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0025-reverse-nodes-in-k-group/0025-reverse-nodes-in-k-group.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+        if k == 1:
+            return head
+
+        dummyHead = ListNode(-1, head)
+
+        beforeSeg = dummyHead
+        firstInSeg = head
+        lastInSeg = None
+        afterSeg = None
+
+        # Get initial values for lastInSeg and afterSeg
+        curr = head
+        for i in range(k):
+            if i == k - 1:
+                lastInSeg = curr
+                afterSeg = curr.next
+            elif curr is None:
+                return head
+
+            curr = curr.next
+        
+
+        # hold one before segment
+        # hold last in segment
+        # hold one after segment
+
+        while lastInSeg is not None:
+            # in segment, flip all links
+            prev = None
+            curr = beforeSeg.next
+            next = beforeSeg.next.next
+
+            for i in range(k):
+                newPrev = curr
+                newCurr = next
+                newNext = next.next if next is not None else None
+
+                curr.next = prev
+
+                prev, curr, next = newPrev, newCurr, newNext
+
+            
+            beforeSeg.next = lastInSeg
+            firstInSeg.next = afterSeg
+
+            beforeSeg = firstInSeg
+            firstInSeg = afterSeg
+
+            curr = firstInSeg
+            lastInSeg = None
+            afterSeg = None
+            for i in range(k):
+                if curr is None:
+                    break
+                elif i == k - 1:
+                    lastInSeg = curr
+                    afterSeg = curr.next
+
+                curr = curr.next
+
+        return dummyHead.next

0025-reverse-nodes-in-k-group/README.md

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+<h2><a href="https://leetcode.com/problems/reverse-nodes-in-k-group">25. Reverse Nodes in k-Group</a></h2><h3>Hard</h3><hr><p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>
+
+<p><code>k</code> is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of <code>k</code> then left-out nodes, in the end, should remain as it is.</p>
+
+<p>You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5], k = 2
+<strong>Output:</strong> [2,1,4,3,5]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5], k = 3
+<strong>Output:</strong> [3,2,1,4,5]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the list is <code>n</code>.</li>
+	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
+	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>

0133-clone-graph/0133-clone-graph.py

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+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val = 0, neighbors = None):
+        self.val = val
+        self.neighbors = neighbors if neighbors is not None else []
+"""
+
+from typing import Optional
+from collections import deque
+class Solution:
+    def cloneGraph(self, node: Optional['Node']) -> Optional['Node']:
+        if not node:
+            return node
+
+        newHead = Node(node.val)
+
+        frontier = deque([ node ])
+        nodeMapping = { node.val: newHead }
+        while len(frontier) > 0:
+            current = frontier.popleft()
+
+            for n in current.neighbors:
+                if n.val not in nodeMapping:
+                    newNode = Node(n.val)
+                    nodeMapping[newNode.val] = newNode
+                    frontier.append(n)
+                nodeMapping[current.val].neighbors.append(nodeMapping[n.val])
+
+        return newHead
+        

0133-clone-graph/README.md

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+<h2><a href="https://leetcode.com/problems/clone-graph">133. Clone Graph</a></h2><h3>Medium</h3><hr><p>Given a reference of a node in a <strong><a href="https://en.wikipedia.org/wiki/Connectivity_(graph_theory)#Connected_graph" target="_blank">connected</a></strong> undirected graph.</p>
+
+<p>Return a <a href="https://en.wikipedia.org/wiki/Object_copying#Deep_copy" target="_blank"><strong>deep copy</strong></a> (clone) of the graph.</p>
+
+<p>Each node in the graph contains a value (<code>int</code>) and a list (<code>List[Node]</code>) of its neighbors.</p>
+
+<pre>
+class Node {
+    public int val;
+    public List&lt;Node&gt; neighbors;
+}
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Test case format:</strong></p>
+
+<p>For simplicity, each node&#39;s value is the same as the node&#39;s index (1-indexed). For example, the first node with <code>val == 1</code>, the second node with <code>val == 2</code>, and so on. The graph is represented in the test case using an adjacency list.</p>
+
+<p><b>An adjacency list</b> is a collection of unordered <b>lists</b> used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.</p>
+
+<p>The given node will always be the first node with <code>val = 1</code>. You must return the <strong>copy of the given node</strong> as a reference to the cloned graph.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/11/04/133_clone_graph_question.png" style="width: 454px; height: 500px;" />
+<pre>
+<strong>Input:</strong> adjList = [[2,4],[1,3],[2,4],[1,3]]
+<strong>Output:</strong> [[2,4],[1,3],[2,4],[1,3]]
+<strong>Explanation:</strong> There are 4 nodes in the graph.
+1st node (val = 1)&#39;s neighbors are 2nd node (val = 2) and 4th node (val = 4).
+2nd node (val = 2)&#39;s neighbors are 1st node (val = 1) and 3rd node (val = 3).
+3rd node (val = 3)&#39;s neighbors are 2nd node (val = 2) and 4th node (val = 4).
+4th node (val = 4)&#39;s neighbors are 1st node (val = 1) and 3rd node (val = 3).
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/01/07/graph.png" style="width: 163px; height: 148px;" />
+<pre>
+<strong>Input:</strong> adjList = [[]]
+<strong>Output:</strong> [[]]
+<strong>Explanation:</strong> Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> adjList = []
+<strong>Output:</strong> []
+<strong>Explanation:</strong> This an empty graph, it does not have any nodes.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the graph is in the range <code>[0, 100]</code>.</li>
+	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
+	<li><code>Node.val</code> is unique for each node.</li>
+	<li>There are no repeated edges and no self-loops in the graph.</li>
+	<li>The Graph is connected and all nodes can be visited starting from the given node.</li>
+</ul>

