diff --git a/javascript/LeetCode/Array/1200.js b/javascript/LeetCode/Array/1200.js
new file mode 100644
index 0000000..14409e1
--- /dev/null
+++ b/javascript/LeetCode/Array/1200.js
@@ -0,0 +1,31 @@
+/**
+ * 1200. Minimum Absolute Difference
+ * 
+ * 給一個無重複元素的數字陣列，找出兩個元素間相差最小的元素並將它們歸類為一組，回傳陣列為二維陣列，必須遞增方式排序[a,b]
+ *   a < b
+ *   b - a = 每組最小相等值
+ * 
+ * @param {number[]} arr
+ * @return {number[][]}
+ */
+var minimumAbsDifference = function(arr) {
+    arr.sort((a,b) => a - b);
+    let min = Infinity;
+    let res = [];
+    for(let i = 1;i < arr.length;++i) {
+      let diff = arr[i] - arr[i-1];
+      if(diff < min){
+        min = diff;
+        res = [i - 1];
+      }else if(diff === min){
+        res.push(i - 1);
+      }
+    }
+    return res.map(i => [arr[i], arr[i + 1]]);
+};
+// let nums = [4,2,1,3];
+// Output: [[1,2],[2,3],[3,4]]
+// Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.
+let nums = [-17,46,63,81,-101,-91,121,-2,112,-15,-65,-96,6,-139];
+// [[-17,15]]
+console.log(minimumAbsDifference(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/2974.js b/javascript/LeetCode/Array/2974.js
new file mode 100644
index 0000000..ae69ec9
--- /dev/null
+++ b/javascript/LeetCode/Array/2974.js
@@ -0,0 +1,38 @@
+/**
+ * 2974. Minimum Number Game
+ * 
+ * nums.length = even(偶數)，每一輪選手(Alice and Bob )皆須：
+ * 第一步，Alice先從陣列中移除最小的元素一次，之後換Bob做同樣的事情(移除第一小＆第二小的元素)
+ * 第二步，Bob將移除的元素加進空陣列arr中，再換Alice做同樣的事情(添加第二小元素＆第一小元素)
+ * 重複上述步驟直到nums變成空陣列為止，回傳最終的arr
+ * 
+ * @param {number[]} nums
+ * @return {number[]} arr
+ */
+var numberGame = function(nums) {
+  let arr = [];
+  nums.sort((a,b) => a - b);
+  // 兩個兩個比較並交換
+  for(let i = 0;i < nums.length;i+=2) {
+    // 有可能有同樣的元素，所以得<=
+    if(nums[i] <= nums[i+1]){
+      let temp = nums[i];
+      arr[i] = nums[i + 1];
+      arr[i + 1] = temp;
+    }
+  }
+  return arr;
+
+  // solution 2.
+  // nums.sort((a,b) => a - b);
+  // // 兩個兩個比較並交換
+  // for(let i = 0;i < nums.length;i+=2) {
+  //   [nums[i],nums[i+1]] = [nums[i+1],nums[i]]
+  // }
+  // return nums;
+};
+let nums = [5,4,2,3];
+// Output: [3,2,5,4]
+// Explanation: In round one, first Alice removes 2 and then Bob removes 3. Then in arr firstly Bob appends 3 and then Alice appends 2. So arr = [3,2].
+// At the begining of round two, nums = [5,4]. Now, first Alice removes 4 and then Bob removes 5. Then both append in arr which becomes [3,2,5,4].
+console.log(numberGame(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3314.js b/javascript/LeetCode/Array/3314.js
new file mode 100644
index 0000000..0dfc5a2
--- /dev/null
+++ b/javascript/LeetCode/Array/3314.js
@@ -0,0 +1,34 @@
+/**
+ * 3314. Construct the Minimum Bitwise Array I
+ * 
+ * prime number = 1 & itself.
+ * ans[i] 與 ans[i] + 1 的位元或運算等於 nums[i]，即 ans[i] OR (ans[i] + 1) == nums[i]。
+ * 
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var minBitwiseArray = function(nums) {
+    let ans = new Array(nums.length);
+
+    for(let i = 0;i < nums.length;++i) {
+      let maybe = -1;
+      for(let j = 1;j < nums[i];++j) {
+        if((j | (j + 1)) === nums[i]){
+          maybe = j;
+          break;
+        }
+      }
+      ans[i] = maybe;
+    }
+    return ans;
+};
+let nums = [2,3,5,7];
+/**
+ *  [-1,1,4,3]
+Explanation:
+For i = 0, as there is no value for ans[0] that satisfies ans[0] OR (ans[0] + 1) = 2, so ans[0] = -1.
