Merge pull request #162 from BrianLusina/feat/algorithms-dynamic-programming-coin-change

feat(algorithms, dynamic programming): coin change

Author: BrianLusina

DIRECTORY.md

@@ -57,6 +57,8 @@
       * [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)
+    * Coin Change
+      * [Test Coin Change](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/coin_change/test_coin_change.py)
     * Countingbits
       * [Test Counting Bits](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/dynamic_programming/countingbits/test_counting_bits.py)
     * Decodeways
@@ -1076,7 +1078,6 @@
     * [Test Bowling](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_bowling.py)
     * [Test Cake Is Not A Lie](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_cake_is_not_a_lie.py)
     * [Test Chess Board Cell Color](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_chess_board_cell_color.py)
-    * [Test Coin Change](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_coin_change.py)
     * [Test Coin Flip](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_coin_flip.py)
     * [Test Count Vegetables](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_count_vegetables.py)
     * [Test Doomsday Fuel](https://github.com/BrianLusina/PythonSnips/blob/master/tests/puzzles/test_doomsday_fuel.py)

algorithms/dynamic_programming/climb_stairs/__init__.py

@@ -13,24 +13,40 @@ def climb_stairs(n: int) -> int:
     return second
 
 
-def climb(n: int) -> int:
+def climb_stairs_dp_bottom_up(n: int) -> int:
+    if n < 0:
+        return 0
+    if n <= 1:
+        return 1
+
+    dp = [0] * (n + 1)
+    dp[1] = 1
+    dp[2] = 2
+
+    for idx in range(3, n + 1):
+        dp[idx] = dp[idx - 1] + dp[idx - 2]
+
+    return dp[n]
+
+
+def climb_stairs_dp_top_down(n: int) -> int:
     """
     Finds the number of possible ways to climb up n steps given the steps can be climbed wither 1 at a time or 2 at a
     time
     :param n: number of steps
     :return: number of possible ways to climb up the steps
 
-    >>> climb(2)
+    >>> climb_stairs_dp_top_down(2)
     2
 
-    >>> climb(1)
+    >>> climb_stairs_dp_top_down(1)
     1
 
-    >>> climb(3)
+    >>> climb_stairs_dp_top_down(3)
     3
     """
     if n < 0:
         return 0
     if n == 1 or n == 0:
         return 1
-    return climb(n - 1) + climb(n - 2)
+    return climb_stairs_dp_top_down(n - 1) + climb_stairs_dp_top_down(n - 2)

algorithms/dynamic_programming/climb_stairs/test_climb_stairs.py

@@ -1,23 +1,34 @@
 import unittest
-
-from . import climb_stairs
+from parameterized import parameterized
+from algorithms.dynamic_programming.climb_stairs import (
+    climb_stairs,
+    climb_stairs_dp_bottom_up,
+    climb_stairs_dp_top_down,
+)
+
+CLIMB_STAIRS_TEST_CASES = [
+    (2, 2),
+    (3, 3),
+    (4, 5),
+    (4, 5),
+    (5, 8),
+]
 
 
 class ClimbStairsTestCase(unittest.TestCase):
-    def test_1(self):
-        """should return 2 for n = 2"""
-        n = 2
-        expected = 2
+    @parameterized.expand(CLIMB_STAIRS_TEST_CASES)
+    def test_climb_stairs(self, n: int, expected: int):
         actual = climb_stairs(n)
-
         self.assertEqual(expected, actual)
 
-    def test_2(self):
-        """should return 3 for n = 3"""
-        n = 3
-        expected = 3
-        actual = climb_stairs(n)
+    @parameterized.expand(CLIMB_STAIRS_TEST_CASES)
+    def test_climb_stairs_dp_bottom_up(self, n: int, expected: int):
+        actual = climb_stairs_dp_bottom_up(n)
+        self.assertEqual(expected, actual)
 
+    @parameterized.expand(CLIMB_STAIRS_TEST_CASES)
+    def test_climb_stairs_dp_top_down(self, n: int, expected: int):
+        actual = climb_stairs_dp_top_down(n)
         self.assertEqual(expected, actual)
 
 

