diff --git a/OJ notes/pages/Bit-Manipulation count true trick.md b/OJ notes/pages/Bit-Manipulation count true trick.md
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+# Bit-Manipulation count true trick
+
+#### 2022-07-22 15:11
+
+> ##### Algorithms:
+> #algorithm #bit_manipulation 
+> ##### Data Structure:
+> #DS #bitset 
+> ##### Difficulty:
+> #CS_analysis #difficulty-easy
+> ##### Additional tags:
+> 
+
+##### Related problems:
+```expander
+tag:#coding_problem tag:#bit_manipulation -tag:#template_remove_me 
+```
+ 
+
+___
+
+### What is Bit-Manipulation count true trick?
+
+(n & (n - 1))
+
+### How does Bit-Manipulation count true trick work?
+
+#### validate if the number is power of 2
+
+See [[Leetcode Power-of-Two#Method 1 using masking]]
+
+#### Counting 1
+
+See [[Leetcode Number-of-1-Bits#Method 2 n n - 1 method]]
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diff --git a/OJ notes/pages/Leetcode Number-of-1-Bits.md b/OJ notes/pages/Leetcode Number-of-1-Bits.md
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+# Leetcode Number-of-1-Bits
+
+#### 2022-07-22 14:45
+
+> ##### Algorithms:
+> #algorithm #bit_manipulation 
+> ##### Data structures:
+> #DS #bitset 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#bit_manipulation
+```
+
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/number-of-1-bits/)
+___
+### Problem
+
+Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the [Hamming weight](http://en.wikipedia.org/wiki/Hamming_weight)).
+
+#### Examples
+
+**Example 1:**
+
+**Input:** n = 00000000000000000000000000001011
+**Output:** 3
+**Explanation:** The input binary string **00000000000000000000000000001011** has a total of three '1' bits.
+
+**Example 2:**
+
+**Input:** n = 00000000000000000000000010000000
+**Output:** 1
+**Explanation:** The input binary string **00000000000000000000000010000000** has a total of one '1' bit.
+
+**Example 3:**
+
+**Input:** n = 11111111111111111111111111111101
+**Output:** 31
+**Explanation:** The input binary string **11111111111111111111111111111101** has a total of thirty one '1' bits.
+
+#### Constraints
+
+- The input must be a **binary string** of length `32`.
+
+### Thoughts
+
+> [!summary]
+> This is a #bit_manipulation problem.
+
+Two methods for this problem.
+
+#### Method 1: cpp's STL implementation
+
+simply use [bitset::count](https://en.cppreference.com/w/cpp/utility/bitset/count)
+
+#### Method 2: (n & (n - 1)) method
+
+By using n = (n & (n - 1)), we can remove the last `true` in the original bitset:
+
+```
+5 : 101
+4 : 100
+5 & 4 : 100
+
+10 : 1010
+9  : 1001
+10 & 9 : 1000
+```
+
+n - 1 changes the trailing `false`s to `true`, and change the last `true` to false, and by AND operation, we can remove the last `true` bit.
+
+### Solution
+
+#### CPP STL:
+
+```cpp
+class Solution {
+public:
+  int hammingWeight(uint32_t n) { return bitset<32>(n).count(); }
+};
+````
+
+#### Method 2:
+
+```cpp
+class Solution {
+public:
+  int hammingWeight(uint32_t n) {
+    int count = 0;
+    while (n != 0) {
+      n = (n & (n - 1));
+      count++;
+    }
+    return count;
+  }
+};
+```
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diff --git a/OJ notes/pages/Leetcode Power-of-Two.md b/OJ notes/pages/Leetcode Power-of-Two.md
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+# Leetcode Power-of-Two
+
+#### 2022-07-22 14:30
+
+> ##### Algorithms:
+> #algorithm #bit_manipulation
+> ##### Data structures:
+> #DS #bitset
+> ##### Difficulty:
+> #coding_problem #difficulty-medium 
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#bitset OR tag:#bit_manipulation
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/power-of-two/)
+___
+### Problem
+
+Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
+
+An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** n = 1
+**Output:** true
+**Explanation:** 20 = 1
+
+**Example 2:**
+
+**Input:** n = 16
+**Output:** true
+**Explanation:** 24 = 16
+
+**Example 3:**
+
+**Input:** n = 3
+**Output:** false
+
+#### Constraints
+
+- `-231 <= n <= 231 - 1`
+
+### Thoughts
+
+> [!summary]
+> This is a #bit_manipulation problem.
+> The solution can come from investigating multiple
+> examples and find the common rules
+
+Simple solution using bit masking:
+
+#### Method 1, using masking
+
+- if n <= 0, return false, because it's impossible to be power of 2.
+- else, return !(n & (n - 1))
+
+power of 2 and --1 looks like this
+
+```
+2 : 10
+1 : 01
+
+4 : 100
+3 : 011
+
+8 : 1000
+7 : 0111
+```
+
+so, if it is power of 2,  `!(n & (n - 1))` will produce true.
+
+otherwise, `(n & (n - 1))` will produce something other than 0.
+
+#### Method 2, count bits
+
+power of 2 must have one and only one `true` in the bitset:
+
+```
+2 : 10
+4 : 100
+8 : 1000
+...
+```
+
+So, by counting if the bitset has only one `true`.
