vault backup: 2022-09-05 19:27:02

.obsidian/graph.json

@@ -39,6 +39,6 @@
   "repelStrength": 10,
   "linkStrength": 1,
   "linkDistance": 250,
-  "scale": 0.7463733455586773,
+  "scale": 1.3711753906745523,
   "close": true
 }

OJ notes/pages/Greedy Algorithm.md

@@ -0,0 +1,35 @@
+# Greedy Algorithm
+
+2022-09-05 17:32
+
+> ##### Algorithms:
+>
+> #algorithm #greedy 
+>
+> ##### Difficulty:
+>
+> #CS_analysis
+>
+> ##### Additional tags:
+
+##### Related problems:
+
+```expander
+tag:#coding_problem tag:#<CHANGE_ME> -tag:#template_remove_me
+```
+
+##### Links:
+
+- [cppreference]()
+
+---
+
+### What is Greedy Algorithm?
+
+### How does Greedy Algorithm work?
+
+#### Example code
+
+### Why and when to use it?
+
+#template_remove_me

OJ notes/pages/Leetcode Longest-Palindrome.md

@@ -0,0 +1,108 @@
+# Leetcode Longest-Palindrome
+
+2022-09-05 17:27
+
+> ##### Algorithms:
+>
+> #algorithm #greedy
+>
+> ##### Data structures:
+>
+> #DS #string
+>
+> ##### Difficulty:
+>
+> #coding_problem #difficulty-easy
+>
+> ##### Additional tags:
+>
+> #leetcode
+>
+> ##### Revisions:
+>
+> N/A
+
+##### Links:
+
+- [Link to problem](https://leetcode.com/problems/longest-palindrome/)
+
+---
+
+### Problem
+
+Given a string `s` which consists of lowercase or uppercase letters, return _the length of the **longest palindrome**_ that can be built with those letters.
+
+Letters are **case sensitive**, for example, `"Aa"` is not considered a palindrome here.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** s = "abccccdd"
+**Output:** 7
+**Explanation:** One longest palindrome that can be built is "dccaccd", whose length is 7.
+
+**Example 2:**
+
+**Input:** s = "a"
+**Output:** 1
+**Explanation:** The longest palindrome that can be built is "a", whose length is 1.
+
+#### Constraints
+
+-   `1 <= s.length <= 2000`
+-   `s` consists of lowercase **and/or** uppercase English letters only.
+
+### Thoughts
+
+> [!summary]
+> This is a #greedy problem.
+
+The solution is simple: count the apperances, sort it using
+std::greater<int>(), and select all even number, for odd 
+number `i`:
+- if odd numbers has been used: `ans += i - 1` (make it 
+  even)
+- if not: `ans += i`, and mark odd number used
+
+### Solution
+
+```cpp
+class Solution {
+public:
+  int longestPalindrome(string s) {
+    int ans = 0;
+    bool used = false;
+    vector<int> table(52, 0);
+
+    // record in table.
+    for (char ch : s) {
+      if (ch <= 'Z') {
+        table[ch - 'A' + 26]++;
+      } else {
+        table[ch - 'a']++;
+      }
+    }
+
+    sort(table.begin(), table.end(), greater<int>());
+
+    for (int i : table) {
+      if (i % 2 == 1) {
+        if (used == false) {
+          used = true;
+          ans += i;
+        } else {
+          ans += i - 1;
+        }
+      } else {
+        if (i == 0) {
+          break;
+        }
+        ans += i;
+      }
+    }
+
+    return ans;
+  }
+};
+```