+For i = 1, the smallest ans[1] that satisfies ans[1] OR (ans[1] + 1) = 3 is 1, because 1 OR (1 + 1) = 3.
+For i = 2, the smallest ans[2] that satisfies ans[2] OR (ans[2] + 1) = 5 is 4, because 4 OR (4 + 1) = 5.
+For i = 3, the smallest ans[3] that satisfies ans[3] OR (ans[3] + 1) = 7 is 3, because 3 OR (3 + 1) = 7.
+ */
+console.log(minBitwiseArray(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3379.js b/javascript/LeetCode/Array/3379.js
new file mode 100644
index 0000000..a70903f
--- /dev/null
+++ b/javascript/LeetCode/Array/3379.js
@@ -0,0 +1,30 @@
+/**
+ * 3379. Transformed Array
+ * 
+ * nums is circular，所以不管往左或往右，都有可能回到原點
+ *  
+ * nums[i] > 0 => index i往右移nums[i]步至nums[i]的位置，並將result[i]設成index i值
+ * nums[i] < 0 => index i往左移nums[i]步至abs(nums[i])的位置，並將result[i]設成index i值
+ * nums[i] === 0 => result[i] = nums[i]
+ *
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var constructTransformedArray = function(nums) {
+    let res = [];
+    for(let i = 0;i < nums.length;++i) {
+      // 計算要往左或右，在哪個index
+      res[i] = nums[((i + nums[i]) % nums.length + nums.length) % nums.length];
+    }
+    return res;
+};
+let nums = [3,-2,1,1];
+/*
+Output: [1,1,1,3]
+Explanation:
+For nums[0] that is equal to 3, If we move 3 steps to right, we reach nums[3]. So result[0] should be 1.
+For nums[1] that is equal to -2, If we move 2 steps to left, we reach nums[3]. So result[1] should be 1.
+For nums[2] that is equal to 1, If we move 1 step to right, we reach nums[3]. So result[2] should be 1.
+For nums[3] that is equal to 1, If we move 1 step to right, we reach nums[0]. So result[3] should be 3.
+*/
+console.log(constructTransformedArray(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3634.js b/javascript/LeetCode/Array/3634.js
new file mode 100644
index 0000000..627394c
--- /dev/null
+++ b/javascript/LeetCode/Array/3634.js
@@ -0,0 +1,28 @@
+/**
+ * 3634. Minimum Removals to Balance Array
+ * 
+ * balanced條件 = 最大元素 <= 最小元素 * k值
+ * 可移除任一元素，回傳要移除幾個元素才能達成balanced這條件
+ * 
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var minRemoval = function(nums, k) {
+  nums.sort((a,b) => a - b);
+  let i = 0;
+  let count = 0;
+  for(let j = 0;j < nums.length;++j) {
+    // 2 pointers.i & j
+    while(nums[j] > nums[i] * k){
+      i++;
+    }
+    count = Math.max(count, j - i + 1);
+  }
+  return nums.length - count;
+};
+let nums = [1,6,2,9], k = 3;
+// 2
+// Remove nums[0] = 1 and nums[3] = 9 to get nums = [6, 2].
+// Now max = 6, min = 2 and max <= min * k as 6 <= 2 * 3. Thus, the answer is 2.
+console.log(minRemoval(nums,k));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3736.js b/javascript/LeetCode/Array/3736.js
new file mode 100644
index 0000000..f7f889e
--- /dev/null
+++ b/javascript/LeetCode/Array/3736.js
@@ -0,0 +1,29 @@
+/**
+ * 3736. Minimum Moves to Equal Array Elements III
+ * 
+ * 參數為數值陣列，在一次操作中可將任一元素+1
+ * 回傳須移動幾次才能將所有元素都變得一樣
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minMoves = function(nums) {
+  // 要先知道nums中最大值是多少，這樣就能知道其他元素跟最大值差多少
+  let count = 0;
+  let maxEle = Math.max(...nums);
+  for(let i = 0;i < nums.length;++i) {
+    count += Math.abs(maxEle - nums[i]);
+  }
+  return count;
+};
+let nums = [2,1,3];
+/*
+Output: 3
+Explanation:
+To make all elements equal:
+Increase nums[0] = 2 by 1 to make it 3.
