diff --git a/javascript/LeetCode/Array/1184.js b/javascript/LeetCode/Array/1184.js
new file mode 100644
index 0000000..57f805f
--- /dev/null
+++ b/javascript/LeetCode/Array/1184.js
@@ -0,0 +1,33 @@
+/**
+ * 1184. Distance Between Bus Stops
+ * 
+ * 從圓中(distance)找出從start到destination最短的距離
+ * 
+ * @param {number[]} distance
+ * @param {number} start
+ * @param {number} destination
+ * @return {number}
+ */
+var distanceBetweenBusStops = function(distance, start, destination) {
+  let res = 0,total = distance.reduce((a,b)=>a + b);
+  if(start > destination){
+    let temp = start;
+    start = destination;
+    destination = temp;
+  }
+  for(let i = 0;i < distance.length;++i) {
+    if(i >= start && i < destination){
+      res += distance[i]
+    }
+  }
+  return Math.min(res,total - res);
+};
+//             [0,1,2,3]
+// let distance = [1,2,3,4], start = 0, destination = 2;
+// ans.3 => 2+1
+// 7 => 1+2+4
+// Distance between 0 and 2 is 3 or 7, minimum is 3.
+
+let distance = [7,10,1,12,11,14,5,0], start = 7, destination = 2;
+// 17
+console.log(distanceBetweenBusStops(distance,start,destination));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/169.js b/javascript/LeetCode/Array/169.js
new file mode 100644
index 0000000..69c8648
--- /dev/null
+++ b/javascript/LeetCode/Array/169.js
@@ -0,0 +1,23 @@
+/**
+ * 169. Majority Element
+ *
+ * 找出majority element  = 出現次數大於 nums.length / 2
+ * @param {number[]} nums
+ * @return {number}
+ */
+var majorityElement = function(nums) {
+    let map = new Map();
+    let ans = 0;
+    for(const i of nums){
+      map.has(i) ? map.set(i,map.get(i)+1) : map.set(i,1);
+    }
+    for(const [key,value] of map) {
+      if(value > (nums.length / 2)){
+        ans = key;
+      }
+    }
+    return ans;
+};
+let nums = [2,2,1,1,1,2,2];
+// 2
+console.log(majorityElement(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/1779.js b/javascript/LeetCode/Array/1779.js
new file mode 100644
index 0000000..81541c7
--- /dev/null
+++ b/javascript/LeetCode/Array/1779.js
@@ -0,0 +1,43 @@
+/**
+ * 1779. Find Nearest Point That Has the Same X or Y Coordinate
+ * 
+ * x & y = current location
+ * A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.
+ * 傳回與目前位置曼哈頓距離最小的有效點的索引（從 0 開始索引）。
+ * 如果存在多個有效點，則傳回索引最小的有效點。如果沒有有效點，則傳回 -1。
+ * 
+ * The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).
+ * 
+ * @param {number} x
+ * @param {number} y
+ * @param {number[][]} points
+ * @return {number}
+ */
+var nearestValidPoint = function(x, y, points) {
+  let ans = -1;
+  let smallest = Infinity;
+  for(let i = 0;i < points.length;++i) {
+    let prev = Math.abs(x - parseInt(points[i][0]));
+    let next = Math.abs(y - parseInt(points[i][1]));
+    if(prev * next === 0 && (prev + next) < smallest){
+      smallest = prev + next;
+      ans = i;
+    }
+    
+  }
+  return ans;
+};
+// let x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]];
+// 2
+// Of all the points, only [3,1], [2,4] and [4,4] are valid. 
+// Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. 
+// [2,4] has the smallest index, so return 2.
