Solution Leetcode and Explaination Every single line

Author: Github-Saurabh0

Bonus28 - Longest Harmonious Subsequence/Program Explaination.txt

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+✅ LeetCode 594 – Longest Harmonious Subsequence
+
+public class Solution {
+ public int findLHS(int[] nums) {
+ Map<Integer, Integer> map = new HashMap<>();
+ int maxLen = 0;
+
+ for (int num : nums) {
+ map.put(num, map.getOrDefault(num, 0) + 1);
+ }
+
+ for (int key : map.keySet()) {
+ if (map.containsKey(key + 1)) {
+ maxLen = Math.max(maxLen, map.get(key) + map.get(key + 1));
+ }
+ }
+
+ return maxLen;
+ }
+}
+
+Explanation
+
+Map<Integer, Integer> map = new HashMap<>();
+🔹 Har number ka count store karne ke liye frequency map bana rahe hain.
+
+for (int num : nums)
+ map.put(num, map.getOrDefault(num, 0) + 1);
+
+🔹 Har element ka occurrence count update karte ja rahe hain.
+
+for (int key : map.keySet())
+🔹 Har unique element par loop kar rahe hain.
+
+if (map.containsKey(key + 1))
+ maxLen = Math.max(maxLen, map.get(key) + map.get(key + 1));
+
+🔹 Check kar rahe hain kya key ke next element (key + 1) bhi present hai.
+Agar haan, toh key aur key+1 dono milke ek harmonious subsequence banayenge.
+Unka combined count (frequency) hi subsequence length hoga.
+
+return maxLen;
+🔹 Sabse longest valid subsequence ka length return kar diya.
+
+🧪 Example Dry Run
+Input:
+nums = [1,3,2,2,5,2,3,7]
+Frequency Map:
+1→1, 2→3, 3→2, 5→1, 7→1
+
+Valid Pairs:
+
+1 & 2 → 1 + 3 = 4
+
+2 & 3 → 3 + 2 = 5 ✅
+
+3 & 4 ❌
+
+5 & 6 ❌
+
+7 & 8 ❌
+
+Output: 5
+
+Time & Space Complexity
+Item	Complexity
+Time	O(N)
+Space	O(N)
+
+🔗 Need more help or Java tricks?
+https://www.linkedin.com/in/saurabh884095/

Bonus28 - Longest Harmonious Subsequence/Solution.java

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+import java.util.;
+
+public class Solution {
+ public int findLHS(int[] nums) {
+ MapInteger, Integer map = new HashMap();
+ int maxLen = 0;
+
+ for (int num nums) {
+ map.put(num, map.getOrDefault(num, 0) + 1);
+ }
+
+ for (int key map.keySet()) {
+ if (map.containsKey(key + 1)) {
+ maxLen = Math.max(maxLen, map.get(key) + map.get(key + 1));
+ }
+ }
+
+ return maxLen;
+ }
+}