+Increase nums[1] = 1 by 1 to make it 2.
+Increase nums[1] = 2 by 1 to make it 3.
+Now, all elements of nums are equal to 3. The minimum total moves is 3.
+*/
+console.log(minMoves(nums))
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3838.js b/javascript/LeetCode/Array/3838.js
new file mode 100644
index 0000000..f4101cd
--- /dev/null
+++ b/javascript/LeetCode/Array/3838.js
@@ -0,0 +1,92 @@
+/**
+ * 3838. Weighted Word Mapping
+ * 
+ * 0 = z;1 = y;2 = x....;25 = a，26個字母倒著
+ * 將每個元素字母重量加總後%26得出的A值，將A值與26個字母倒著的value做比對，取對應的key並以字串回傳
+ * 
+ * @param {string[]} words
+ * @param {number[]} weights
+ * @return {string}
+ */
+var mapWordWeights = function(words, weights) {
+    // 解法1，使用map
+    // 可能遇到的狀況：key(字母)、value(數字)重複出現
+    let alp = generateAlphabet();
+    let sumWeights = [];
+    
+    for (const element of words) {
+      let countLen = 0;
+      for(let i = 0;i < element.length;++i) {
+        countLen += weights[element.charCodeAt(i) - 'a'.charCodeAt()]
+      }
+      // modulo 26
+        sumWeights.push(countLen % 26);
+    }
+    // 方法1
+    //   const result = sumWeights.map(num => {
+    //   const found = [...alp.entries()]
+    //     .find(([key, value]) => value === num);
+    //   return found ? found[0] : null;
+    // });
+
+    // 方法2 反向 Map
+    const reverseMap = new Map(
+      [...alp.entries()].map(([k, v]) => [v, k])
+    );
+
+    const result = sumWeights.map(num => reverseMap.get(num) ?? null);
+
+    return result.join("")
+
+
+     /**
+     * 產生26個英文字母
+     * a = 26,b = 25 ....
+     * @returns obj
+     */
+    function generateAlphabet(){
+      let start = "a";
+      let end = "z";
+      let alp = new Map();
+      let range = 25;
+      let i = start.charCodeAt(0), j = end.charCodeAt(0);
+      for (; i <= j; ++i) {
+        alp.set(String.fromCharCode(i),range--);
+      }
+      return alp;
+    }
+
+};
+
+/**
+ * 解法2。沒有另寫涵式產生26個英文字母
+ * 此法較快
+ * 
+ * @param {*} words 
+ * @param {*} weights 
+ * @returns 
+ */
+var mapWordWeights2 = function(words, weights) {
+  let sumWeights = [];
+    
+  for (const element of words) {
+    let countLen = 0;
+    for(let i = 0;i < element.length;++i) {
+      countLen += weights[element.charCodeAt(i) - 'a'.charCodeAt()]
+    }
+    // String.fromCharCode(ascii code) => ascii code to char.
+    sumWeights.push(String.fromCharCode('z'.charCodeAt() - countLen % 26));
+  }
+  return sumWeights.join("")
+}
+let list = ["abcd","def","xyz"], weights = [5,3,12,14,1,2,3,2,10,6,6,9,7,8,7,10,8,9,6,9,9,8,3,7,7,2];
+/*
+Output: "rij"
+Explanation:
+The weight of "abcd" is 5 + 3 + 12 + 14 = 34. The result modulo 26 is 34 % 26 = 8, which maps to 'r'.
+The weight of "def" is 14 + 1 + 2 = 17. The result modulo 26 is 17 % 26 = 17, which maps to 'i'.
+The weight of "xyz" is 7 + 7 + 2 = 16. The result modulo 26 is 16 % 26 = 16, which maps to 'j'.
+Thus, the string formed by concatenating the mapped characters is "rij".
