Sync LeetCode submission Runtime - 367 ms (32.30%), Memory - 40.7 MB (29.35%)

Author: jimit105

0518-coin-change-ii/README.md

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+<p>You are given an integer array <code>coins</code> representing coins of different denominations and an integer <code>amount</code> representing a total amount of money.</p>
+
+<p>Return <em>the number of combinations that make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>0</code>.</p>
+
+<p>You may assume that you have an infinite number of each kind of coin.</p>
+
+<p>The answer is <strong>guaranteed</strong> to fit into a signed <strong>32-bit</strong> integer.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> amount = 5, coins = [1,2,5]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> there are four ways to make up the amount:
+5=5
+5=2+2+1
+5=2+1+1+1
+5=1+1+1+1+1
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> amount = 3, coins = [2]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> the amount of 3 cannot be made up just with coins of 2.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> amount = 10, coins = [10]
+<strong>Output:</strong> 1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= coins.length &lt;= 300</code></li>
+	<li><code>1 &lt;= coins[i] &lt;= 5000</code></li>
+	<li>All the values of <code>coins</code> are <strong>unique</strong>.</li>
+	<li><code>0 &lt;= amount &lt;= 5000</code></li>
+</ul>

0518-coin-change-ii/solution.py

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+# Approach 1: Top-down Dynamic Programming
+
+# Time: O(n * amount)
+# Space: O(n * amount)
+
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        def number_of_ways(i, amount):
+            if amount == 0:
+                return 1
+            if i == len(coins):
+                return 0
+            if memo[i][amount] != -1:
+                return memo[i][amount]
+
+            if coins[i] > amount:
+                memo[i][amount] = number_of_ways(i + 1, amount)
+            else:
+                memo[i][amount] = number_of_ways(i, amount - coins[i]) + number_of_ways(i + 1, amount)
+
+            return memo[i][amount]
+
+        memo = [[-1] * (amount + 1) for _ in range(len(coins))]
+
+        return number_of_ways(0, amount)