Merge pull request #154 from BrianLusina/feat/algorithms-dp-bs

feat(algorithms, two-pointers, dynamic-programming): two-pointers dp and bs problems

Author: BrianLusina

DIRECTORY.md

@@ -59,6 +59,8 @@
       * [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)
+    * Decodeways
+      * [Test Num Decodings](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/decodeways/test_num_decodings.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
@@ -214,6 +216,8 @@
         * [Test Divide Chocolate](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/search/binary_search/divide_chocolate/test_divide_chocolate.py)
       * Maxruntime N Computers
         * [Test Max Runtime](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/search/binary_search/maxruntime_n_computers/test_max_runtime.py)
+      * Split Array Largest Sum
+        * [Test Split Array Largest Sum](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/search/binary_search/split_array_largest_sum/test_split_array_largest_sum.py)
       * [Test Binary Search](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/search/binary_search/test_binary_search.py)
     * Interpolation
       * [Test Interpolation Search](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/search/interpolation/test_interpolation_search.py)
@@ -255,6 +259,8 @@
   * Two Pointers
     * Array 3 Pointers
       * [Test Array 3 Pointers](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/two_pointers/array_3_pointers/test_array_3_pointers.py)
+    * Count Pairs
+      * [Test Count Pairs](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/two_pointers/count_pairs/test_count_pairs.py)
     * Find Sum Of Three
       * [Test Find Sum Of Three](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/two_pointers/find_sum_of_three/test_find_sum_of_three.py)
     * Merge Sorted Arrays

algorithms/dynamic_programming/decodeways/README.md

@@ -0,0 +1,84 @@
+# Decode Ways
+
+You have intercepted a secret message encoded as a string of numbers. The message is decoded via the following mapping:
+
+```text
+'1' -> "A"
+'2' -> "B"
+'3' -> "C"
+...
+'26' -> "Z"
+```
+However, while decoding the message, you realize that there are many different ways you can decode the message because
+some codes are contained in other codes ("2" and "5" vs "25").
+
+For example, "11106" can be decoded into:
+
+- "AAJF" with the grouping (1, 1, 10, 6)
+- "KJF" with the grouping (11, 10, 6)
+- The grouping (1, 11, 06) is invalid because "06" is not a valid code (only "6" is valid).
+
+Note: there may be strings that are impossible to decode.
+
+Given a string s containing only digits, return the number of ways to decode it. If the entire string cannot be decoded
+in any valid way, return 0.
+
+The test cases are generated so that the answer fits in a 32-bit integer.
+
+## Examples
+
+Example 1
+
+```text
+Input: s = "12"
+
+Output: 2
+
+Explanation:
+
+"12" could be decoded as "AB" (1 2) or "L" (12).
+```
+ 
+Example 2
+
+```text
+Input: s = "226"
+
+Output: 3
+
+Explanation:
+
+"226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
+```
+
+Example 3
+
+```text
+Input: s = "06"
+
+Output: 0
+
+Explanation:
+
+"06" cannot be mapped to "F" because of the leading zero ("6" is different from "06"). In this case, the string is not
+a valid encoding, so return 0.
+```
+
+Example 4
+
+```text
+s = 101
+output = 1
+Explanation: The only way to decode it is "JA". "01" cannot be decoded into "A" as "1" and "01" are different.
+```
+
+## Constraints
+
+- 1 <= s.length <= 100
+- s contains only digits and may contain leading zero(s).
+
+## Topics
+
+- String
+- Dynamic Programming
+

algorithms/dynamic_programming/decodeways/__init__.py

@@ -0,0 +1,28 @@
+def num_decodings(s: str) -> int:
+    if not s or s[0] == "0":
+        return 0
+
+    n = len(s)
+    # Initialize the dp array. dp[i] will store the number of ways to decode the string s[:i]
+    dp = [0] * (n + 1)
+    dp[0] = 1
+    dp[1] = 1
+
+    # Iterate through the string
+    for i in range(2, n + 1):
+        # Get the current character
+        current_char = s[i - 1]
+        # Get the previous character
+        prev_char = s[i - 2]
+
+        # If the current character is not a zero, then we can decode it as a single character
+        if current_char != "0":
+            dp[i] += dp[i - 1]
+
+        # If the previous character is 1 or 2 and the current character is less than or equal to 6, then we can decode
+        # it as a two character
+        if prev_char == "1" or (prev_char == "2" and current_char <= "6"):
+            dp[i] += dp[i - 2]
+
+    # Return the number of ways to decode the string
+    return dp[n]