+*/
+console.log(mapWordWeights(list,weights));
+console.log(mapWordWeights2(list,weights));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/345.js b/javascript/LeetCode/String/345.js
new file mode 100644
index 0000000..7ee642b
--- /dev/null
+++ b/javascript/LeetCode/String/345.js
@@ -0,0 +1,37 @@
+/**
+ * 345. Reverse Vowels of a String
+ * 
+ * 找出所有母音（不分大小寫），其餘子音維持原位，唯獨反轉母音
+ * @param {string} s
+ * @return {string}
+ */
+var reverseVowels = function(s) {
+    let vowels = 'aeiouAEIOU';
+    let splitS = s.split("");
+    let i = 0, j = s.length - 1;
+    while(i < j){
+      while(i < j && vowels.indexOf(splitS[i]) == -1){
+        i++;
+      }
+      while(i < j && vowels.indexOf(splitS[j]) == -1) {
+        j--;
+      }
+      // 交換母音
+      let chars = splitS[i];
+      splitS[i] = splitS[j];
+      splitS[j] = chars;
+      
+      // 2 pointers
+      i++;
+      j--;
+    }
+    return splitS.join("");
+};
+let s = "IceCreAm";
+/**
+ * Output: "AceCreIm"
+ * Explanation:
+ * The vowels in s are ['I', 'e', 'e', 'A']. On reversing the vowels, s becomes "AceCreIm".
+ * 
+ */
+console.log(reverseVowels(s));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/3856.js b/javascript/LeetCode/String/3856.js
new file mode 100644
index 0000000..330feb9
--- /dev/null
+++ b/javascript/LeetCode/String/3856.js
@@ -0,0 +1,25 @@
+/**
+ * 3856. Trim Trailing Vowels
+ * 
+ * 移除s中後半部的母音
+ *
+ * @param {string} s
+ * @return {string}
+ */
+var trimTrailingVowels = function(s) {
+  /**
+   * 從後面開始檢查每個字元是否是母音，若是，則移除並繼續往前找直到非母音為止
+   */
+
+  let splitS = s.split("").reverse();
+  let i = 0;
+  while(splitS[i] === 'a' || splitS[i] === 'e' || splitS[i] === 'i' || splitS[i] === 'o' || splitS[i] === 'u'){
+    splitS.shift();
+  }
+  return splitS.reverse().join("");
+};
+// let s = "idea";
+//"id"
+// let s = "day";
+let s = "aeiou";
+console.log(trimTrailingVowels(s));
\ No newline at end of file
diff --git a/javascript/index.js b/javascript/index.js
index 1218a05..f11eb90 100644
--- a/javascript/index.js
+++ b/javascript/index.js
@@ -3,6 +3,7 @@ import { format } from 'node:path';
 import {ExecutionTimer} from './time.js';
 import assert from 'node:assert/strict';
 import { count } from 'node:console';
+import { lchown } from 'node:fs';
 
 /*
 22. Generate Parentheses
@@ -1161,42 +1162,120 @@ var specialTriplets = function(nums) {
 // console.log(specialTriplets(nums));
 
 
-
 /**
- * 345. Reverse Vowels of a String
+ * Alphabet symmetry
  * 
- * 找出所有母音（不分大小寫），其餘子音維持原位，唯獨反轉母音
- * @param {string} s
- * @return {string}
- */
-var reverseVowels = function(s) {
-    let vowels = ["a","e","i","o","u","A","E","I","O","U"];
-    let splitS = s.split("");
-    // 2 pointer?
-    let j = splitS.length - 1,i = 0;
-    while(i < j){
-      if(!vowels.includes(splitS[i],i)){
-        i++;
+ * 參數為有英文字母但大小寫不一定的陣列，依據26個字母順序來看：a - z /A - Z = 1 ~ 26，以陣列型態回傳元素字母與26個字母對應且字母順序正確的有幾個
+ * 
+ * EG.["abode","ABc","xyzD"]) = [4, 3, 1]
+ * 說明：
+ * a,b = 在26個順序中是1,2 且在這也是1,2;
+ * d,e = 在26個順序中是4,5 且在這也是4,5 => 總共有4個字母出現順序正確
+*/
+function solve(arr){
+  // 元素字母有大小寫
+  // 同一元素字串可能會有重複的字母
+  let letterObj = generateAlphabet();
+  // console.log(letterObj)
+  let map = new Map();
+  // let set  = new Set();
+  let result = [];
+  let count = 0;
+  let baseASCIICode = "A".charCodeAt();
+  for(const letter of arr){
+    let element = letter.toLowerCase()
+    for(let i = 0;i < element.length;++i) {
+      let ascii = element.charCodeAt(i);
+      if(ascii+1 === element.charAt(i)){
+        count++;
         continue;
+       
       }
-      if(!vowels.includes(splitS[j],j)){
-        j--;
-        continue;
+      if(count === element.length){
+        count = 0;
+      }
+      result.push(count);
+    }
+    console.log(result)
+    // let set = new Set( [...letter.toLowerCase().split("")].join(''))
+    // console.log([...set].join(""))
+    // let toStrFromSet = [...set].join("");
+    
+    // for(let i = 0;i < toStrFromSet.length;++i) {
+    //   console.log(toStrFromSet[i]);
+
+    // }
+  }
+
+   /**
+     * 產生26個英文字母
+     * a = 26,b = 25 ....
