g2301_2400.s2317_maximum_xor_after_operations.Solution Maven / Gradle / Ivy

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package g2301_2400.s2317_maximum_xor_after_operations;

// #Medium #Array #Math #Bit_Manipulation #2022_06_26_Time_1_ms_(100.00%)_Space_53.3_MB_(100.00%)

/**
 * 2317 - Maximum XOR After Operations\.
 *
 * Medium
 *
 * You are given a **0-indexed** integer array `nums`. In one operation, select **any** non-negative integer `x` and an index `i`, then **update** `nums[i]` to be equal to `nums[i] AND (nums[i] XOR x)`.
 *
 * Note that `AND` is the bitwise AND operation and `XOR` is the bitwise XOR operation.
 *
 * Return _the **maximum** possible bitwise XOR of all elements of_ `nums` _after applying the operation **any number** of times_.
 *
 * **Example 1:**
 *
 * **Input:** nums = [3,2,4,6]
 *
 * **Output:** 7
 *
 * **Explanation:** Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
 *
 * Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
 *
 * It can be shown that 7 is the maximum possible bitwise XOR.
 *
 * Note that other operations may be used to achieve a bitwise XOR of 7.
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,2,3,9,2]
 *
 * **Output:** 11
 *
 * **Explanation:** Apply the operation zero times.
 *
 * The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
 *
 * It can be shown that 11 is the maximum possible bitwise XOR.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   0 <= nums[i] <= 10⁸
**/
public class Solution {
    public int maximumXOR(int[] nums) {
        int max = 0;
        for (int n : nums) {
            max |= n;
        }
        return max;
    }
}