algorithms/dynamic_programming/decodeways/test_num_decodings.py

@@ -0,0 +1,28 @@
+import unittest
+from parameterized import parameterized
+from algorithms.dynamic_programming.decodeways import num_decodings
+
+NUM_DECODINGS_TEST_CASES = [
+    ("11106", 2),
+    ("12", 2),
+    ("226", 3),
+    ("06", 0),
+    ("101", 1),
+    ("12", 2),
+    ("012", 0),
+    ("0", 0),
+    ("30", 0),
+    ("10", 1),
+    ("27", 1),
+]
+
+
+class NumDecodingsTestCase(unittest.TestCase):
+    @parameterized.expand(NUM_DECODINGS_TEST_CASES)
+    def test_num_decodings(self, s: str, expected: int):
+        actual = num_decodings(s)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()

algorithms/dynamic_programming/painthouse/test_min_cost_to_paint_houses.py

@@ -1,7 +1,9 @@
 import unittest
 from typing import List
 from parameterized import parameterized
-from algorithms.dynamic_programming.painthouse import min_cost_to_paint_houses_alternate_colors
+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),
@@ -16,5 +18,5 @@ def test_min_cost_to_paint_houses(self, cost: List[List[int]], expected: int):
         self.assertEqual(expected, actual)
 
 
-if __name__ == '__main__':
+if __name__ == "__main__":
     unittest.main()

algorithms/search/binary_search/split_array_largest_sum/README.md

@@ -0,0 +1,106 @@
+# Split Array Largest Sum
+
+Given an integer list nums and an integer k, split nums into k non-empty subarrays such that the largest sum among these
+subarrays is minimized. The task is to find the minimized largest sum by choosing the split such that the largest sum of
+every split of subarrays is the minimum among the sum of other splits.
+
+## Constraints
+
+- 1 <= nums.length <= 1000
+- 0 <= nums[i] <= 10^4
+- 1 <= k <= nums.length
+
+## Examples
+
+![Example 1](images/examples/split_array_largest_sum_example_1.png)
+![Example 2](images/examples/split_array_largest_sum_example_2.png)
+![Example 3](images/examples/split_array_largest_sum_example_3.png)
+![Example 4](images/examples/split_array_largest_sum_example_4.png)
+
+## Topics
+
+- Array
+- Binary Search
+- Dynamic Programming
+- Greedy
+- Prefix Sum
+
+## Solution
+
+In a brute force method, you would try all possible ways to split the array into k subarrays, calculate the largest sum
+for each split, and then find the smallest among those largest sums. This approach is extremely inefficient because the
+number of ways to split the array grows exponentially as the size increases.
+
+We reverse the process by guessing a value for the minimum largest sum and checking if it’s feasible:
+
+- Instead of iterating through all splits, we only focus on testing whether a specific value allows us to split the
+  array into k subarrays.
+
+- But wait—just because one value works doesn’t mean it’s the smallest feasible value. We keep exploring smaller values
+  to achieve the most optimized result.
+
+The solution uses the binary search approach to find the optimal largest subarray sum without testing all possible splits.
+The binary search finds the smallest possible value of the largest subarray sum and applies searching over the range of
+possible values for this largest sum. But how do we guess this value? We guess the value using a certain range.
+Here’s how:
+
+- **Left boundary**: The maximum element in the array is the minimum possible value for the largest subarray sum. This
+  is because any valid subarray must have a sum at least as large as the largest element.
+
+- **Right boundary**: The maximum possible value for the largest subarray sum is the sum of all elements in the array.
+  You would get this sum if the entire array were one single subarray.
+
+The binary search iteratively tests midpoints in the above ranges. It determines whether dividing the array results in
+at most k subarrays will result in the smallest largest sum. If it does, the search shifts to lower values to minimize
+the largest sum. Otherwise, it shifts to higher values. Still, there might be subarrays whose sum could be smaller, so
+the search keeps going until the search range crosses each other, i.e., **left boundary** > **right boundary**.
+
+Here’s the step-by-step implementation of the solution:
+
+- Start by initializing the ranges for search. The left will be the largest number in the array, and the right will be
+  the sum of all numbers.
+- Use a guessing approach. Start by considering a mid value between the left and right as a test value.
+- Check if it is possible to divide the array into k subarrays so that the sum of no subarray is greater than mid.
+  - Start with an empty sum and add numbers from the array. If adding the next number exceeds mid:
+    - Start a new subarray with that number and increment the count of the subarrays.
+  - Return FALSE if the count exceeds k. Otherwise, return TRUE.
+- Adjust the guessing range by checking if the number of subarrays needed is within the k and reduce the mid to see if
+  a smaller largest sum is possible.
+- Otherwise, if the count of subarrays is more than k:
+  - Increase the mid to make larger groups possible.
+- Continue adjusting the mid until left < right. Return left as it contains the minimized largest possible sum.
+
+- Let’s look at the following illustrations to get a better understanding of the solution:
+
+![Solution 1](images/solutions/split_array_largest_sum_solution_1.png)
+![Solution 2](images/solutions/split_array_largest_sum_solution_2.png)
+![Solution 3](images/solutions/split_array_largest_sum_solution_3.png)
+![Solution 4](images/solutions/split_array_largest_sum_solution_4.png)
+![Solution 5](images/solutions/split_array_largest_sum_solution_5.png)
+![Solution 6](images/solutions/split_array_largest_sum_solution_6.png)
+![Solution 7](images/solutions/split_array_largest_sum_solution_7.png)
+![Solution 8](images/solutions/split_array_largest_sum_solution_8.png)
+![Solution 9](images/solutions/split_array_largest_sum_solution_9.png)
+![Solution 10](images/solutions/split_array_largest_sum_solution_10.png)
+![Solution 11](images/solutions/split_array_largest_sum_solution_11.png)
+![Solution 12](images/solutions/split_array_largest_sum_solution_12.png)
+![Solution 13](images/solutions/split_array_largest_sum_solution_13.png)
+![Solution 14](images/solutions/split_array_largest_sum_solution_14.png)
+![Solution 15](images/solutions/split_array_largest_sum_solution_15.png)
+![Solution 16](images/solutions/split_array_largest_sum_solution_16.png)
+![Solution 17](images/solutions/split_array_largest_sum_solution_17.png)
+![Solution 18](images/solutions/split_array_largest_sum_solution_18.png)
+![Solution 19](images/solutions/split_array_largest_sum_solution_19.png)
+![Solution 20](images/solutions/split_array_largest_sum_solution_20.png)
+
+### Time Complexity
+
+The time complexity of this solution is O(n log(m)), where n is the length of the input array, and m is the difference
+between `max(nums)` and `sum(nums)` because the range of possible sums considered during the binary search is from
+`max(nums)` to `sum(nums)`. This range size determines the number of iterations in the binary search. The tighter this
+range, the fewer iterations are needed. However, in the worst case, it spans the full difference: `sum(nums) - max(nums)`.
+The time complexity becomes `n log(m)` because the `can_split` function is called `n` times for each iteration.
+
+### Space Complexity
+
+The space complexity of this solution is O(1) because only constant space is used.

