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Formatted question description: https://leetcode.ca/all/611.html

611. Valid Triangle Number (Medium)

Given an array consists of non-negative integers, your task is to count 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: [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


Note:

1. The length of the given array won’t exceed 1000.
2. The integers in the given array are in the range of [0, 1000].

Solution 1.

The naive solution is of O(N^3) time complexity, that is, for each triplet, detect if it can form a triangle. This solution will get TLE.

To optimize it, I first sort nums in ascending order. And for each doublet a and b, use binary search to find the count of numbers greater than a + b and less than a - b (a >= b).

// OJ: https://leetcode.com/problems/valid-triangle-number
// Time: O(N^2logN)
// Space: O(1)
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cnt = 0, N = nums.size();
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
int lb = nums[j] - nums[i], rb = nums[i] + nums[j];
int L = j + 1, R = N - 1, left = 0;
while (L <= R) {
int M = (L + R) / 2;
if (nums[M] > lb) R = M - 1;
else L = M + 1;
}
left = L;
L = j + 1, R = N - 1;
while (L <= R) {
int M = (L + R) / 2;
if (nums[M] >= rb) R = M - 1;
else L = M + 1;
}
if (R >= left) cnt += R - left + 1;
}
}
return cnt;
}
};


Solution 2.

Same as solution 1, just uses built-in functions lower_bound and upper_bound.

// OJ: https://leetcode.com/problems/valid-triangle-number
// Time: O(N^2logN)
// Space: O(1)
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cnt = 0, N = nums.size();
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
auto left = upper_bound(nums.begin() + j + 1, nums.end(), nums[j] - nums[i]);
auto right = lower_bound(nums.begin() + j + 1, nums.end(), nums[i] + nums[j]);
if (right > left) cnt += right - left;
}
}
return cnt;
}
};

• class Solution {
public int triangleNumber(int[] nums) {
int length = nums.length;
if (length < 3)
return 0;
int count = 0;
Arrays.sort(nums);
int startIndex = 0;
while (startIndex < length && nums[startIndex] == 0)
startIndex++;
int end1 = length - 2;
int end2 = length - 1;
for (int i = startIndex; i < end1; i++) {
int num1 = nums[i];
for (int j = i + 1; j < end2; j++) {
int num2 = nums[j];
for (int k = j + 1; k < length; k++) {
int num3 = nums[k];
if (num1 + num2 > num3)
count++;
else
break;
}
}
}
return count;
}
}

############

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;
}
}


• // OJ: https://leetcode.com/problems/valid-triangle-number
// Time: O(N^2logN)
// Space: O(1)
class Solution {
public:
int triangleNumber(vector<int>& A) {
int N = A.size(), ans = 0;
sort(begin(A), end(A));
for (int i = 0; i < N; ++i) {
int a = A[i];
for (int j = i + 1; j < N; ++j) {
int b = A[j];
auto begin = upper_bound(A.begin() + j + 1, A.end(), b - a);
auto end = lower_bound(A.begin() + j + 1, A.end(), a + b);
ans += max(0, (int)(end - begin));
}
}
return ans;
}
};

• 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

############

class Solution(object):
def triangleNumber(self, nums):
ans = 0
nums.sort()
for i in range(2, len(nums)):
start = 0
end = i - 1
while start < end:
if nums[start] + nums[end] > nums[i]:
ans += end - start
end -= 1
else:
start += 1
return ans


• 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
}

• 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;
}


• 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
}
}