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;
}
};


Java

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;
}
}