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Solution: Longest Harmonious Subsequence

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

Leetcode Problem #594 (Easy): Longest Harmonious Subsequence


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


  • 1 <= nums.length <= 2 * 10^4
  • -10^9 <= nums[i] <= 10^9


Since our target harmonious array is dealing with the absolute value of its elements and since it's a subsequence of our numbers array (N), we don't need to worry about the order of numbers or their index in N.

If all we care about is what numbers appear in N and not their order or index, then it means that we should start by building a frequency map from N.

Then we can just iterate through the entries in our frequency map (fmap) and keep track of the largest value found by adding each number's (key) frequency (val) with the frequency of key+1.

We should then return the best result (ans).


Since javascript's Map() stores its keys as strings, you need to use some method of converting the key back into a number before adding 1. The normal way to do this is with parseInt(), but applying a double bitwise NOT (~) does the same thing far more efficiently, as long as the number is greater than -2^31 and less than 2^31.

Javascript Code:

var findLHS = function(N) {
    let fmap = new Map(), ans = 0
    for (let num of N)
        fmap.set(num, (fmap.get(num) || 0) + 1)
    for (let [key,val] of fmap) 
        if (fmap.has(~~key+1))
            ans = Math.max(ans, val + fmap.get(~~key+1))
    return ans
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