Valid Triangle Number – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
Example 1:
Input: nums = [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) 2,2,3
Example 2:
Input: nums = [4,2,3,4] Output: 4
Constraints:
1 <= nums.length <= 10000 <= nums[i] <= 1000
C++ Valid Triangle Number LeetCode Solution
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size(), ans = 0;
for (int k = 2; k < n; ++k) {
int i = 0, j = k – 1;
while (i < j) {
if (nums[i] + nums[j] > nums[k]) {
ans += j – i;
j -= 1;
} else {
i += 1;
}
}
}
return ans;
}
};
Java Valid Triangle Number LeetCode Solution
class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int n = nums.length, ans = 0;
for (int k = 2; k < n; ++k) {
int i = 0, j = k – 1;
while (i < j) {
if (nums[i] + nums[j] > nums[k]) {
ans += j – i;
j -= 1;
} else {
i += 1;
}
}
}
return ans;
}
}
Python 3 Valid Triangle Number LeetCode Solution
class Solution:
def triangleNumber(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
ans = 0
for k in range(2, n):
i = 0
j = k – 1
while i < j:
if nums[i] + nums[j] > nums[k]:
ans += j – i
j -= 1
else:
i += 1
return ans
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