+
+// let x = 3, y = 4, points = [[3,4]];
+// 0
+// let x = 3, y = 4, points = [[2,3]];
+// -1
+
+let x = 5, y = 1, points = [[1,1],[6,2],[1,5],[3,1]];
+// 3
+console.log(nearestValidPoint(x,y,points));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/2154.js b/javascript/LeetCode/Array/2154.js
new file mode 100644
index 0000000..d09247f
--- /dev/null
+++ b/javascript/LeetCode/Array/2154.js
@@ -0,0 +1,35 @@
+/**
+ * 2154. Keep Multiplying Found Values by Two
+ * 
+ * 初次先尋找original是否在nums內，若不在直接回傳original
+ * 若在，則original * 2 = 新的original （重複此步驟直到找不到original為止）
+ * 
+ * @param {number[]} nums
+ * @param {number} original
+ * @return {number}
+ */
+var findFinalValue = function(nums, original) {
+    nums.sort((a,b) => a - b);
+    for(let i = 0;i < nums.length;++i) {
+      if(nums.includes(original)){
+        original *= 2;
+      }
+    }
+    return original;
+
+    // solution 2
+    // nums.sort((a,b) => a - b);
+    // for(const a of nums){
+    //   if(a === original){
+    //     original*=2;
+    //   }
+    // }
+    // return original;
+};
+// let nums = [5,3,6,1,12], original = 3;
+// 24
+// let nums = [2,7,9], original = 4;
+// 4
+let nums = [8,19,4,2,15,3], original = 2;
+// 16
+console.log(findFinalValue(nums,original));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/219.js b/javascript/LeetCode/Array/219.js
index f41e4a5..913180d 100644
--- a/javascript/LeetCode/Array/219.js
+++ b/javascript/LeetCode/Array/219.js
@@ -76,6 +76,18 @@ var containsNearbyDuplicate = function (nums, k) {
         toMap.set(nums[i], i);
     }
     return false;
+
+    // solution 3.
+    // i & j => nums[i] == nums[j] & abs(i - j) <= k.
+    // nums[i]跟nums[j]一樣
+    // let map = new Map();
+    // for(let i = 0;i < nums.length;++i){
+    //   if(map.has(nums[i]) && i - map.get(nums[i]) <= k){
+    //     return true
+    //   }
+    //   map.set(nums[i],i);
+    // }
+    // return false;
 };
 const nums = [1, 2, 3, 1], k = 3; // return true
 //Input: nums = [1,0,1,1], k = 1 return true
diff --git a/javascript/LeetCode/Array/3151.js b/javascript/LeetCode/Array/3151.js
new file mode 100644
index 0000000..aed5cc4
--- /dev/null
+++ b/javascript/LeetCode/Array/3151.js
@@ -0,0 +1,27 @@
+/**
+ * 3151. Special Array I
+ * 
+ * 檢查陣列任兩個相鄰元素的奇偶性是否都不同。若是回傳true
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var isArraySpecial = function(nums) {
+  if(nums.length === 1){
+    return true;
+  }
+  for(let i = 0;i < nums.length;++i) {
+    let previous = nums[i];
+    let next = nums[i + 1];
+    if(previous % 2 !== next % 2){
+      continue;
+    }else{
+      return false;
+    }
+  }
+  return true;
+};
+// let nums = [4,3,1,6];
+// false
+let nums = [2,1];
+// true
+console.log(isArraySpecial(nums))
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3190.js b/javascript/LeetCode/Array/3190.js
new file mode 100644
index 0000000..08c268f
--- /dev/null
+++ b/javascript/LeetCode/Array/3190.js
@@ -0,0 +1,20 @@
+/**
+ * 3190. Find Minimum Operations to Make All Elements Divisible by Three
+ * 
+ * 計算能將所有元素被3整除的步驟為幾，一次操作能將元素+1或-1。
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumOperations = function(nums) {
+  let answer = 0;
+  for(let i = 0;i < nums.length;++i) {
+    if(nums[i] % 3 !== 0){
+      answer++;
+    }
+  }
+  return answer;
+};
+let nums = [1,2,3,4];
+// 3
+console.log(minimumOperations(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/Array/3573.js b/javascript/LeetCode/Array/3573.js
new file mode 100644
index 0000000..68fcefe
--- /dev/null
+++ b/javascript/LeetCode/Array/3573.js
@@ -0,0 +1,37 @@
+/**
+ * 3573. Best Time to Buy and Sell Stock V
+ * 
+ * 正常交易：i天買，j天賣(i < j)，利潤：proces[j] - prices[i]
+ * 炒短線：i天賣，j天買回來(i < j)，利潤：prices[i] - prices[j]
+ * 不能同一天買賣單張股票
+ * 回傳在交易k次後最多能賺多少
+ * 
+ * @param {number[]} prices 第i天的股價
+ * @param {number} k 最多只能交易k次
+ * @return {number}
+ */
+var maximumProfit = function(prices, k) {
+    const firstPrice = prices[0];
+    const dp = Array(k + 1).fill(null).map(() => ({
+        maxProfit: 0,
+        buyHold: -firstPrice,
+        sellHold: firstPrice
+    }));
+    for(let i = 0;i < prices.length;++i) {
+      const current = prices[i];
+      for(let j = k;j > 0;--j) {
+          const prevProfit = dp[j - 1].maxProfit;
+          dp[j].maxProfit = Math.max(dp[j].maxProfit, dp[j].buyHold + current, dp[j].sellHold - current);
+          dp[j].buyHold = Math.max(dp[j].buyHold, prevProfit - current);
+          dp[j].sellHold = Math.max(dp[j].sellHold, prevProfit + current);
+      }
+    }
+    return dp[k].maxProfit;
+};
+let prices = [1,7,9,8,2], k = 2;
+// 14
+// can make $14 of profit through 2 transactions:
+// A normal transaction: buy the stock on day 0 for $1 then sell it on day 2 for $9. 
