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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2301_2400.s2354_number_of_excellent_pairs;

// #Hard #Array #Hash_Table #Binary_Search #Bit_Manipulation
// #2022_08_07_Time_80_ms_(86.77%)_Space_109.2_MB_(29.50%)

import java.util.HashSet;
import java.util.Set;

/**
 * 2354 - Number of Excellent Pairs\.
 *
 * Hard
 *
 * You are given a **0-indexed** positive integer array `nums` and a positive integer `k`.
 *
 * A pair of numbers `(num1, num2)` is called **excellent** if the following conditions are satisfied:
 *
 * *   **Both** the numbers `num1` and `num2` exist in the array `nums`.
 * *   The sum of the number of set bits in `num1 OR num2` and `num1 AND num2` is greater than or equal to `k`, where `OR` is the bitwise **OR** operation and `AND` is the bitwise **AND** operation.
 *
 * Return _the number of **distinct** excellent pairs_.
 *
 * Two pairs `(a, b)` and `(c, d)` are considered distinct if either `a != c` or `b != d`. For example, `(1, 2)` and `(2, 1)` are distinct.
 *
 * **Note** that a pair `(num1, num2)` such that `num1 == num2` can also be excellent if you have at least **one** occurrence of `num1` in the array.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,2,3,1], k = 3
 *
 * **Output:** 5
 *
 * **Explanation:** The excellent pairs are the following:
 *
 * - (3, 3). (3 AND 3) and (3 OR 3) are both equal to (11) in binary. The total number of set bits is 2 + 2 = 4, which is greater than or equal to k = 3.
 *
 * - (2, 3) and (3, 2). (2 AND 3) is equal to (10) in binary, and (2 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3.
 *
 * - (1, 3) and (3, 1). (1 AND 3) is equal to (01) in binary, and (1 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3.
 *
 * So the number of excellent pairs is 5.
 *
 * **Example 2:**
 *
 * **Input:** nums = [5,1,1], k = 10
 *
 * **Output:** 0
 *
 * **Explanation:** There are no excellent pairs for this array. 
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   1 <= nums[i] <= 10⁹
 * *   `1 <= k <= 60`
**/
public class Solution {
    public long countExcellentPairs(int[] nums, int k) {
        long[] cnt = new long[30];
        long res = 0L;
        Set set = new HashSet<>();
        for (int a : nums) {
            set.add(a);
        }
        for (int a : set) {
            cnt[Integer.bitCount(a)]++;
        }
        for (int i = 1; i < 30; ++i) {
            for (int j = 1; j < 30; ++j) {
                if (i + j >= k) {
                    res += cnt[i] * cnt[j];
                }
            }
        }
        return res;
    }
}