feat(algorithms, dynamic-programming): counting bits

Author: BrianLusina

puzzles/climb_stairs/README.md …namic_programming/climb_stairs/README.mdpuzzles/climb_stairs/README.md renamed to algorithms/dynamic_programming/climb_stairs/README.md puzzles/climb_stairs/README.md renamed to algorithms/dynamic_programming/climb_stairs/README.md

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+# Counting Bits
+
+Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's
+in the binary representation of i.
+
+## Examples
+
+```text
+Example 1:
+
+Input: n = 2
+Output: [0,1,1]
+Explanation:
+0 --> 0
+1 --> 1
+2 --> 10
+```
+
+```
+Example 2:
+
+Input: n = 5
+Output: [0,1,1,2,1,2]
+Explanation:
+0 --> 0
+1 --> 1
+2 --> 10
+3 --> 11
+4 --> 100
+5 --> 101
+```
+
+## Constraints
+
+- 0 <= `n` <= 10^5
+
+## Solution
+
+This solution uses a bottom-up dynamic programming approach to solve the problem.
+The key to this problem lies in the fact that any binary number can be broken down into two parts: the least-significant
+(rightmost bit), and the rest of the bits. The rest of the bits can be expressed as the binary number divided by 2
+(rounded down), or `i >> 1`.
+
+For example:
+- 4 in binary = 100
+- rightmost bit = 0
+- rest of bits = 10, which is also (4 // 2) = 2 in binary.
+
+When the number is odd,
+- 5 in binary = 101
+- rightmost bit = 1
+- rest of bits = 10, which is also (5 // 2) = 2 in binary.
+
+in the binary representation of i is that number plus 1 if the rightmost bit is 1. We can tell if the last significant
+bit is 1 by checking if it is odd.
+
+As an example, we know that the number of 1's in 2 is 1. This information can be used to calculate the number of 1's in 4.
+The number of 1's in 4 is the number of 1's in 2 plus 0, because 4 is even.
+
+This establishes a recurrence relationship between the number of 1's in the binary representation of i and the number of
+1's in the binary representation of i // 2: if count[i] is the number of 1's in the binary representation of i, then
+count[i] = count[i // 2] + (i % 2).
+
+With the recurrence relationship established, we can now solve the problem using a bottom-up dynamic programming approach.
+We start with the base case dp[0] = 0, and then build up the solution for dp[i] for i from 1 to n.
+
+![Solution 1](./images/solutions/counting_bits_solution_1.png)
+![Solution 2](./images/solutions/counting_bits_solution_2.png)
+![Solution 3](./images/solutions/counting_bits_solution_3.png)
+![Solution 4](./images/solutions/counting_bits_solution_4.png)
+![Solution 5](./images/solutions/counting_bits_solution_5.png)
+![Solution 6](./images/solutions/counting_bits_solution_6.png)
+![Solution 7](./images/solutions/counting_bits_solution_7.png)
+![Solution 8](./images/solutions/counting_bits_solution_8.png)
+![Solution 9](./images/solutions/counting_bits_solution_9.png)
+![Solution 10](./images/solutions/counting_bits_solution_10.png)
+![Solution 11](./images/solutions/counting_bits_solution_11.png)
+![Solution 12](./images/solutions/counting_bits_solution_12.png)

puzzles/climb_stairs/__init__.py …mic_programming/climb_stairs/__init__.pypuzzles/climb_stairs/__init__.py renamed to algorithms/dynamic_programming/climb_stairs/__init__.py puzzles/climb_stairs/__init__.py renamed to algorithms/dynamic_programming/climb_stairs/__init__.py

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+from typing import List
+
+
+def count_bits(n: int) -> List[int]:
+    """
+    Counts the number of 1 bits in the given integer provided returning a list where count[i] stores the count of '1'
+    bits in the binary form of i.
+    Args:
+        n(int): integer
+    Returns:
+        list: count of 1 bits where each index contains the count of 1 bits
+    """
+    dp = [0] * (n + 1)
+
+    for i in range(1, n + 1):
+        # this can also be solved as which is faster in Python
+        # dp[i] = dp[i >> 1] + (i & 1)
+        dp[i] = dp[i // 2] + (i % 2)
+
+    return dp