+// A short selling transaction: sell the stock on day 3 for $8 then buy back on day 4 for $2.
+// 9 - 1 = 8 ; 8 - 2 = 6; 8 + 6 = 14
+// console.log(maximumProfit(prices,k))
diff --git a/javascript/LeetCode/Array/3731.js b/javascript/LeetCode/Array/3731.js
new file mode 100644
index 0000000..3f1e742
--- /dev/null
+++ b/javascript/LeetCode/Array/3731.js
@@ -0,0 +1,20 @@
+/**
+ * 3731. Find Missing Elements
+ * 
+ * 找出參數陣列中缺少的數字，並以陣列型態回傳
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var findMissingElements = function(nums) {
+  let ans = [];
+  let min = Math.min(...nums),max = Math.max(...nums);
+  for(let i = min;i <= max;++i){
+    if(nums.includes(i) === false){
+      ans.push(i);
+    }
+  }
+  return ans;
+};
+let nums = [5,1];
+// [2,3,4]
+console.log(findMissingElements(nums));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/168.js b/javascript/LeetCode/String/168.js
new file mode 100644
index 0000000..55ed841
--- /dev/null
+++ b/javascript/LeetCode/String/168.js
@@ -0,0 +1,20 @@
+/**
+ * 168. Excel Sheet Column Title
+ * 
+ * 給定一個整數 columnNumber，傳回其在 Excel 表格中對應的列標題。
+ * @param {number} columnNumber
+ * @return {string}
+ */
+var convertToTitle = function(columnNumber) {
+    let ans = "";
+    let baseASCIICode = "A".charCodeAt();
+    while(columnNumber > 0){
+      let offset = (columnNumber - 1) % 26;
+      ans = String.fromCharCode(baseASCIICode + offset) + ans;
+      columnNumber = parseInt((parseInt(columnNumber - 1) / 26));
+    }
+    return ans;
+};
+let columnNumber = 701;
+// "ZY";
+console.log(convertToTitle(columnNumber));
diff --git a/javascript/LeetCode/String/1859.js b/javascript/LeetCode/String/1859.js
new file mode 100644
index 0000000..d9d5956
--- /dev/null
+++ b/javascript/LeetCode/String/1859.js
@@ -0,0 +1,28 @@
+/**
+ * 1859. Sorting the Sentence
+ * 
+ * 字串句子中每個單字最後面都會有個數字，利用該數字重新將句子做排序，但結果每個單字不能有數字。
+ * 
+ * @param {string} s
+ * @return {string}
+ */
+var sortSentence = function(s) {
+    let res = "";
+    let splitS = s.split(" ");
+    let map = new Map();
+    for(let i = 0;i < splitS.length;++i){
+      // 取得每個字的最後一個數字，並把它放入new map => key = int,value = str
+      const lastInt = parseInt(splitS[i].substring(splitS[i].length - 1));
+      map.set(lastInt,splitS[i].substring(0,splitS[i].length - 1));
+    }
+    // sort map
+    var mapAsc = new Map([...map.entries()].sort());
+    for(const [key,value] of mapAsc) {
+      res += " "+value;
+    }
+    return res.trim()
+};
+let s = "is2 sentence4 This1 a3";
+// Output: "This is a sentence"
+// Explanation: Sort the words in s to their original positions "This1 is2 a3 sentence4", then remove the numbers.