+     * @returns obj
+     */
+    function generateAlphabet(){
+      let start = "a";
+      let end = "z";
+      let alp = new Map();
+      let range = 26;
+      let i = start.charCodeAt(0), j = end.charCodeAt(0);
+      for (; i <= j; ++i) {
+        // alp[String.fromCharCode(i)] = range--;
+        alp.set(String.fromCharCode(i),range--);
       }
-      let char = splitS[i];
-      splitS[i] = splitS[j];
-      splitS[j] = char;
+      return alp;
+    }
+
+  // console.log(letterObj)
+};
+let arr = ["IAMDEFANDJKL","thedefgh","xyzDEFghijabc"];
+// describe("Basic tests", () => {
+//   it("Fixed tests", () => {
+//     assert.deepEqual(solve(["abode","ABc","xyzD"]),[4,3,1]);
+//     assert.deepEqual(solve(["abide","ABc","xyz"]),[4,3,0]);
+//     assert.deepEqual(solve(["IAMDEFANDJKL","thedefgh","xyzDEFghijabc"]),[6, 5, 7]);
+//     assert.deepEqual(solve(["encode","abc","xyzD","ABmD"]),[1, 3, 1, 3]);
+//   });
+// });
+// console.log(solve(arr))
+
+var minRemoval = function(nums, k) {
+  nums.sort((a,b) => a - b);
+  let i = 0;
+  let count = 0;
+  for(let j = 0;j < nums.length;++j) {
+    // 2 pointers.i & j
+    while(nums[j] > nums[i] * k){
       i++;
-      j--;
     }
-    return splitS.join("");
+    count = Math.max(count, j - i + 1);
+  }
+  return nums.length - count;
 };
-let s = "IceCreAm";
+// let nums = [1,6,2,9], k = 3;
+// 2
+// Remove nums[0] = 1 and nums[3] = 9 to get nums = [6, 2].
+// Now max = 6, min = 2 and max <= min * k as 6 <= 2 * 3. Thus, the answer is 2.
+// console.log(minRemoval(nums,k));
+
+
 /**
- * Output: "AceCreIm"
- * Explanation:
- * The vowels in s are ['I', 'e', 'e', 'A']. On reversing the vowels, s becomes "AceCreIm".
