feat(algorithms, dynamic-programming): counting bits

DIRECTORY.md

@@ -55,6 +55,10 @@
       * [Test Max Profit Three](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/buy_sell_stock/test_max_profit_three.py)
       * [Test Max Profit Two](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/buy_sell_stock/test_max_profit_two.py)
       * [Test Max Profit With Fee](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/buy_sell_stock/test_max_profit_with_fee.py)
+    * Climb Stairs
+      * [Test Climb Stairs](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/climb_stairs/test_climb_stairs.py)
+    * Countingbits
+      * [Test Counting Bits](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/countingbits/test_counting_bits.py)
     * Domino Tromino Tiling
       * [Test Domino Tromino Tiling](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/domino_tromino_tiling/test_domino_tromino_tiling.py)
     * Duffle Bug Value
@@ -69,6 +73,8 @@
       * [Test Min Distance](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/min_distance/test_min_distance.py)
     * Min Path Sum
       * [Test Min Path Sum](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/min_path_sum/test_min_path_sum.py)
+    * Painthouse
+      * [Test Min Cost To Paint Houses](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/painthouse/test_min_cost_to_paint_houses.py)
     * Palindromic Substring
       * [Longest Palindromic Substring](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/palindromic_substring/longest_palindromic_substring.py)
       * [Test Longest Palindromic Substring](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/palindromic_substring/test_longest_palindromic_substring.py)
@@ -138,12 +144,16 @@
     * Spread Stones
       * [Test Minimum Moves To Spread Stones](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/spread_stones/test_minimum_moves_to_spread_stones.py)
   * Heap
+    * Kclosestelements
+      * [Test Find K Closest Elements](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/kclosestelements/test_find_k_closest_elements.py)
     * Longest Happy String
       * [Test Longest Happy String](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/longest_happy_string/test_longest_happy_string.py)
     * Maximal Score After K Operations
       * [Test Maximal Score](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/maximal_score_after_k_operations/test_maximal_score.py)
     * Maximum Subsequence Score
       * [Test Max Subsequence Score](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/maximum_subsequence_score/test_max_subsequence_score.py)
+    * Mergeksortedlists
+      * [Test Merge K Sorted Lists](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/mergeksortedlists/test_merge_k_sorted_lists.py)
     * Min Cost Hire K Workers
       * [Test Min Cost To Hire Workers](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/heap/min_cost_hire_k_workers/test_min_cost_to_hire_workers.py)
     * Min Cost To Connect Sticks
@@ -676,8 +686,6 @@
     * [Battleship](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/battleship/battleship.py)
   * Beeramid
     * [Test Bearamid](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/beeramid/test_bearamid.py)
-  * Climb Stairs
-    * [Test Climb Stairs](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/climb_stairs/test_climb_stairs.py)
   * Find Missing Elem
     * [Test Find Missing Elem](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/find_missing_elem/test_find_missing_elem.py)
   * Hashmap

algorithms/dynamic_programming/countingbits/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)

algorithms/dynamic_programming/countingbits/__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

algorithms/dynamic_programming/countingbits/test_counting_bits.py

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+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.dynamic_programming.countingbits import count_bits
+
+COUNTING_BITS_TEST_CASES = [
+    (6, [0, 1, 1, 2, 1, 2, 2]),
+    (2, [0, 1, 1]),
+    (5, [0, 1, 1, 2, 1, 2]),
+    (0, [0]),
+    (7, [0, 1, 1, 2, 1, 2, 2, 3]),
+    (10, [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2]),
+]
+
+
+class CountingBitsTestCase(unittest.TestCase):
+    @parameterized.expand(COUNTING_BITS_TEST_CASES)
+    def test_counting_bits(self, n: int, expected: List[int]):
+        actual = count_bits(n)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()

