LeetCode 2419: Longest Subarray With Maximum Bitwise AND
LeetCode 2419 Solution Explanation
Explanation:
To solve this problem, we can iterate through the array nums
to find the maximum bitwise AND value k
. Once we have k
, we can iterate through the array again to find the longest subarray that has a bitwise AND equal to k
. We can achieve this by keeping track of the start index of the current subarray and updating the length of the longest subarray when we encounter a bitwise AND equal to k
.
Algorithmic Steps:
- Iterate through
nums
to find the maximum bitwise AND valuek
. - Iterate through
nums
again to find the longest subarray with a bitwise AND equal tok
. - Keep track of the start index of the current subarray and update the length of the longest subarray when necessary.
Time Complexity:
The time complexity of this approach is O(n), where n is the length of the input array nums
.
Space Complexity:
The space complexity of this approach is O(1) as we are using a constant amount of extra space.
:
LeetCode 2419 Solutions in Java, C++, Python
class Solution {
public int longestSubarray(int[] nums) {
int maxBitwiseAND = 0;
for (int num : nums) {
maxBitwiseAND = Math.max(maxBitwiseAND, num);
}
int start = 0;
int maxLength = 0;
for (int i = 0; i < nums.length; i++) {
if ((nums[i] & maxBitwiseAND) != maxBitwiseAND) {
maxLength = Math.max(maxLength, i - start);
start = i;
}
}
return Math.max(maxLength, nums.length - start);
}
}
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