Here, we will see how to solve Number of Unequal Triplets in Array Solution of leet code 2475 problem.
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], andnums[k]are pairwise distinct.- In other words,
nums[i] != nums[j],nums[i] != nums[k], andnums[j] != nums[k].
- In other words,
You have to return the number of triplets that meet the conditions.
Example 1:
Input: nums = [5,5,3,5,4] Output: 3 Explanation: The following triplets meet the conditions: - (0, 2, 4) because 5 != 3 != 4 - (1, 2, 4) because 5 != 3 != 4 - (2, 3, 4) because 3 != 5 != 4 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 = [2,2,2,2,2] Output: 0 Explanation: No triplets meet the conditions so we return 0.
Number of Unequal Triplets in Array Solution code in C++ and Go lang:
Here, we will be solving problem in multiple ways with code.
C++ code 1:
class Solution {
public:
int unequalTriplets(vector<int>& nums) {
int n = nums.size();
int ret = n * (n - 1) * (n - 2) / 6;
map<int, int> count;
for (int i: nums) {
count[i] += 1;
}
for (auto p: count) {
int f = p.second;
ret -= f * (f - 1) * (n - f) / 2;
ret -= f * (f - 1) * (f - 2) / 6;
}
return ret;
}
};C++ code 2:
class Solution {
public:
int unequalTriplets(vector<int>& nums) {
int n = nums.size();
int ret = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j]) {
continue;
}
for (int k = j + 1; k < n; ++k) {
if (nums[k] == nums[j] || nums[k] == nums[i]) {
continue;
}
++ret;
}
}
}
return ret;
}
};C++ code 3:
class Solution {
public:
int unequalTriplets(vector<int>& nums) {
int count = 0;
for(int i = 0; i < nums.size() - 2; i++) {
for(int j = i + 1; j < nums.size() - 1; j++) {
for(int k = j + 1; k < nums.size(); k++) {
if ((nums[i] != nums[j]) && (nums[i] != nums[k]) && (nums[j] != nums[k])) {
count++;
}
}
}
}
return count;
}
};Go code 1:
func unequalTriplets(nums []int) int {
var i, j, k, count = 0, 0, 0, 0
for ; i < len(nums) - 2; i++ {
for j = i + 1; j < len(nums) - 1; j++ {
for k = j + 1; k < len(nums); k++ {
if (nums[i] != nums[j]) && (nums[i] != nums[k]) && (nums[j] != nums[k]) {
count++;
}
}
}
}
return count
}Output:
Input: nums = [5,5,3,5,4] Output: 3
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