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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0611_valid_triangle_number;

// #Medium #Array #Sorting #Greedy #Binary_Search #Two_Pointers #Binary_Search_II_Day_1
// #2022_03_21_Time_10_ms_(100.00%)_Space_45_MB_(5.06%)

/**
 * 611 - Valid Triangle Number\.
 *
 * Medium
 *
 * Given an integer array `nums`, return _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:** nums = [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
 *
 * **Example 2:**
 *
 * **Input:** nums = [4,2,3,4]
 *
 * **Output:** 4
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 1000`
 * *   `0 <= nums[i] <= 1000`
**/
public class Solution {
    public int triangleNumber(int[] nums) {
        int n;
        int max = 0;
        int[] count = new int[1001];
        for (int i : nums) {
            count[i]++;
            max = Math.max(max, i);
        }
        count[0] = 0;
        int idx = 0;
        for (int i = 1; i <= max; ++i) {
            for (int j = 0; j < count[i]; ++j, ++idx) {
                nums[idx] = i;
            }
            count[i] += count[i - 1];
        }
        n = idx;
        int r = 0;
        for (int i = 0; i < n - 2; ++i) {
            for (int j = i + 1; j < n - 1; ++j) {
                if (nums[i] + nums[j] > max) {
                    r += (n - j) * (n - j - 1) / 2;
                    break;
                }
                r += count[nums[i] + nums[j] - 1] - j - 1;
            }
        }
        return r;
    }
}