+ * 1653. Minimum Deletions to Make String Balanced
  * 
+ * 參數s中只有'a' & 'b'這兩個字母。
+ * 刪除任一字母使s balanced，若不存在一對index (i,j) 使得 i < j 且 s[i] = 'b' 且 s[j] = 'a'，則s 是balanced。
+ * 回傳最小須刪除幾次才能使s balanced
+ * 
+ * @param {string} s
+ * @return {number}
  */
-// console.log(reverseVowels(s));
+var minimumDeletions = function(s) {
+    
+};
+// let s = "aababbab";
+/*Output: 2
+Explanation: You can either:
+Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
+Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
+*/
+// console.log(minimumDeletions(s));
 
diff --git a/javascript/nodejs/app.js b/javascript/nodejs/app.js
deleted file mode 100644
index 02003ad..0000000
--- a/javascript/nodejs/app.js
+++ /dev/null
@@ -1,12 +0,0 @@
-const { createServer } = require('node:http');
-const hostname = '127.0.0.1';
-const port = 3000;
-
-const server = createServer((req, res) => {
-  res.statusCode = 200;
-  res.setHeader('Content-Type', 'text/plain');
-  res.end('Hello World');
-});
-server.listen(port, hostname, () => {
-  console.log(`Server running at http://${hostname}:${port}/`);
-});
\ No newline at end of file
diff --git a/javascript/nodejs/app.mjs b/javascript/nodejs/app.mjs
new file mode 100644
index 0000000..17a33ca
--- /dev/null
+++ b/javascript/nodejs/app.mjs
@@ -0,0 +1,24 @@
+// const { createServer } = require('node:http');
+// const hostname = '127.0.0.1';
+// const port = 3000;
+
+// const server = createServer((req, res) => {
+//   res.statusCode = 200;
+//   res.setHeader('Content-Type', 'text/plain');
+//   res.end('Hello World');
+// });
+// server.listen(port, hostname, () => {
+//   console.log(`Server running at http://${hostname}:${port}/`);
+// });
+
+// ESM
+import http from "http";
+
+const server = http.createServer((req, res) => {
+  res.writeHead(200, { "Content-Type": "text/plain; charset=utf-8" });
+  res.end("Hello Node Server 👋 (ESM)");
+});
+
+server.listen(3000, () => {
+  console.log("Server running at http://localhost:3000");
+});
diff --git a/python/lo.py b/python/lo.py
new file mode 100644
index 0000000..ca363db
--- /dev/null
+++ b/python/lo.py
@@ -0,0 +1,47 @@
+# import pandas as pd
+
+# # 建立範例資料 (第 114000079 期到 115000007 期)
+# data = {
+#     "期別": ["114000079","114000080","114000081","114000082","114000083",
+#             "115000001","115000002","115000003","115000004","115000005","115000006","115000007"],
+#     "開獎日期": ["2025-10-02","2025-10-06","2025-10-09","2025-10-13","2025-10-16",
+#              "2026-01-01","2026-01-05","2026-01-08","2026-01-12","2026-01-15","2026-01-19","2026-01-22"],
+#     "第一區號碼": [
+#         [3,5,12,24,27,30],[5,6,9,14,15,37],[1,6,11,20,34,35],
+#         [2,8,12,18,28,33],[4,9,16,21,25,36],[7,14,22,23,31,35],
+#         [11,14,19,25,34,37],[7,17,25,26,27,33],[1,9,14,17,33,38],
+#         [8,10,16,26,31,38],[10,16,20,23,35,37],[11,17,29,30,34,35]
+#     ],
+#     "第二區": [7,5,8,3,4,1,4,3,3,5,5,6],
+#     "備註": ["頭獎","無頭獎","無頭獎","無頭獎","無頭獎","無頭獎",
+#            "無頭獎","無頭獎","無頭獎","無頭獎","無頭獎","無頭獎"]
+# }
+
+# df = pd.DataFrame(data)
+
+# # 計算第一區 01~38 次數與頻率
+# freq_zone1 = pd.DataFrame({"號碼": list(range(1,39))})
+# freq_zone1["出現次數"] = freq_zone1["號碼"].apply(lambda x: sum(df["第一區號碼"].apply(lambda y: x in y)))
+# freq_zone1["頻率(%)"] = freq_zone1["出現次數"]/freq_zone1["出現次數"].sum()*100
+
+# # 計算第二區 01~08 次數與頻率
+# freq_zone2 = pd.DataFrame({"號碼": list(range(1,9))})
+# freq_zone2["出現次數"] = freq_zone2["號碼"].apply(lambda x: sum(df["第二區"]==x))
+# freq_zone2["頻率(%)"] = freq_zone2["出現次數"]/freq_zone2["出現次數"].sum()*100
+
+# # 將資料寫入 Excel，多 sheet
+# with pd.ExcelWriter("威力彩分析_114000079_115000007.xlsx") as writer:
+#     df.to_excel(writer, sheet_name="RawData", index=False)
+#     freq_zone1.to_excel(writer, sheet_name="Freq_Zone1", index=False)
+#     freq_zone2.to_excel(writer, sheet_name="Freq_Zone2", index=False)
+
+# print("Excel 檔案已生成：威力彩分析_114000079_115000007.xlsx")
+
+
+import random
+
+# 模擬一次威力彩開獎
+main_numbers = random.sample(range(1, 39), 6)  # 從1~38中抽6個
+special_number = random.randint(1, 8)          # 從1~8中抽1個
+print("第一區:", sorted(main_numbers))
+print("特別號:", special_number)