algorithms/dynamic_programming/coin_change/README.md

@@ -0,0 +1,137 @@
+# Coin Change
+
+Correctly determine the fewest number of coins to be given to a customer such that the sum of the coins' value would
+equal the correct amount of change.
+
+## For example
+
+- An input of 15 with [1, 5, 10, 25, 100] should return one nickel (5)
+  and one dime (10) or [0, 1, 1, 0, 0]
+- An input of 40 with [1, 5, 10, 25, 100] should return one nickel (5)
+  and one dime (10) and one quarter (25) or [0, 1, 1, 1, 0]
+
+## Edge cases
+
+- Does your algorithm work for any given set of coins?
+- Can you ask for negative change?
+- Can you ask for a change value smaller than the smallest coin value?
+
+## Exception messages
+
+Sometimes it is necessary to raise an exception. When you do this, you should include a meaningful error message to
+indicate what the source of the error is. This makes your code more readable and helps significantly with debugging. Not
+every exercise will require you to raise an exception, but for those that do, the tests will only pass if you include a
+message.
+
+To raise a message with an exception, just write it as an argument to the exception type. For example, instead of
+`raise Exception`, you should write:
+
+```python
+raise Exception("Meaningful message indicating the source of the error")
+```
+
+## Source
+
+Software Craftsmanship - Coin Change
+Kata [https://web.archive.org/web/20130115115225/http://craftsmanship.sv.cmu.edu:80/exercises/coin-change-kata](https://web.archive.org/web/20130115115225/http://craftsmanship.sv.cmu.edu:80/exercises/coin-change-kata)
+
+## Solution
+
+If we look at the problem, we might immediately think that it could be solved through a greedy approach. However, if we
+look at it closely, we’ll know that it’s not the correct approach here. Let’s take a look at an example to understand why
+this problem can’t be solved with a greedy approach.
+
+Let's suppose we have coins = [1, 3, 4, 5] and we want to find the total = 7 and we try to solve the problem with a greedy
+approach. In a greedy approach, we always start from the very end of a sorted array and traverse backward to find our
+solution because that allows us to solve the problem without traversing the whole array. However, in this situation, we
+start off with a 5 and add that to our total. We then check if it’s possible to get a 7 with the help of either 4 or 3,
+but as expected, that won't be the case, and we would need to add 1 twice to get our required total.
+
+The problem seems to be solved, and we have concluded that we need maximum 3 coins to get to the total of 7. However, if
+we take a look at our array, that isn’t the case. In fact, we could have reached the total of 7 with just 2 coins: 4 and 3.
+So, the problem needs to be broken down into subproblems, and an optimal solution can be reached from the optimal 
+solutions of its subproblems.
+
+To split the problem into subproblems, let's assume we know the number of coins required for some total value and the
+last coin denomination is C. Because of the optimal substructure property, the following equation will be true:
+
+Min(total)=Min(total−C)+1
+
+But, we don't know what is the value of C yet, so we compute it for each element of the coins array and select the
+minimum from among them. This creates the following recurrence relation:
+
+Min(total)=mini=0.....n-1(Min(total−Ci)+1), such that
+Min(total)=0, for total=0
+Min(total)= -1, for n=0
+
+> Note: The problem can also be solved with the help of a simple recursive tree without any backtracking, but that would
+> take extra memory and time complexity, as we can see in the illustration below.
+
+![Coin Change Recursive Tree](images/solutions/coin_change_recursive_tree_1.png)
+
+> Recursive tree for finding minimum number of coins for the total 5 with the coins [1,2,3]
+
+### Step-by-step solution construction
+
+The idea is to solve the problem using the top-down technique of dynamic programming. If the required total is less than
+the number that’s being evaluated, the algorithm doesn’t make any more recursive calls. Moreover, the recursive tree
+calculates the results of many subproblems multiple times. Therefore, if we store the result of each subproblem in a
+table, we can drastically improve the algorithm’s efficiency by accessing the required value at a constant time. This
+massively reduces the number of recursive calls we need to make to reach our results.
+
+We start our solution by creating a helper function that assists us in calculating the number of coins we need. It has
+three base cases to cover about what to return if the remaining amount is:
+
+- Less than zero
+- Equal to zero
+- Neither less than zero nor equal to zero
+
+> The top-down solution, commonly known as the memoization technique, is an enhancement of the recursive solution. It
+> solves the problem of calculating redundant solutions over and over by storing them in an array.
+
+In the last case, when the remaining amount is neither of the base cases, we traverse the coins array, and at each
+element, we recursively call the calculate_minimum_coins() function, passing the updated remaining amount remaining_amount
+minus the value of the current coin. This step effectively evaluates the number of coins needed for each possible
+denomination to make up the remaining amount. We store the return value of the base cases for each subproblem in a
+variable named result. We then add 1 to the result variable indicating that we're using this coin denomination in the
+process of making up the corresponding total. Now, we assign this value to minimum, which is initially set to infinity
+at the start of each path.
+
+To avoid recalculating the minimum values for subproblems, we utilize the counter array, which serves as a memoization
+table. This array stores the minimum number of coins required to make up each specific amount of money up to the given
+total. At the end of each path traversal, we update the corresponding index of the counter array with the calculated
+minimum value. Finally, we return the minimum number of coins needed for the given total amount.
+
+![Solution 1](./images/solutions/coin_change_solution_1.png)
+![Solution 2](./images/solutions/coin_change_solution_2.png)
+![Solution 3](./images/solutions/coin_change_solution_3.png)
+![Solution 4](./images/solutions/coin_change_solution_4.png)
+![Solution 5](./images/solutions/coin_change_solution_5.png)
+![Solution 6](./images/solutions/coin_change_solution_6.png)
+![Solution 7](./images/solutions/coin_change_solution_7.png)
+![Solution 8](./images/solutions/coin_change_solution_8.png)
+
+### Summary
+
+To recap, the solution to this problem can be divided into the following parts:
+
+1. We first check the base cases, if total is either 0 or less than 0:
+   - 0 means no new coins need to be added because we have reached a viable solution.
+   - Less than 0 means our path can’t lead to this solution, so we need to backtrack.
+2. After this, we use the top-down approach and traverse the given coin denominations.
+3. At each iteration, we either pick a coin or we don’t. 
+   - If we pick a coin, we move on to solve a new subproblem based on the reduced total value. 
+   - If we don’t pick a coin, then we look up the answer from the counter array if it is already computed to avoid
+     recalculation.
+4. Finally, we return the minimum number of coins required for the given total.
+
+### Time Complexity
+
+The time complexity for the above algorithm is O(n*m). Here, 
+n represents the total and m represents the number of coins we have. In the worst case, the height of the recursive tree
+is n as the subproblems solved by the algorithm will be n because we're storing precalculated solutions in a table. Each
+subproblem takes m iterations, one per coin value. So, the time complexity is O(n*m).
+
+### Space Complexity
+
+The space complexity for this algorithm is O(n) because we’re using the counter array which is the size of total.