+console.log(sortSentence(s));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/1930.js b/javascript/LeetCode/String/1930.js
new file mode 100644
index 0000000..294d1f9
--- /dev/null
+++ b/javascript/LeetCode/String/1930.js
@@ -0,0 +1,63 @@
+/**
+ * 1930. Unique Length-3 Palindromic Subsequences
+ * 
+ * 回傳屬於s迴文的子字串
+ *
+ * @param {string} s
+ * @return {number}
+ */
+var countPalindromicSubsequence = function(s) {
+    // 只能有3個字母，第一個字母和最後一個字母是一樣的，唯獨中間字母不同 = 迴文
+    // 會有一個字母出現至少2次 
+    
+    // solution 1.
+    // let ans = 0;
+    // let set = new Set(s);
+    // for(const char of set) {
+    //   let start = s.indexOf(char);
+    //   let end = s.lastIndexOf(char);
+
+    //   if(start < end){
+    //     ans += new Set(s.slice(start + 1, end)).size;
+    //   }
+    // }
+    // return ans;
+
+    // solution 2.
+    let map = new Map();
+    let ans =0;
+    for(let i = 0;i < s.length;++i) {
+      if (!map.has(s[i])) {
+        map.set(s[i], []);
+      }
+      map.get(s[i]).push(i);
+    }
+    console.log(map)
+    for(const [char,index] of map) {
+        const start = index[0];
+        const end = index[index.length - 1];
+        if (end - start <= 1) {
+          continue;
+        }
+
+        const set = new Set();
+        for (let i = start + 1; i < end; i++) {
+          set.add(s[i]);
+        }
+        ans += set.size;
+    }
+    return ans;
+};
+// let s = "aabca";
+/**
+ * 3
+ * The 3 palindromic subsequences of length 3 are:
+ * "aba" (subsequence of "aabca")
+ * "aaa" (subsequence of "aabca")
+ * "aca" (subsequence of "aabca")
+ */
+// let s = "uuuuu";
+// 1
+let s ="ckafnafqo"
+// 4
+console.log(countPalindromicSubsequence(s));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/28.js b/javascript/LeetCode/String/28.js
index 51a685c..30ae897 100644
--- a/javascript/LeetCode/String/28.js
+++ b/javascript/LeetCode/String/28.js
@@ -1,70 +1,17 @@
-
-/**
- * 28. Implement strStr()
- * Difficulty:Easy(459延伸題)
- *
- * Implement strStr().
- *
- * Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
- *
- * Clarification:
- * What should we return when needle is an empty string? This is a great question to ask during an interview.
- * For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
- *
- * ---------------------------------------------------
- * Input: haystack and needle are string
- * Output:在haystack中第一個出現的needle index值
- * 若haystack沒有needle，return -1;
- * 或needle是空的，return 0
- * ------------------------------------------------------
- * Example 1:
- * Input: haystack = "hello", needle = "ll"
- * Output: 2
- *
- * Example 2:
- * Input: haystack = "aaaaa", needle = "bba"
- * Output: -1
- *
- * Example 3:
- * Input: haystack = "", needle = ""
- * Output: 0
- *
- * Constraints:
- * 0 <= haystack.length, needle.length <= 5 * 104
- * haystack and needle consist of only lower-case English characters.
- */
 /**
+ * 28. Find the Index of the First Occurrence in a String
+ 
+ * 回傳needle第一次出現在haystack的index，找不到回傳-1
+ * 
  * @param {string} haystack
  * @param {string} needle
  * @return {number}
  */
-var strStr = function (haystack, needle) {
-  /**
-  * haystack中:
-  * 有needle，return haystack index
-  * 沒有，return -1
-  * $haystack或$needle是空，retunr 0
-  * 
-  * $haystack、$needle 都是英文小寫
-  */
-
-  if (needle === "") {
-    return 0;
-  }
-  for (let i = 0; i < haystack.length; i++) {
-
-    // Return part of string:
-    // for JS: string.substr(start,end)
-    // for PHP: substr(string,start,end)
-    let slice = haystack.substr(i, needle.length);
-    if (slice === needle) {
-      return i;
-    }
-  }
-  return -1;
-
+var strStr = function(haystack, needle) {
+  return haystack.indexOf(needle);
 };
-let h = "hello";
-let nee = "ll";
-// 2
-console.log(strStr(h, nee));
\ No newline at end of file
+// let haystack = "leetcode", needle = "leeto";
+// -1
+let haystack = "sadbutsad", needle = "sad";
+// 0
+console.log(strStr(haystack,needle));
\ No newline at end of file
diff --git a/javascript/LeetCode/String/58.js b/javascript/LeetCode/String/58.js
new file mode 100644
index 0000000..34f7940
--- /dev/null
+++ b/javascript/LeetCode/String/58.js
@@ -0,0 +1,34 @@
+/**
+ * 58. Length of Last Word
+ * 
+ * 取得最後一個單字的長度，參數內會有空白格
+ * 
+ * @param {string} s
+ * @return {number}
+ */
+var lengthOfLastWord = function(s) {
+  // solution 1.
