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2419. Longest Subarray With Maximum Bitwise AND

Description

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 k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should 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 <= 105
• 1 <= nums[i] <= 106

Solutions

Solution 1: Quick Thinking

Due to the bitwise AND operation, the number will not get larger, so the maximum value is the maximum value in the array.

The problem can be transformed into finding the maximum number of consecutive occurrences of the maximum value in the array.

First, traverse the array once to find the maximum value, then traverse the array again to find the number of consecutive occurrences of the maximum value, and finally return this count.

The time complexity is $O(n)$, where $n$ is the length of the array.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      public int longestSubarray(int[] nums) {
          int mx = 0;
          for (int v : nums) {
              mx = Math.max(mx, v);
          }
          int ans = 0, cnt = 0;
          for (int v : nums) {
              if (v == mx) {
                  ++cnt;
                  ans = Math.max(ans, cnt);
              } else {
                  cnt = 0;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int longestSubarray(vector<int>& nums) {
          int mx = *max_element(nums.begin(), nums.end());
          int ans = 0, cnt = 0;
          for (int v : nums) {
              if (v == mx) {
                  ++cnt;
                  ans = max(ans, cnt);
              } else {
                  cnt = 0;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def longestSubarray(self, nums: List[int]) -> int:
          mx = max(nums)
          ans = cnt = 0
          for v in nums:
              if v == mx:
                  cnt += 1
                  ans = max(ans, cnt)
              else:
                  cnt = 0
          return ans
  
  

• func longestSubarray(nums []int) int {
  	mx := slices.Max(nums)
  	ans, cnt := 0, 0
  	for _, v := range nums {
  		if v == mx {
  			cnt++
  			ans = max(ans, cnt)
  		} else {
  			cnt = 0
  		}
  	}
  	return ans
  }
  

• function longestSubarray(nums: number[]): number {
      const mx = Math.max(...nums);
      let [ans, cnt] = [0, 0];
  
      for (const x of nums) {
          if (x === mx) {
              cnt++;
              ans = Math.max(ans, cnt);
          } else {
              cnt = 0;
          }
      }
  
      return ans;
  }
  
  

• function longestSubarray(nums) {
      const mx = Math.max(...nums);
      let [ans, cnt] = [0, 0];
  
      for (const x of nums) {
          if (x === mx) {
              cnt++;
              ans = Math.max(ans, cnt);
          } else {
              cnt = 0;
          }
      }
  
      return ans;
  }
  
  

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