algorithms/dynamic_programming/coin_change/__init__.py

@@ -0,0 +1,100 @@
+"""
+Finds the minimum number of coins that sum up to the total change given the total change due and the list of coins with
+denominations
+"""
+
+from typing import List
+from itertools import combinations_with_replacement
+
+
+def find_minimum_coins(total_change: int, coins: List[int]) -> List[int]:
+    """
+    Finds the minimum number coins that add up to the total change given a list of coins. This should return a list of
+    the coins that add up to the total change
+    :param total_change: Total change due
+    :type total_change int
+    :param coins: list of denominations
+    :type coins list
+    :return: list with the least amount of coins that sum up to the total change
+    :rtype list
+    """
+    # return early when there is no change
+    if total_change == 0:
+        return []
+
+    if total_change < 0:
+        raise ValueError("Cannot find change of negative values")
+
+    if total_change < min(coins):
+        raise ValueError(
+            "Cannot find change if total change is smaller than smallest coin"
+        )
+
+    result = None
+
+    for n in range(total_change):
+        for combination in combinations_with_replacement(coins, n):
+            if sum(combination) == total_change:
+                return list(combination)
+    if result is None:
+        raise ValueError("No combination can add up to target")
+    return []
+
+
+def coin_change(coins: List[int], total: int) -> int:
+    if total == 0:
+        return 0
+    # Initialize dimensions: number of coin types and target amount
+    num_coins = len(coins)
+
+    # Create 2D DP table
+    # dp[i][j] represents minimum coins needed to make amount j using first i coin types
+    dp = [[float("inf")] * (total + 1) for _ in range(num_coins + 1)]
+
+    # Base case: 0 coins needed to make amount 0
+    dp[0][0] = 0
+
+    # Fill the DP table
+    for coin_idx in range(1, num_coins + 1):
+        current_coin_value = coins[coin_idx - 1]
+
+        for current_amount in range(total + 1):
+            # Option 1: Don't use the current coin type
+            dp[coin_idx][current_amount] = dp[coin_idx - 1][current_amount]
+
+            # Option 2: Use the current coin if possible
+            if current_amount >= current_coin_value:
+                # Compare with using one more of the current coin
+                dp[coin_idx][current_amount] = min(
+                    dp[coin_idx][current_amount],
+                    dp[coin_idx][current_amount - current_coin_value] + 1,
+                )
+
+    # Return result: -1 if impossible, otherwise the minimum number of coins
+    return -1 if dp[num_coins][total] == float("inf") else dp[num_coins][total]
+
+
+def coin_change_dp(coins: List[int], total: int) -> int:
+    if total < 1:
+        return 0
+    counter: List[int | float] = [float("inf")] * total
+
+    def calculate_minimum_coins(remaining_amount: int) -> int:
+        if remaining_amount < 0:
+            return -1
+        if remaining_amount == 0:
+            return 0
+        if counter[remaining_amount - 1] != float("inf"):
+            return counter[remaining_amount - 1]
+
+        minimum = float("inf")
+
+        for coin in coins:
+            result = calculate_minimum_coins(remaining_amount - coin)
+            if 0 <= result < minimum:
+                minimum = 1 + result
+
+        counter[remaining_amount - 1] = minimum if minimum != float("inf") else -1
+        return counter[remaining_amount - 1]
+
+    return calculate_minimum_coins(remaining_amount=total)