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# Longest Harmonious Subsequence

Updated: Mar 25, 2021

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

Solution:

Brute Force O(N^2):

```public class Solution {
public int findLHS(int[] nums) {
int res = 0;
for (int i = 0; i < nums.length; i++) {
int count = 0;
boolean flag = false;
for (int j = 0; j < nums.length; j++) {
if (nums[j] == nums[i])
count++;
else if (nums[j] + 1 == nums[i]) {
count++;
flag = true;
}
}
if (flag)
res = Math.max(count, res);
}
return res;
}
}```

Sorting O(NlogN):

```public class Solution {
public int findLHS(int[] nums) {
Arrays.sort(nums);
int prev_count = 1, res = 0;
for (int i = 0; i < nums.length; i++) {
int count = 1;
if (i > 0 && nums[i] - nums[i - 1] == 1) {
while (i < nums.length - 1 && nums[i] == nums[i + 1]) {
count++;
i++;
}
res = Math.max(res, count + prev_count);
prev_count = count;
} else {
while (i < nums.length - 1 && nums[i] == nums[i + 1]) {
count++;
i++;
}
prev_count = count;
}
}
return res;
}
}
```

HashMap O(N):

```class Solution {
public int findLHS(int[] nums) {
Map<Integer,Integer> map = new HashMap<>();
int count=0;
for(int i:nums)
{
map.put(i,map.getOrDefault(i,0)+1);
}
for(int key:map.keySet())
{
if(map.containsKey(key+1))
count = Math.max(count,map.get(key)+map.get(key+1));
}
return count;
}
}```