feat(algorithms, hash-maps): powerful integers

DIRECTORY.md

@@ -217,6 +217,8 @@
       * [Test First Unique Character](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/hash_table/first_unique_character/test_first_unique_character.py)
     * Jewels And Stones
       * [Test Jewels And Stones](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/hash_table/jewels_and_stones/test_jewels_and_stones.py)
+    * Powerful Integers
+      * [Test Powerful Integers](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/hash_table/powerful_integers/test_powerful_integers.py)
     * Ransom Note
       * [Test Ransom Note](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/hash_table/ransom_note/test_ransom_note.py)
   * Heap

algorithms/dynamic_programming/max_subarray/README.md

@@ -20,3 +20,36 @@ find_subarr_maxsum([4, -1, 2, 1, -40, 1, 2, -1, 4]) == [[[4, -1, 2, 1], [1, 2, -
 ```
 
 If the array does not have any sub-array with a positive sum of its terms, the function will return [[], 0].
+
+## Solution
+
+We’ll solve this problem using Kadane’s Algorithm. It’s a dynamic programming approach. The key idea is that we can
+efficiently find the maximum subarray ending at any position based on the maximum subarray ending at the previous 
+position.
+
+The subproblem here is finding the maximum subarray sum that ends at a specific index i. We need to calculate this for
+every index i in the array. The base case is the first element of the array, where both current subarray sum and maximum
+subarray sum are initialized with the first element’s value. This is the starting point for solving the subproblems. At
+each step, we reuse the previously computed maximum subarray sum to find the solution for the current subproblem.
+
+The steps of the algorithm are given below:
+
+1. Initialize two variables, current_sum and max_sum, with the value of the first element in the input list. These 
+   variables are used to keep track of the current sum of the subarray being considered, and the maximum sum found so
+   far.
+2. Iterate through the input list, starting from the second element to the end of the list. Within the loop, perform the
+   following steps:
+   - If current_sum + nums[i] is smaller than nums[i], this indicates that starting a new subarray with nums[i] would
+     yield a larger sum than extending the previous subarray. In such cases, reset current_sum to nums[i].
+   - Compare the current max_sum with the updated current_sum. This step ensures that the max_sum always stores the
+     maximum sum encountered so far.
+3. After the loop completes, the function returns the max_sum, which represents the maximum sum of any contiguous
+   subarray within the input list.
+
+### Time Complexity
+The time complexity of this solution is O(n) because we are iterating the array once, where n is the total number of
+elements in the array.
+
+### Space Complexity
+
+The space complexity of this solution is O(1).

algorithms/hash_table/powerful_integers/README.md

@@ -0,0 +1,127 @@
+# Powerful Integers
+
+Given three integers x, y, and bound, return a list of all the powerful integers that have a value less than or equal to
+bound.
+
+An integer is powerful if it can be represented as xi + yj for some integers i >= 0 and j >= 0.
+
+You may return the answer in any order. In your answer, each value should occur at most once.
+
+## Examples
+
+Example 1:
+
+```text
+Input: x = 2, y = 3, bound = 10
+Output: [2,3,4,5,7,9,10]
+Explanation:
+2 = 20 + 30
+3 = 21 + 30
+4 = 20 + 31
+5 = 21 + 31
+7 = 22 + 31
+9 = 23 + 30
+10 = 20 + 32
+```
+
+Example 2:
+```text
+Input: x = 3, y = 5, bound = 15
+Output: [2,4,6,8,10,14]
+```
+
+## Constraints
+
+- 1 <= x, y <= 100
+- 0 <= bound <= 10^6
+
+## Topics
+
+- Hash Table
+- Math
+- Enumeration
+
+## Solution(s)
+
+1. [Logarithmic Bounds](#logarithmic-bounds)
+2. [Logarithmic Bound 2](#logarithmic-bound-2)
+
+### Logarithmic Bounds
+
+Our approach here will only focus on finding the bounds for numbers x and y. One way to get the bounds on the powers is
+to have nested loops that iterate from [0⋯bound]. However, this is very inefficient because the bound can be an extremely
+large value and a nested-loop over this bound will take forever to finish. Also, we don't need to iterate over all of the
+values and combinations. There is a way to find a much smaller bound for the powers.
