**611. Valid Triangle Number three pointer O(n^3) -> square(binary search larget number smaller than target)

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].

reference -- https://leetcode.com/problems/valid-triangle-number/solution/

Idea: it is not like permutation problem(for each position, there are many cases to fill in), because we choose some elements from the array.

sort: to check only one case a+b >c instead of three of them

a +b > c and sorting the array

accepted solution 1: binary search

class Solution {
    int res = 0;
    public int binarySearch(int[] nums, int left, int right, int target){//[]
        while(right >= left && right < nums.length ){//stop comdition is right >= left: there is onl one element
            int mid  = (right - left)/2 + left;
            if(nums[mid] >= target){
                right = mid -1;
            }else {
                left = mid + 1;
            }
        }
        return left;// it is on the right of the number we want: real target, left(index)
    }
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        for(int i = 0 ; i <= n-3 ; i++){
            int a = nums[i];
            for(int j = i+1 ; j <= n-2 ; j++){
                int b = nums[j];
                //System.out.println(j+1);
                int index = binarySearch(nums,j+1,n-1, a+b ); //[]// find the largest element smaller than a+b
                //System.out.println(index);
                //System.out.println(nums[index]);
                res  = res + index - j-1; //why - 1??
            }
        }
        return res;
    }
}

Accepted solution 2 -- O(n^2)

class Solution {
    public int triangleNumber(int[] nums) {
        int res = 0;
        //int n = nums.length;
        Arrays.sort(nums);
        for(int i = 0; i < nums.length-2; i++){
            if(nums[i] == 0) continue;
            int k = i + 2; // why here
            if(nums[k] == 0) continue;
            for(int j = i+1 ; j<nums.length-1; j++){
                if(nums[j] == 0) continue;
                while(k < nums.length && nums[k]< nums[i]+nums[j] && nums[k]!=0){
                    k++;// move k
                }
                res = res + k - j -1;
            }
        }
        return res;
    }
}

traverse k and j only n^2 -> square time complexixity