditions is $a \times b \times c$, which is added to the answer. Then update $a = a + b$ and continue to enumerate the number of middle elements $b$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

class Solution:
  def unequalTriplets(self, nums: List[int]) -> int:
    cnt = Counter(nums)
    n = len(nums)
    ans = a = 0
    for b in cnt.values():
      c = n - a - b
      ans += a * b * c
      a += b
    return ans

class Solution {
  public int unequalTriplets(int[] nums) {
    Map<Integer, Integer> cnt = new HashMap<>();
    for (int v : nums) {
      cnt.merge(v, 1, Integer::sum);
    }
    int ans = 0, a = 0;
    int n = nums.length;
    for (int b : cnt.values()) {
      int c = n - a - b;
      ans += a * b * c;
      a += b;
    }
    return ans;
  }
}

class Solution {
public:
  int unequalTriplets(vector<int>& nums) {
    unordered_map<int, int> cnt;
    for (int& v : nums) {
      ++cnt[v];
    }
    int ans = 0, a = 0;
    int n = nums.size();
    for (auto& [_, b] : cnt) {
      int c = n - a - b;
      ans += a * b * c;
      a += b;
    }
    return ans;
  }
};

func unequalTriplets(nums []int) (ans int) {
  cnt := map[int]int{}
  for _, v := range nums {
    cnt[v]++
  }
  a, n := 0, len(nums)
  for _, b := range cnt {
    c := n - a - b
    ans += a * b * c
    a += b
  }
  return
}

use std::collections::HashMap;

impl Solution {
  pub fn unequal_triplets(nums: Vec<i32>) -> i32 {
    let cnt = nums.iter().fold(HashMap::new(), |mut map, &n| {
      *map.entry(n).or_insert(0) += 1;
      map
    });

    let mut ans = 0;
    let n = nums.len();
    let mut a = 0;
    for &b in cnt.values() {
      let c = n - a - b;
      ans += a * b * c;
      a += b;
    }

    ans as i32
  }
}

Solution 4

impl Solution {
  pub fn unequal_triplets(nums: Vec<i32>) -> i32 {
    let mut ans = 0;
    let mut nums = nums;
    nums.sort();
    let n = nums.len();

    for i in 1..n - 1 {
      let mut l = 0;
      let mut r = i;
      while l < r {
        let mid = (l + r) >> 1;
        if nums[mid] >= nums[i] {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      let j = r;

      let mut l = i + 1;
      let mut r = n;
      while l < r {
        let mid = (l + r) >> 1;
        if nums[mid] > nums[i] {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      let k = r;

      ans += j * (n - k);
    }

    ans as i32
  }
}