algorithms/dynamic_programming/painthouse/README.md

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+# Paint House
+
+You are a renowned painter who is given a task to paint n houses in a row. You can paint each house with one of three colors: Red, Blue, or Green. The cost of painting each house with each color is different and given in a 2D array costs:
+
+- costs[i][0] = cost of painting house `i` Red
+- costs[i][1] = cost of painting house `i` Blue
+- costs[i][2] = cost of painting house `i` Green
+
+No two neighboring houses can have the same color. Return the minimum cost to paint all houses.
+
+## Constraints
+
+- 1 ≤ n ≤ 100
+- costs[i].length == 3
+- 1 ≤ costs[i][j] ≤ 1000
+
+## Examples
+
+Example 1
+
+```text
+costs = [[8, 4, 15], [10, 7, 3], [6, 9, 12]]
+Output: 13
+
+Explanation:
+
+House 0: Blue (cost = 4)
+House 1: Green (cost = 3)
+House 2: Red (cost = 6)
+Total = 4 + 3 + 6 = 13
+
+```
+
+Example 2
+```text
+Input:
+
+costs = [[5, 8, 6], [19, 14, 13], [7, 5, 12], [14, 5, 9]]
+Output: 30
+
+Explanation: Red(5) → Green(13) → Red(7) → Blue(5) = 30
+```
+
+## Solution
+
+The solution implements a space-optimized dynamic programming approach using three variables to track the minimum costs.
+
+Variable Initialization:
+
+prev_min_cost_red = 0: Minimum cost if the current house is painted red (color 0)
+prev_min_cost_blue = 0: Minimum cost if the current house is painted blue (color 1)
+prev_min_cost_green = 0: Minimum cost if the current house is painted green (color 2)
+
+These start at 0 because before painting any house, the cost is 0.
+
+Main Loop: The algorithm iterates through each house's costs using tuple unpacking:
+
+```python
+for cost_red, cost_blue, cost_green in costs:
+    # 
+```
+
+Where cost_red, cost_blue, cost_green represent the costs of painting the current house with red, blue, and green
+respectively.
+
+State Transition: For each house, we simultaneously update all three variables:
+
+```python
+prev_min_cost_red, prev_min_cost_blue, prev_min_cost_green = min(prev_min_cost_blue, prev_min_cost_green) + cost_red, 
+min(prev_min_cost_red, prev_min_cost_green) + cost_blue, min(prev_min_cost_red, prev_min_cost_blue) + cost_green
+```
+
+Breaking this down:
+
+- New `prev_min_cost_red` (cost if current house is red): `min(prev_min_cost_blue, prev_min_cost_green) + cost_red` 
+  We take the minimum of the previous costs where the house was NOT red (either blue or green)
+  Add the cost of painting the current house red
+
+- New `prev_min_cost_blue` (cost if current house is blue): `min(prev_min_cost_red, prev_min_cost_green) + cost_blue`
+  We take the minimum of the previous costs where the house was NOT blue (either red or green)
+  Add the cost of painting the current house blue
+
+- New `prev_min_cost_green` (cost if current house is green): `min(prev_min_cost_red, prev_min_cost_blue) + cost_green`
+  We take the minimum of the previous costs where the house was NOT green (either red or blue)
+  Add the cost of painting the current house green
+
+The simultaneous assignment is crucial here - all three values are calculated using the old values before any updates occur.
+
+Final Result: After processing all houses, `prev_min_cost_red`, `prev_min_cost_blue`, and `prev_min_cost_green` contain
+the minimum costs to paint all houses with the last house being red, blue, or green respectively. The answer is the
+minimum among these three values:
+
+`return min(prev_min_cost_red, prev_min_cost_blue, prev_min_cost_green)`
+
+### Complexity Analysis
+
+#### Time Complexity: O(n)
+
+Where n is the number of houses (length of the costs array). The algorithm iterates through the costs array exactly once,
+performing constant-time operations (comparisons and additions) for each house.
+
+#### Space Complexity: O(1)
+
+The algorithm uses only three variables (`prev_min_cost_red`, `prev_min_cost_blue`, `prev_min_cost_green`) to track the
+minimum costs regardless of the input size. The space usage remains constant as it doesn't create any additional data
+structures that scale with the input. The variables are reused and updated in each iteration rather than storing all
+intermediate results.

