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# Valid Triangle Number

## Description

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].

## Solutions

### My solutions

#### Solution 1

```cpp
class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int count = 0, n = nums.size();
        for(int i = 0; i + 2 < n; ++i){
            for(int j = i + 1; j + 1 < n; ++j){
                int k = j + 1;
                while(k < n && nums[k] < nums[i] + nums[j]) ++k;
                count += k - j - 1;
            }
        }
        return count;
    }
};
```

#### Solution 2

```cpp
class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int count = 0, n = nums.size();
        int i = 0;
        while(i < n && nums[i] == 0) ++i;
        for( ; i + 2 < n; ++i){
            int k = i + 2;
            for(int j = i + 1; j + 1 < n; ++j){
                while(k < n && nums[k] < nums[i] + nums[j]) ++k;
                count += k - j - 1;
            }
        }
        return count;
    }
};
```

### A better solution

The previous solutions fix two shorter sides. This solution fixes the longest side and then use binary search to find the other two sides.

```cpp
class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        int count = 0;
        int n = nums.size();
        sort(nums.begin(), nums.end());
        for(int i = n - 1; i >= 2; --i){
            int l = 0, r = i - 1;
            while(l < r){
                if(nums[l] + nums[r] > nums[i]){
                    count += r - l;
                    --r;
                }else{
                    ++l;
                }
            }
        }
        return count;
    }
};
```