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594. Longest Harmonious Subsequence
Description
We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1.
Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.
A subsequence of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [1,3,2,2,5,2,3,7] Output: 5 Explanation: The longest harmonious subsequence is [3,2,2,2,3].
Example 2:
Input: nums = [1,2,3,4] Output: 2
Example 3:
Input: nums = [1,1,1,1] Output: 0
Constraints:
1 <= nums.length <= 2 * 104-109 <= nums[i] <= 109
Solutions
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class Solution { public int findLHS(int[] nums) { Map<Integer, Integer> counter = new HashMap<>(); for (int num : nums) { counter.put(num, counter.getOrDefault(num, 0) + 1); } int ans = 0; for (int num : nums) { if (counter.containsKey(num + 1)) { ans = Math.max(ans, counter.get(num) + counter.get(num + 1)); } } return ans; } } -
class Solution { public: int findLHS(vector<int>& nums) { unordered_map<int, int> counter; for (int num : nums) { ++counter[num]; } int ans = 0; for (int num : nums) { if (counter.count(num + 1)) { ans = max(ans, counter[num] + counter[num + 1]); } } return ans; } }; -
class Solution: def findLHS(self, nums: List[int]) -> int: counter = Counter(nums) ans = 0 for num in nums: if num + 1 in counter: ans = max(ans, counter[num] + counter[num + 1]) return ans -
func findLHS(nums []int) int { counter := make(map[int]int) for _, num := range nums { counter[num]++ } ans := 0 for _, num := range nums { if counter[num+1] > 0 { ans = max(ans, counter[num]+counter[num+1]) } } return ans }