algorithms/dynamic_programming/painthouse/__init__.py

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+from typing import List
+
+
+def min_cost_to_paint_houses_alternate_colors(costs: List[List[int]]) -> int:
+    """
+    Finds the minimum cost to paint houses in a street without repeating colors consecutively
+    Args:
+        costs(list): Costs of painting houses as a 2D list
+    Returns:
+        int: minimum cost of painting houses
+    """
+    # If there are no costs of painting the houses, then the minimum cost is 0
+    if not costs:
+        return 0
+
+    # initially, the cost of painting any house any color is 0
+    # Initialize minimum costs for painting up to previous house with each color
+    # prev_min_cost_red: min cost if previous house painted red
+    # prev_min_cost_blue: min cost if previous house painted blue
+    # prev_min_cost_green: min cost if previous house painted green
+    prev_min_cost_red = 0
+    prev_min_cost_blue = 0
+    prev_min_cost_green = 0
+
+    # Iterate through each house cost
+    for current_cost in costs:
+        cost_red, cost_blue, cost_green = current_cost
+
+        # Calculate minimum cost for current house with each color
+        # Current house painted red: add red cost to min of (prev blue, prev green)
+        # Current house painted blue: add blue cost to min of (prev red, prev green)
+        # Current house painted green: add green cost to min of (prev red, prev blue)
+        curr_min_cost_red = min(prev_min_cost_blue, prev_min_cost_green) + cost_red
+        curr_min_cost_blue = min(prev_min_cost_red, prev_min_cost_green) + cost_blue
+        curr_min_cost_green = min(prev_min_cost_red, prev_min_cost_blue) + cost_green
+
+        # Update previous costs for next iteration
+        prev_min_cost_red = curr_min_cost_red
+        prev_min_cost_blue = curr_min_cost_blue
+        prev_min_cost_green = curr_min_cost_green
+
+    # Return minimum cost among all three color options for the last house
+    return min(prev_min_cost_red, prev_min_cost_blue, prev_min_cost_green)

algorithms/dynamic_programming/painthouse/test_min_cost_to_paint_houses.py

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+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.dynamic_programming.painthouse import min_cost_to_paint_houses_alternate_colors
+
+MIN_COST_PAINT_HOUSE = [
+    ([[8, 4, 15], [10, 7, 3], [6, 9, 12]], 13),
+    ([[5, 8, 6], [19, 14, 13], [7, 5, 12], [14, 5, 9]], 30),
+]
+
+
+class MinCostToPaintHouseTestCase(unittest.TestCase):
+    @parameterized.expand(MIN_COST_PAINT_HOUSE)
+    def test_min_cost_to_paint_houses(self, cost: List[List[int]], expected: int):
+        actual = min_cost_to_paint_houses_alternate_colors(cost)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == '__main__':
+    unittest.main()

algorithms/heap/kclosestelements/README.md

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+# Find K Closest Elements
+
+Given a sorted array nums, a target value target, and an integer k, find the k closest elements to target in the array,
+where "closest" is the absolute difference between each element and target. Return these elements in an array, sorted in
+ascending order.
+
+## Examples
+
+```text
+nums = [-1, 0, 1, 4, 6]
+target = 1
+k = 3
+
+Output
+[-1, 0, 1]
+
+Explanation
+-1 is 2 away from 1, 0 is 1 away from 1, and 1 is 0 away from 1. All other elements are more than 2 away. 
+Since we need to return the elements in ascending order, the answer is [-1, 0, 1]
+```
+
+```text
+nums = [5, 6, 7, 8, 9]
+target = 10
+k = 2
+
+Output:
+[8, 9]
+```

algorithms/heap/kclosestelements/__init__.py

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+from typing import List
+import heapq
+
+
+def k_closest(nums: List[int], k: int, target: int):
+    heap = []
+
+    for num in nums:
+        diff = abs(num - target)
+
+        if len(heap) < k:
+            heapq.heappush(heap, (-diff, num))
+        elif diff < -heap[0][0]:
+            heapq.heappushpop(heap, (-diff, num))
+
+    distances = [pair[1] for pair in heap]
+    distances.sort()
+    return distances