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Formatted question description: https://leetcode.ca/all/2419.html

2419. Longest Subarray With Maximum Bitwise AND

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Number of Different Integers in a String, Remove Colored Pieces if Both Neighbors are the Same Color, Count Number of Maximum Bitwise-OR Subsets, Smallest Subarrays With Maximum Bitwise OR.

Problem

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 <= 10^5

  • 1 <= nums[i] <= 10^6

Solution (Java, C++, Python)

  • 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 := 0
	for _, v := range nums {
		mx = max(mx, v)
	}
	ans, cnt := 0, 0
	for _, v := range nums {
		if v == mx {
			cnt++
			ans = max(ans, cnt)
		} else {
			cnt = 0
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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