+  let temp = [];
+  let splitToArr = s.split(" ");
+  for(let i = 0;i < splitToArr.length;++i) {
+    if(splitToArr[i] !== ""){
+      temp.push(splitToArr[i]);
+    }
+  }
+  return temp[temp.length - 1].length;
+
+  // solution 2.
+//   let ans = 0;
+//   let trim = s.trim();
+//   for(let i = trim.length - 1;i >= 0;i--){
+//     if(trim[i] !== " "){
+//       ans++;
+//     }else if(ans > 0){
+//       break;
+//     }
+//   }
+//   return ans;
+};
+let s = "   fly me   to   the moon  ";
+// 4
+console.log(lengthOfLastWord(s))
\ No newline at end of file
diff --git a/javascript/LeetCode/math/2119.js b/javascript/LeetCode/math/2119.js
new file mode 100644
index 0000000..e610fed
--- /dev/null
+++ b/javascript/LeetCode/math/2119.js
@@ -0,0 +1,24 @@
+/**
+ * 2119. A Number After a Double Reversal
+ * 
+ * 將參數nums反轉2次後，檢查是否還是跟原參數一樣
+ * @param {number} num
+ * @return {boolean}
+ */
+var isSameAfterReversals = function(num) {
+  // 可能遇到的狀況：數字反轉後，前面是0
+  let countReverseTimes = 0;
+  let copyNum = num;
+
+  while(countReverseTimes !== 2){
+    let reverse = copyNum.toString().split("").reverse().join("");
+    copyNum = parseInt(reverse);
+    countReverseTimes++;
+  }
+  return copyNum === num;
+};
+let num = 526;
+// true
+// let num = 1800;
+// false
+console.log(isSameAfterReversals(num));
\ No newline at end of file
diff --git a/javascript/LeetCode/math/3432.js b/javascript/LeetCode/math/3432.js
index 6040cb5..92333a4 100644
--- a/javascript/LeetCode/math/3432.js
+++ b/javascript/LeetCode/math/3432.js
@@ -44,22 +44,37 @@
  */
 var countPartitions = function (nums) {
   // 左右皆不斷更新
-  let right = 0;
-  let left = 0;
-  // 加總後的值
-  let sum = 0;
-  // 操作次數
+  // let right = 0;
+  // let left = 0;
+  // // 加總後的值
+  // let sum = 0;
+  // // 操作次數
+  // let count = 0;
+  // for (let i = 0; i < nums.length; i++) {
+  //   sum += nums[i];
+  // }
+  // for (let i = 0; i < nums.length - 1; i++) {
+  //   left += nums[i];
+  //   // 總值 - 左邊 = 右邊
+  //   right = sum - left;
+
+  //   // 檢查兩邊是否都是偶數
+  //   if ((left % 2) === (right % 2)) {
+  //     count++;
+  //   }
+  // }
+  // return count;
+
+  // solution 2.
+  let sum = nums.reduce((a,b) => a + b);
   let count = 0;
-  for (let i = 0; i < nums.length; i++) {
-    sum += nums[i];
-  }
-  for (let i = 0; i < nums.length - 1; i++) {
+  let right = 0, left = 0;
+  for(let i = 0;i < nums.length;++i) {
+    // 左邊的元素會越來越多，右邊則會變少
     left += nums[i];
-    // 總值 - 左邊 = 右邊
     right = sum - left;
 
-    // 檢查兩邊是否都是偶數
-    if ((left % 2) === (right % 2)) {
+    if(right % 2 === 0 && left % 2 === 0){
       count++;
     }
   }
diff --git a/javascript/LeetCode/math/69.js b/javascript/LeetCode/math/69.js
new file mode 100644
index 0000000..1f01843
--- /dev/null
+++ b/javascript/LeetCode/math/69.js
@@ -0,0 +1,17 @@
+/**
+ * 69. Sqrt(x)
+ * 
+ * @param {number} x
+ * @return {number}
+ */
+var mySqrt = function(x) {
+    // 不能使用內建的pow() or sqrt()
+    // rounded down to the nearest integer.
