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0322-coin-change/0322-coin-change.py

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+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        """
+        Complexities:
+        Time:  O(mn)
+        Space: O(m)
+
+        where n = len(coins), m = amount
+        """
+
+        dp = [-1] * (amount + 1)
+
+        dp[0] = 0
+        for i in range(amount + 1):
+            for coinValue in coins:
+                if dp[i] == -1 or i + coinValue > amount:
+                    continue
+                elif dp[i + coinValue] == -1:
+                    dp[i + coinValue] = dp[i] + 1
+                else:
+                    dp[i + coinValue] = min(dp[i + coinValue], dp[i] + 1)
+
+        print(dp)
+        return dp[amount]

0322-coin-change/README.md

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+<h2><a href="https://leetcode.com/problems/coin-change">322. Coin Change</a></h2><h3>Medium</h3><hr><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 fewest number of coins that you need to make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>
+
+<p>You may assume that you have an infinite number of each kind of coin.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> coins = [1,2,5], amount = 11
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 11 = 5 + 5 + 1
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> coins = [2], amount = 3
+<strong>Output:</strong> -1
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> coins = [1], amount = 0
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= coins.length &lt;= 12</code></li>
+	<li><code>1 &lt;= coins[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>0 &lt;= amount &lt;= 10<sup>4</sup></code></li>
+</ul>

2483-minimum-penalty-for-a-shop/2483-minimum-penalty-for-a-shop.py

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+class Solution:
+    def bestClosingTime(self, customers: str) -> int:
+        penalty = 0
+        minPenalty = 0
+        bestHour = 0
+
+        for hour, hasCustomers in enumerate(customers):
+            if hasCustomers == "Y":
+                penalty -= 1
+            else:
+                penalty += 1
+
+            if penalty < minPenalty:
+                minPenalty = penalty
+                bestHour = hour + 1
+
+        return bestHour

2483-minimum-penalty-for-a-shop/README.md

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+<h2><a href="https://leetcode.com/problems/minimum-penalty-for-a-shop">2483. Minimum Penalty for a Shop</a></h2><h3>Medium</h3><hr><p>You are given the customer visit log of a shop represented by a <strong>0-indexed</strong> string <code>customers</code> consisting only of characters <code>&#39;N&#39;</code> and <code>&#39;Y&#39;</code>:</p>
+
+<ul>
+	<li>if the <code>i<sup>th</sup></code> character is <code>&#39;Y&#39;</code>, it means that customers come at the <code>i<sup>th</sup></code> hour</li>
+	<li>whereas <code>&#39;N&#39;</code> indicates that no customers come at the <code>i<sup>th</sup></code> hour.</li>
+</ul>
+
+<p>If the shop closes at the <code>j<sup>th</sup></code> hour (<code>0 &lt;= j &lt;= n</code>), the <strong>penalty</strong> is calculated as follows:</p>
+
+<ul>
+	<li>For every hour when the shop is open and no customers come, the penalty increases by <code>1</code>.</li>
+	<li>For every hour when the shop is closed and customers come, the penalty increases by <code>1</code>.</li>
+</ul>
+
+<p>Return<em> the <strong>earliest</strong> hour at which the shop must be closed to incur a <strong>minimum</strong> penalty.</em></p>
+
+<p><strong>Note</strong> that if a shop closes at the <code>j<sup>th</sup></code> hour, it means the shop is closed at the hour <code>j</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> customers = &quot;YYNY&quot;
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 
+- Closing the shop at the 0<sup>th</sup> hour incurs in 1+1+0+1 = 3 penalty.
+- Closing the shop at the 1<sup>st</sup> hour incurs in 0+1+0+1 = 2 penalty.
+- Closing the shop at the 2<sup>nd</sup> hour incurs in 0+0+0+1 = 1 penalty.
+- Closing the shop at the 3<sup>rd</sup> hour incurs in 0+0+1+1 = 2 penalty.
+- Closing the shop at the 4<sup>th</sup> hour incurs in 0+0+1+0 = 1 penalty.
+Closing the shop at 2<sup>nd</sup> or 4<sup>th</sup> hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> customers = &quot;NNNNN&quot;
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> It is best to close the shop at the 0<sup>th</sup> hour as no customers arrive.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> customers = &quot;YYYY&quot;
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> It is best to close the shop at the 4<sup>th</sup> hour as customers arrive at each hour.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= customers.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>customers</code> consists only of characters <code>&#39;Y&#39;</code> and <code>&#39;N&#39;</code>.</li>
+</ul>