algorithms/search/binary_search/split_array_largest_sum/__init__.py

@@ -0,0 +1,87 @@
+from typing import List
+
+
+def split_array(nums: List[int], k: int) -> int:
+    if not nums:
+        return -1
+
+    left = max(nums)
+    right = sum(nums)
+    first_true_index = -1
+
+    def feasible(max_sum: int) -> bool:
+        """
+        Check if we can split the array into at most k sub arrays with each sub array sum less than or equal to max_sum
+        """
+        current_sum = 0
+        # Start with one
+        sub_array_count = 1
+
+        for num in nums:
+            if current_sum + num > max_sum:
+                current_sum = num
+                sub_array_count += 1
+            else:
+                current_sum += num
+
+        return sub_array_count <= k
+
+    while left <= right:
+        mid = (right + left) // 2
+        if feasible(mid):
+            first_true_index = mid
+            # Find a smaller valid value
+            right = mid - 1
+        else:
+            # Find a larger valid value
+            left = mid + 1
+
+    return first_true_index
+
+
+def split_array_2(nums, k):
+    if not nums:
+        return -1
+
+    # Set the initial search range for the largest sum:
+    # Minimum is the largest number in the array, and maximum is the sum of all numbers
+    left, right = max(nums), sum(nums)
+
+    def can_split(middle: int):
+        # Initialize the count of subarrays and the current sum of the current subarray
+        subarrays = 1
+        current_sum = 0
+
+        for num in nums:
+            # Check if adding the current number exceeds the allowed sum (mid)
+            if current_sum + num > middle:
+                # Increment the count of subarrays
+                subarrays += 1
+                # Start a new subarray with the current number
+                current_sum = num
+
+                # If the number of subarrays exceeds the allowed k, return False
+                if subarrays > k:
+                    return False
+            else:
+                # Otherwise, add the number to the current subarray
+                current_sum += num
+
+        # Return True if the array can be split within the allowed subarray count
+        return True
+
+    # Perform binary search to find the minimum largest sum
+    while left < right:
+        # Find the middle value of the current range
+        mid = (left + right) // 2
+
+        # Check if the array can be split into k or fewer subarrays with this maximum sum
+        if can_split(mid):
+            # If possible, try a smaller maximum sum
+            right = mid
+        else:
+            # Otherwise, increase the range to allow larger sums
+            left = mid + 1
+
+    # Return the smallest maximum sum that satisfies the condition
+    return left