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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2601_2700.s2680_maximum_or;

// #Medium #Array #Greedy #Bit_Manipulation #Prefix_Sum
// #2023_09_11_Time_2_ms_(100.00%)_Space_55.2_MB_(56.19%)

/**
 * 2680 - Maximum OR\.
 *
 * Medium
 *
 * You are given a **0-indexed** integer array `nums` of length `n` and an integer `k`. In an operation, you can choose an element and multiply it by `2`.
 *
 * Return _the maximum possible value of_ `nums[0] | nums[1] | ... | nums[n - 1]` _that can be obtained after applying the operation on nums at most_ `k` _times_.
 *
 * Note that `a | b` denotes the **bitwise or** between two integers `a` and `b`.
 *
 * **Example 1:**
 *
 * **Input:** nums = [12,9], k = 1
 *
 * **Output:** 30
 *
 * **Explanation:** If we apply the operation to index 1, our new array nums will be equal to [12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.
 *
 * **Example 2:**
 *
 * **Input:** nums = [8,1,2], k = 2
 *
 * **Output:** 35
 *
 * **Explanation:** If we apply the operation twice on index 0, we yield a new array of [32,1,2]. Thus, we return 32|1|2 = 35.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   1 <= nums[i] <= 10⁹
 * *   `1 <= k <= 15`
**/
public class Solution {
    public long maximumOr(int[] nums, int k) {
        int[] suffix = new int[nums.length];
        for (int i = nums.length - 2; i >= 0; i--) {
            suffix[i] = suffix[i + 1] | nums[i + 1];
        }
        long prefix = 0L;
        long max = 0L;
        for (int i = 0; i <= nums.length - 1; i++) {
            long num = nums[i];
            max = Math.max(max, prefix | (num << k) | suffix[i]);
            prefix |= num;
        }
        return max;
    }
}