+  if (x < 0) {
+    return;
+  }
+  return Math.floor(x ** 0.5); 
+};
+let x = 8;
+// 2
+console.log(mySqrt(x));
\ No newline at end of file
diff --git a/javascript/LeetCode/math/9.js b/javascript/LeetCode/math/9.js
new file mode 100644
index 0000000..083284e
--- /dev/null
+++ b/javascript/LeetCode/math/9.js
@@ -0,0 +1,17 @@
+/**
+ * 9. Palindrome Number
+ * 
+ * 檢查x是否是迴文（倒著念也一樣）
+ * @param {number} x
+ * @return {boolean}
+ */
+var isPalindrome = function(x) {
+    // 數值不適合顛倒，轉字串
+    let toStr = x.toString().split("").reverse();
+    // console.log(toStr.join(""))
+    
+    return x.toString() === toStr.join("");
+};
+let x = -121;
+// false
+console.log(isPalindrome(x))
\ No newline at end of file
diff --git a/javascript/index.js b/javascript/index.js
index 2d5a115..b9f992c 100644
--- a/javascript/index.js
+++ b/javascript/index.js
@@ -1086,3 +1086,76 @@ var findXSum = function(nums, k, x) {
 // For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.
 // console.log(findXSum(nums,k,x));
 
+/**
+ * 1015. Smallest Integer Divisible by K
+ * 
+ * 找出最小能被k整除了數字，且該數字只有1位數
+ * @param {number} k
+ * @return {number}
+ */
+var smallestRepunitDivByK = function(k) {
+    
+};
+// let k = 2;
+// -1
+// console.log(smallestRepunitDivByK(k));
+
+
+/**
+ * 3583. Count Special Triplets
+ * 
+ * special triplet = index i,j,k
+ * 0 <= i < j < k < nums.length
+ * nums[i] === nums[j] * 2
+ * nums[k] === nums[j] * 2
+ * return it modulo 10的9次方 + 7.
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var specialTriplets = function(nums) {
+    let ans = 0;
+    // j as the middle of the triplet.
+    // For each j, you only need:
+    // how many values equal to 2 * nums[j] appear before j
+    // how many appear after j
+    // Then the contribution from index j is just:
+    // leftCount * rightCount
+    // let j = Math.floor(nums[nums.length % 2 ]);
+    // // console.log(j)
+    // for(let i = 0;i < nums.length;++i) {
+
+    // }
+
+    /**
+     * j = middle index
+     * 在j之前，檢查nums[i] === nums[j] * 2 的有幾個
+     * 在j之後，檢查nums[k] === nums[j] * 2 的有幾個
+     */
+    // let left = new Map();
+    // let right = new Map();
+    let j = Math.round(nums.length % 2);
+    for(let i = 0;i < nums.length;++i) {
+      if(nums[i] === nums[j] * 2 && i < j){
+        ans++;
+      }
+    }
+    console.log(ans)
+};
+// let nums = [8,4,2,8,4];
+/**
+ * 2
+ * 
+ * There are exactly two special triplets:
+ * (i, j, k) = (0, 1, 3)
+ * nums[0] = 8, nums[1] = 4, nums[3] = 8
+ * nums[0] = nums[1] * 2 = 4 * 2 = 8
+ * nums[3] = nums[1] * 2 = 4 * 2 = 8
+ * 
+ * (i, j, k) = (1, 2, 4)
+ * nums[1] = 4, nums[2] = 2, nums[4] = 4
+ * nums[1] = nums[2] * 2 = 2 * 2 = 4
+ * nums[4] = nums[2] * 2 = 2 * 2 = 4
+ */
+// console.log(specialTriplets(nums));
+
diff --git a/python/exercise.py b/python/exercise.py
index 78dd450..afe159f 100644
--- a/python/exercise.py
+++ b/python/exercise.py
@@ -237,4 +237,8 @@ def checkExist(arr:list,item:int):
 
 array = [1,2,3,4]
 item = 3
-checkExist(array,item)
\ No newline at end of file
+checkExist(array,item)
+
+
+r = lambda a: "True" if a > 5 else "False"
+print(r(1))
