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.lengthnums[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 <= 1001 <= 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 $O(1)$.
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| 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 $nums$.
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| 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 $nums$.
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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
}
}
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Solution 4#
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| 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
}
}
|
