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2760. Longest Even Odd Subarray With Threshold

Description

You are given a 0-indexed integer array nums and an integer threshold.

Find the length of the longest subarray of nums starting at index l and ending at index r (0 <= l <= r < nums.length) that satisfies the following conditions:

• nums[l] % 2 == 0
• For all indices i in the range [l, r - 1], nums[i] % 2 != nums[i + 1] % 2
• For all indices i in the range [l, r], nums[i] <= threshold

Return an integer denoting the length of the longest such subarray.

Note: A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [3,2,5,4], threshold = 5
Output: 3
Explanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 3 => [2,5,4]. This subarray satisfies the conditions.
Hence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.

Example 2:

Input: nums = [1,2], threshold = 2
Output: 1
Explanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 1 => [2]. 
It satisfies all the conditions and we can show that 1 is the maximum possible achievable length.

Example 3:

Input: nums = [2,3,4,5], threshold = 4
Output: 3
Explanation: In this example, we can select the subarray that starts at l = 0 and ends at r = 2 => [2,3,4]. 
It satisfies all the conditions.
Hence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.

 

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= threshold <= 100

Solutions

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

• class Solution {
      public int longestAlternatingSubarray(int[] nums, int threshold) {
          int ans = 0, n = nums.length;
          for (int l = 0; l < n; ++l) {
              if (nums[l] % 2 == 0 && nums[l] <= threshold) {
                  int r = l + 1;
                  while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
                      ++r;
                  }
                  ans = Math.max(ans, r - l);
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int longestAlternatingSubarray(vector<int>& nums, int threshold) {
          int ans = 0, n = nums.size();
          for (int l = 0; l < n; ++l) {
              if (nums[l] % 2 == 0 && nums[l] <= threshold) {
                  int r = l + 1;
                  while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
                      ++r;
                  }
                  ans = max(ans, r - l);
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:
          ans, n = 0, len(nums)
          for l in range(n):
              if nums[l] % 2 == 0 and nums[l] <= threshold:
                  r = l + 1
                  while r < n and nums[r] % 2 != nums[r - 1] % 2 and nums[r] <= threshold:
                      r += 1
                  ans = max(ans, r - l)
          return ans
  
  

• func longestAlternatingSubarray(nums []int, threshold int) int {
  	ans, n := 0, len(nums)
  	for l := range nums {
  		if nums[l]%2 == 0 && nums[l] <= threshold {
  			r := l + 1
  			for r < n && nums[r]%2 != nums[r-1]%2 && nums[r] <= threshold {
  				r++
  			}
  			ans = max(ans, r-l)
  		}
  	}
  	return ans
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• function longestAlternatingSubarray(nums: number[], threshold: number): number {
      const n = nums.length;
      let ans = 0;
      for (let l = 0; l < n; ++l) {
          if (nums[l] % 2 === 0 && nums[l] <= threshold) {
              let r = l + 1;
              while (r < n && nums[r] % 2 !== nums[r - 1] % 2 && nums[r] <= threshold) {
                  ++r;
              }
              ans = Math.max(ans, r - l);
          }
      }
      return ans;
  }
  
  

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