\ No newline at end of file
diff --git a/python/index.py b/python/index.py
index f532269..850a37c 100644
--- a/python/index.py
+++ b/python/index.py
@@ -265,4 +265,36 @@ def replaceDigits(s: str) -> str:
   return res
 s = "a1c1e1"
 # Output: "abcdef"
-# print(replaceDigits(s))
\ No newline at end of file
+# print(replaceDigits(s))
+
+class So:
+  def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
+    ans = -1
+    smallest = float('inf')
+    for i in range(0,len(points)):
+      prev = x - int(points[i][0])
+      nextOne = y - int(points[1][0])
+      if (prev * nextOne) == 0 and abs(prev + nextOne) < smallest:
+        smallest = abs(prev + nextOne)
+        ans = i
+    return ans
+
+c = So()
+x = 3
+y = 4
+points = [[1,2],[3,1],[2,4],[2,3],[4,4]]
+# 2
+# print(c.nearestValidPoint(x,y,points))
+
+def showResult(func):
+  def wrap():
+    print("run this one first")
+    func()
+    print("run this at the end")
+  return wrap
+
+@showResult
+def say():
+  print("then run this function")
+
+say()
diff --git a/python/leetcode/list/136.py b/python/leetcode/list/136.py
new file mode 100644
index 0000000..5f13412
--- /dev/null
+++ b/python/leetcode/list/136.py
@@ -0,0 +1,21 @@
+def singleNumber(nums: List[int]) -> int:
+  '''
+  136. Single Number
+  '''
+  # 找出只出現一次的元素
+  # ans = 0
+  # for key,value in Counter(nums).items():
+  #   if(value == 1):
+  #     ans = key
+  # return ans
+
+  # solution 2
+  ans = 0
+  for i in nums:
+    ans ^= i
+  return ans
+
+
+nums = [4,1,2,1,2]
+# 4
+print(singleNumber(nums))
\ No newline at end of file
diff --git a/python/leetcode/math/168.py b/python/leetcode/math/168.py
new file mode 100644
index 0000000..ced705a
--- /dev/null
+++ b/python/leetcode/math/168.py
@@ -0,0 +1,24 @@
+def convertToTitle(columnNumber: int) -> str:
+  '''
+  168. Excel Sheet Column Title
+
+  :param columnNumber: 給定一個整數 columnNumber，傳回其在 Excel 表格中對應的列標題。
+  :type columnNumber: int
+  :return: 回傳對應的列
+  :rtype: str
+  '''
+  ans = ""
+  # ASCII 
+  # ord('alphabet') => number
+  # chr(number) => alphabet
+  baseChar = ord('A')
+  while(columnNumber > 0):
+    offset = (columnNumber - 1) % 26
+    # print(offset)
+    ans = str(chr(baseChar + offset)) + ans
+
+    columnNumber = int((int(columnNumber - 1) / 26))
+  return ans
+columnNumber = 701
+# ZY
+print(convertToTitle(columnNumber))
\ No newline at end of file
diff --git a/python/leetcode/math/1925.py b/python/leetcode/math/1925.py
new file mode 100644
index 0000000..f0cfe61
--- /dev/null
+++ b/python/leetcode/math/1925.py
@@ -0,0 +1,30 @@
+from math import sqrt
+class So:
+  '''
+  So 的 Docstring
+ 
+  '''
+  def countTriples(self, n: int) -> int:
+    '''
+    1925. Count Square Sum Triples
+
+    triple = a 二次方 + b 二次方 = c 二次方
+
+    :param self: 說明
+    :param n: 數值n
+    :type n: int
+    :return: 說明
+    :rtype: int
+    '''
+    ans = 0
+    for a in range(1,n + 1):
+      for b in range(1,n + 1):
+        pairs = int(sqrt(a**2 + b**2 + 1))
+        if pairs <= n and pairs **2 == a**2 + b**2:
+          ans += 1
+    return ans
+
+ans = So()
+n = 5
+# 2
+print(ans.countTriples(n))
\ No newline at end of file