611. Valid Triangle Number

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

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.


// solution 1 : 
class Solution {
    public int triangleNumber(int[] nums) {
        // 219 / 220 test cases passed.
        // tle , time : n ^ 3 
        // brute force solution: try all possible combination of any three numbers 
        // check if its valid , 
        int res = 0;
        for(int i = 0; i < nums.length - 2; i++){
            for(int j = i + 1; j < nums.length - 1; j++){
                for(int k = j + 1; k < nums.length; k++){
                    int a = nums[i], b = nums[j], c = nums[k];
                    if(a + b > c && b + c > a && a + c > b) res++;
                }
            }
        }
        return res;
        
    }
}


// solution 2 

https://www.youtube.com/watch?v=bojX9bdra-w
Sort it first 
Greedy: use math tricks to avoid trying all possible solutions , still guarantee the correctness of the result 


?? How to guarantee that  a + b > c 
Also means a + c > b and b + c > a , if only one ineuqkaity is valid and we change b to d , how to ensure the other two are also valid ?? 

class Solution {
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);
        //  use some space or tricks to trade time (binary search)
        // what's some inefficeincy about this algorithms 
        int res = 0;
        for(int i = nums.length - 1; i >= 2; i--){ // fix i 
            int l = 0;
            int r = i - 1;
            while(l < r){
                if(nums[l] + nums[r] > nums[i]){
                    res += r - l ; // 
                    r--; // fix one r , try all possible l 
                }else{
                    // nums[l] + nums[r] <= nums[i]
                    // need to move l to the right 
                    l++;
                }
            }
        }
        return res;
        
    }
}