Description#
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 <= 1000
0 <= nums[i] <= 1000
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
| class Solution:
def triangleNumber(self, nums: List[int]) -> int:
nums.sort()
ans, n = 0, len(nums)
for i in range(n - 2):
for j in range(i + 1, n - 1):
k = bisect_left(nums, nums[i] + nums[j], lo=j + 1) - 1
ans += k - j
return ans
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
| class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int res = 0;
for (int i = n - 1; i >= 2; --i) {
int l = 0, r = i - 1;
while (l < r) {
if (nums[l] + nums[r] > nums[i]) {
res += r - l;
--r;
} else {
++l;
}
}
}
return res;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
| class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 0, n = nums.size();
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
int k = lower_bound(nums.begin() + j + 1, nums.end(), nums[i] + nums[j]) - nums.begin() - 1;
ans += k - j;
}
}
return ans;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
| func triangleNumber(nums []int) int {
sort.Ints(nums)
ans := 0
for i, n := 0, len(nums); i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
left, right := j+1, n
for left < right {
mid := (left + right) >> 1
if nums[mid] >= nums[i]+nums[j] {
right = mid
} else {
left = mid + 1
}
}
ans += left - j - 1
}
}
return ans
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
| function triangleNumber(nums: number[]): number {
nums.sort((a, b) => a - b);
let n = nums.length;
let ans = 0;
for (let i = n - 1; i >= 2; i--) {
let left = 0,
right = i - 1;
while (left < right) {
if (nums[left] + nums[right] > nums[i]) {
ans += right - left;
right--;
} else {
left++;
}
}
}
return ans;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| impl Solution {
pub fn triangle_number(mut nums: Vec<i32>) -> i32 {
nums.sort();
let n = nums.len();
let mut res = 0;
for i in (2..n).rev() {
let mut left = 0;
let mut right = i - 1;
while left < right {
if nums[left] + nums[right] > nums[i] {
res += right - left;
right -= 1;
} else {
left += 1;
}
}
}
res as i32
}
}
|
Solution 2#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int ans = 0;
for (int i = 0, n = nums.length; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
int left = j + 1, right = n;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] >= nums[i] + nums[j]) {
right = mid;
} else {
left = mid + 1;
}
}
ans += left - j - 1;
}
}
return ans;
}
}
|