2419 - Longest Subarray With Maximum Bitwise AND (Medium)
Problem Link
https://leetcode.com/problems/longest-subarray-with-maximum-bitwise-and/
Problem Statement
You are given an integer array nums of size n.
Consider a non-empty subarray from nums that has the maximum possible bitwise AND.
- In other words, let
kbe the maximum value of the bitwise AND of any subarray ofnums. Then, only subarrays with a bitwise AND equal tokshould be considered.
Return the length of the longest such subarray.
The bitwise AND of an array is the bitwise AND of all the numbers in it.
A subarray is a contiguous sequence of elements within an array.
Example 1:
Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.
Example 2:
Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.
Constraints:
1 <= nums.length <= 1e51 <= nums[i] <= 10^6
Approach 1: Consecutive Max Element
- C++
- Java
class Solution {
public:
int longestSubarray(vector<int>& nums) {
// the bitwise AND of two different numbers will always be strictly less than the maximum of those two numbers
// so the longest subarray with max bitwise AND would be the subarray containing only the max numbers
int mx = *max_element(nums.begin(), nums.end());
int ans, cnt = 0;
for (auto &x : nums) {
// increase the count by 1 if it is same as the max number
if (x == mx) cnt += 1;
// else reset it
else cnt = 0;
// update ans
ans = max(ans, cnt);
}
return ans;
}
};
class Solution {
public int longestSubarray(int[] nums) {
int max = 0, longest = 1, current = 0;
for (int num : nums) {
max = Math.max(max, num);
}
for (int num : nums) {
if (num == max) {
longest = Math.max(longest, ++current);
} else {
current = 0;
}
}
return longest;
}
}