+
+### Solution
+
+#### Method 1:
+
+```cpp
+class Solution {
+public:
+  bool isPowerOfTwo(int n) {
+    // n == 0: return 0
+    // n != 0: return !(n & (n - 1))
+    if (n <= 0) {
+      return false;
+    } else {
+      return !(n & (n - 1));
+    }
+  }
+};
+```
+
+#### Method 2:
+
+```cpp
+class Solution {
+public:
+  bool isPowerOfTwo(int n) {
+    if (n <= 0) {
+      return false;
+    } else {
+      return (bitset<32>(n).count() == 1);
+    }
+  }
+};
+```
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diff --git a/OJ notes/pages/Leetcode Reverse-Bits.md b/OJ notes/pages/Leetcode Reverse-Bits.md
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+# Leetcode Reverse-Bits
+
+#### 2022-07-22 15:15
+
+> ##### Algorithms:
+> #algorithm #bit_manipulation 
+> ##### Data structures:
+> #DS #bitset 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#bitset
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/reverse-bits/)
+- [Solution and detailed explanation](https://leetcode.com/problems/reverse-bits/discuss/1232842/JAVA-C%2B%2B-%3A-0ms-or-O(1)-Time-Complexity-or-in-place-or-Detailed-Explanation)
+- [Shifting solution an explanation](https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operation-C++-solution-(8ms)/301342)
+___
+### Problem
+
+Reverse bits of a given 32 bits unsigned integer.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** n = 00000010100101000001111010011100
+**Output:**    964176192 (00111001011110000010100101000000)
+**Explanation:** The input binary string **00000010100101000001111010011100** represents the unsigned integer 43261596, so return 964176192 which its binary representation is **00111001011110000010100101000000**.
+
+**Example 2:**
+
+**Input:** n = 11111111111111111111111111111101
+**Output:**   3221225471 (10111111111111111111111111111111)
+**Explanation:** The input binary string **11111111111111111111111111111101** represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is **10111111111111111111111111111111**.
+
+#### Constraints
+
+- The input must be a **binary string** of length `32`
+
+### Thoughts
+
+> [!summary]
+> This is a #bit_manipulation problem.
+
+There are two methods:
+- swapping bits inside the bitset
+- shifting
+
+#### Method 1: Swapping
+
+This is the most simple and intuitive one, all you have to do is swap the first and last numbers, and continue like that.
+
+#### Method 2: Shifting
+
+The bitset has operations like this `<<` `>>`, which can be used to shift, and reverse the array.
+
+The operation is already documented in the link above:
+[[Leetcode Reverse-Bits#Links]]
+
+### Solution
+
+#### Method 1:
+
+```cpp
+class Solution {
+public:
+  uint32_t reverseBits(uint32_t n) {
+    // swap bits inside the function
+    bitset<32> nb(n);
+    bool tmp;
+    int l = 0, r = 31;
+
+    while (l < r) {
+      tmp = nb[l];
+      nb[l] = nb[r];
+      nb[r] = tmp;
+      l++;
+      r--;
+    }
+
+    return nb.to_ulong();
+  }
+};
+```
+
+#### Shifting:
+
+==#TODO solve it using shifting==
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diff --git a/OJ notes/pages/Leetcode Single-Number.md b/OJ notes/pages/Leetcode Single-Number.md
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+# Leetcode Single-Number
+
+#### 2022-07-22 15:31
+
+> ##### Algorithms:
+> #algorithm #bit_manipulation 
+> ##### Data structures:
+> #DS #bitset
+> ##### Difficulty:
+> #coding_problem #difficulty-easy 
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#bit_manipulation
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/single-number/)
+___
+### Problem
+
+Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
+
+You must implement a solution with a linear runtime complexity and use only constant extra space.
+
+#### Examples
+
+#### Constraints
+
+### Thoughts
+
+> [!summary]
+> This is a #template_remove_me
+
+### Solution
diff --git a/OJ notes/pages/Leetcode Triangle.md b/OJ notes/pages/Leetcode Triangle.md
index ef91671..47dbfb2 100644
--- a/OJ notes/pages/Leetcode Triangle.md	
+++ b/OJ notes/pages/Leetcode Triangle.md	
@@ -67,3 +67,41 @@ Same as in [[Leetcode House-Robber]], there are four stages to optimization:
 #### Stage 2: recursion with cachinqg
 
 ### Solution
+
+#### Stage 2:
+
+```cpp
+class Solution {
+  vector<vector<int>> cache;
+  int minimum(vector<vector<int>> &triangle, int level, int l, int r) {
+    if (level == 0) {
+      return triangle[0][0];
+    } else {
+      int minLen = INT_MAX;
+      for (int i = l; i <= r; i++) {
+        if (i < 0 || i > level) {
+          continue;
+        }
+        if (cache[level][i] != -1) {
+          minLen = min(cache[level][i], minLen);
+          // cout<<"Using cache: "<<minLen<<" for "<<level<<", "<<i<<'\n';
+        } else {
+          cache[level][i] =
+              triangle[level][i] + minimum(triangle, level - 1, i - 1, i);
+          minLen = min(cache[level][i], minLen);
+        }
+      }
+      // cout<<minLen<<", "<<level<<'\n';
+      return minLen;
+    }
+  }
+
+public:
+  int minimumTotal(vector<vector<int>> &triangle) {
+    // Stage one: recursive
+    cache =
+        vector<vector<int>>(triangle.size(), vector<int>(triangle.size(), -1));
+    return minimum(triangle, triangle.size() - 1, 0, triangle.size() - 1);
+  }
+};
+```
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