+
+m^n <= bound
+
+This formula implies that
+
+n<=log(m) bound
+
+> We can use the log function to determine the bounds for the powers of "x" and "y".
+
+#### Algorithm
+
+1. Let's define `a` as the power bound for the number `x`. Thus `a=log(x)bound`.
+2. Similarly, let's define `b` as the power bound for the number `y`. Thus `b=log(y)bound`.
+3. Now we will have our nested for-loop structure where the outer loop will iterate from [0⋯a] and the inner loop will
+   iterate from [0⋯b].
+4. We will use a set to store our results. This is because we might generate the same value multiple times. E.g. 
+   `2^1 + 3^2 = 11` and `2^3 + 3^1 = 11`. We only need to include the value 11 once and hence, we will use a set called
+   `resultSet` to store our answers.
+5. At each step, we calculate `x^a + y^b` and check if this value is less than or equal to bound. If it is, then this is
+   a powerful integer and we add it to our set of answers.
+6. We need special break conditions to handle the scenario when x or y is 1. This is because if the number x or y is 1,
+   then their power-bound will be equal to bound itself. Also, it doesn't matter what their power-bound is because 1^N
+   is always 1. Thus, when the number is 1, we don't need to loop from [0⋯N] and we can break early.
+7. Finally, convert the set to a list and return.
+
+#### Time Complexity
+
+Time Complexity: Let `N` be `log(x)bound` and `M` be `log(y)bound`. Then the overall time complexity is `O(N×M)` because
+we used a nested loop structure to calculate all of the powerful integers.
+
+#### Space Complexity
+
+`O(N×M)` because we use a set to omit duplicates. We could just use our result list to check membership before adding
+values. However, that would be costly in terms of time complexity because it would require a full scan of the result list
+to see if the value already exists.
+
+### Logarithmic Bound 2
+
+The key insight is that since x^i and y^j grow exponentially, the number of distinct powers of x and y that remain within
+bound is at most O(log(bound) each. We can enumerate all pairs (i,j) by iterating through powers of x in an outer loop
+and powers of y in an inner loop, adding each sum x^i + y^j that is less than or equal to bound into a hash set called
+`resultSet`. The set automatically handles deduplication, ensuring each powerful integer appears at most once. A special
+case arises when x or y equals 1, because 1^i is always 1 regardless of the exponent, which would cause an infinite loop.
+We handle this by breaking out of the respective loop after the first iteration when x or y is 1.
+
+Solution steps:
+
+1. Initialize an empty set resultSet to store unique powerful integers.
+2. Initialize `powX` to 1, representing x^0
+3. Start an outer while loop that continues as long as `powX ≤ bound`.
+   - Inside the outer loop, initialize `powY` to 1, representing y^0
+   - Start an inner while loop that continues as long as `powX + powY ≤ bound`.
+     - Add the value `powX` + `powY` to `resultSet`.
+     - If y equals 1, break out of the inner loop immediately, since y^j will always be 1 and further iterations would
+       not produce new values.
+     - Otherwise, multiply `powY` by y to advance to the next power of y.
+   - After the inner loop, if x equals 1, break out of the outer loop immediately, since x^i will always be 1 and further
+     iterations would not produce new values.
+   - Otherwise, multiply `powX` by x to advance to the next power of x.
+4. Convert resultSet to a list and return it.
+
+#### Time Complexity
+
+The time complexity of the solution is O(logx(bound)) * logy(bound) because the outer loop runs at most O(logx(bound))
+times and the inner loop runs at most O(logy(bound)) times for each iteration of the outer loop. Since bound ≤10^6 and
+the minimum base greater than 1 is 2, each loop runs at most about log2(10^6)≈20 iterations, making this very efficient.
