diff --git a/OJ notes/pages/Leetcode Partition-Labels.md b/OJ notes/pages/Leetcode Partition-Labels.md
index ab1cdb2..6e94b1d 100644
--- a/OJ notes/pages/Leetcode Partition-Labels.md	
+++ b/OJ notes/pages/Leetcode Partition-Labels.md	
@@ -16,7 +16,7 @@
 >
 > ##### Additional tags:
 >
-> #leetcode 
+> #leetcode #CS_list_need_practicing 
 >
 > ##### Revisions:
 >
@@ -57,6 +57,70 @@ A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits
 ### Thoughts
 
 > [!summary]
-> This is a #template_remove_me
+> This is a #greedy problem.
+
+#### Why is it a greedy problem?
+
+Consider this situation:
+```
+s: ababc
+partition: abab|c
+```
+
+Note that, for every appearance `ch`, it has be in one 
+interval.
+
+```
+ababc
+|-|
+ |-|
+    |
+```
+
+if the interval for a element `b` is bigger than `a`, 
+merge it.
+
+```
+acdcdabb
+|----|
+ |-|
+  |
+      ||
+```
+
+We have to make interval from one side of string to another,
+to save space and do it efficiently.
 
 ### Solution
+
+```cpp
+class Solution {
+public:
+  vector<int> partitionLabels(string s) {
+    // greedy hash table, O(N)
+
+    // We don't have to init values, as every element in
+    // s must have a value here.
+    vector<int> last(26);
+
+    int segEnd = 0, segStart = 0;
+
+    vector<int> ans;
+
+    for (int i = 0, size = s.size(); i < size; i++) {
+      last[s[i] - 'a'] = i;
+    }
+
+    for (int i = 0, size = s.size(); i < size; i++) {
+      segEnd = max(segEnd, last[s[i] - 'a']);
+
+      if (i == segEnd) {
+        ans.push_back(segEnd - segStart + 1);
+        segStart = segEnd + 1;
+      }
+    }
+
+    return ans;
+  }
+};
+```
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