+When x or y is 1, the corresponding loop runs only once.
+
+#### Space Complexity
+
+The space complexity is `O(logx(bound)) * logy(bound)`  because in the worst case, every combination of powers produces
+a unique sum, and all of these are stored in the resultSet. This is bounded by the total number of pairs enumerated,
+which is at most `O(logx(bound)) * logy(bound)`.

algorithms/hash_table/powerful_integers/__init__.py

@@ -0,0 +1,55 @@
+from typing import List, Set
+from math import log
+
+
+def powerful_integers(x: int, y: int, bound: int) -> List[int]:
+    # Use a set to store unique powerful integers
+    result_set: Set[int] = set()
+
+    # Compute powers of x up to bound
+    # If x == 1, x^i is always 1, so only need i=0
+    pow_x = 1  # x^0 = 1
+    while pow_x <= bound:
+        # For each power of x, iterate over powers of y
+        pow_y = 1  # y^0 = 1
+        while pow_x + pow_y <= bound:
+            # Add the powerful integer to the set
+            result_set.add(pow_x + pow_y)
+            # If y is 1, y^j is always 1, so break after first iteration
+            if y == 1:
+                break
+            # Move to next power of y
+            pow_y *= y
+        # If x is 1, x^i is always 1, so break after first iteration
+        if x == 1:
+            break
+        # Move to next power of x
+        pow_x *= x
+
+    # Convert set to list and return
+    return list(result_set)
+
+
+def powerful_integers_logarithmic_bounds(x: int, y: int, bound: int) -> List[int]:
+    if bound == 0:
+        return []
+
+    a = bound if x == 1 else int(log(bound, x))
+    b = bound if y == 1 else int(log(bound, y))
+
+    result_set = set([])
+
+    for i in range(a + 1):
+        for j in range(b + 1):
+            value = x**i + y**j
+
+            if value <= bound:
+                result_set.add(value)
+
+            if y == 1:
+                break
+
+        if x == 1:
+            break
+
+    return list(result_set)

algorithms/hash_table/powerful_integers/test_powerful_integers.py

@@ -0,0 +1,37 @@
+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.hash_table.powerful_integers import (
+    powerful_integers,
+    powerful_integers_logarithmic_bounds,
+)
+
+POWERFUL_INTEGERS_TEST_CASE = [
+    (2, 2, 20, [2, 3, 4, 5, 6, 8, 9, 10, 12, 16, 17, 18, 20]),
+    (1, 1, 5, [2]),
+    (5, 3, 50, [2, 4, 6, 8, 10, 14, 26, 28, 32, 34]),
+    (100, 100, 1000000, [2, 101, 200, 10001, 10100, 20000]),
+    (2, 5, 0, []),
+    (2, 3, 10, [2, 3, 4, 5, 7, 9, 10]),
+    (3, 5, 15, [2, 4, 6, 8, 10, 14]),
+]
+
+
+class PowerfulIntegersTestCase(unittest.TestCase):
+    @parameterized.expand(POWERFUL_INTEGERS_TEST_CASE)
+    def test_powerful_integers(self, x: int, y: int, bound: int, expected: List[int]):
+        actual = powerful_integers(x, y, bound)
+        actual.sort()
+        self.assertEqual(expected, actual)
+
+    @parameterized.expand(POWERFUL_INTEGERS_TEST_CASE)
+    def test_powerful_integers_logarithmic_bounds(
+        self, x: int, y: int, bound: int, expected: List[int]
+    ):
+        actual = powerful_integers_logarithmic_bounds(x, y, bound)
+        actual.sort()
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()

algorithms/trie/index_pairs_of_a_string/__init__.py

@@ -1,6 +1,7 @@
 from typing import List
 from datastructures.trees.trie import Trie
 
+
 def index_pairs(text: str, words: List[str]) -> List[List[int]]:
     trie = Trie()