0227-basic-calculator-ii/0227-basic-calculator-ii.py

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+class Solution:
+    def calculate(self, s: str) -> int:
+        """
+        Complexities:
+        Time:  O(n)
+        Space: O(n)
+
+        where n = len(s)
+        """
+
+
+        numbers = "0123456789"
+
+        # Read from left to right: save in stack (full number or operator)
+        # - save number
+        # - on operator:
+        # -- if * or /, evaluate
+        # -- if + or -, wait and continue
+        currentNumber = 0
+        stack = []
+
+        # Evaluate numbers and create stack (of additions + subtractions)
+        for c in s + "\0":
+            if c == " ":
+                continue
+            elif c in numbers:
+                digit = int(c)
+                currentNumber = (currentNumber * 10) + digit
+            else:
+                stack.append(currentNumber)
+                currentNumber = 0
+
+                if len(stack) >= 3 and stack[-2] in "*/":
+                    rOperand = stack.pop()
+                    operator = stack.pop()
+                    lOperand = stack.pop()
+
+                    if operator == "*":
+                        stack.append(lOperand * rOperand)
+                    elif operator == "/":
+                        stack.append(int(lOperand / rOperand))
+
+                stack.append(c)
+
+        # Pop off null terminator
+        stack.pop()
+
+        # Evaluate all additions and subtractions from stack
+        current = stack[0]
+        for i in range(1, len(stack) - 1, 2):
+            operator = stack[i]
+            rOperand = stack[i + 1]
+
+            if operator == "+":
+                current += rOperand
+            elif operator == "-":
+                current -= rOperand
+
+        return current
+            

0227-basic-calculator-ii/README.md

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+<h2><a href="https://leetcode.com/problems/basic-calculator-ii">227. Basic Calculator II</a></h2><h3>Medium</h3><hr><p>Given a string <code>s</code> which represents an expression, <em>evaluate this expression and return its value</em>.&nbsp;</p>
+
+<p>The integer division should truncate toward zero.</p>
+
+<p>You may assume that the given expression is always valid. All intermediate results will be in the range of <code>[-2<sup>31</sup>, 2<sup>31</sup> - 1]</code>.</p>
+
+<p><strong>Note:</strong> You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as <code>eval()</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<pre><strong>Input:</strong> s = "3+2*2"
+<strong>Output:</strong> 7
+</pre><p><strong class="example">Example 2:</strong></p>
+<pre><strong>Input:</strong> s = " 3/2 "
+<strong>Output:</strong> 1
+</pre><p><strong class="example">Example 3:</strong></p>
+<pre><strong>Input:</strong> s = " 3+5 / 2 "
+<strong>Output:</strong> 5
+</pre>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 3 * 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of integers and operators <code>(&#39;+&#39;, &#39;-&#39;, &#39;*&#39;, &#39;/&#39;)</code> separated by some number of spaces.</li>
+	<li><code>s</code> represents <strong>a valid expression</strong>.</li>
+	<li>All the integers in the expression are non-negative integers in the range <code>[0, 2<sup>31</sup> - 1]</code>.</li>
+	<li>The answer is <strong>guaranteed</strong> to fit in a <strong>32-bit integer</strong>.</li>
+</ul>

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>

2050-parallel-courses-iii/2050-parallel-courses-iii.py

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+class Solution:
+    def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
+        """
+        Complexities:
+        Time:  O(n^2)
+        Space: O(r)
+
+        where n = n, r = len(relations)
+        """
+
+        # courses been taken array
+        # - O(r) time / O(r) space
+        hasPrereq = [False] * n
+        prereqMap = [set() for i in range(n)]
+        leadsToMap = [set() for i in range(n)]
+        for prevCourse, nextCourse in relations:
+            hasPrereq[nextCourse - 1] = True
+            prereqMap[nextCourse - 1].add(prevCourse)
+            leadsToMap[prevCourse - 1].add(nextCourse)
+
+        # O(n) space
+        timeUntilComplete = [-1] * n
+        coursesCalculated = 0
+        maxOverallTimeFound = 0
+        toResolve = set()
+
+        # calculate timeUntilComplete for no prereq courses
+        # - O(n) time
+        for i in range(n):
+            if not hasPrereq[i]:
+                timeUntilComplete[i] = time[i]
+                maxOverallTimeFound = max(maxOverallTimeFound, time[i])
+                coursesCalculated += 1
+                toResolve.update(leadsToMap[i])
+
+        # find all timeUntilComplete
+        # - if all prereq times are known, then calc
+        # - O(n^2) time
+        while len(toResolve) > 0:
+            nextCourse = toResolve.pop()
+            maxPrereqTime = -1
+            for prereq in prereqMap[nextCourse - 1]:
+                if timeUntilComplete[prereq - 1] == -1:
+                    maxPrereqTime = -1
+                    break
+                maxPrereqTime = max(maxPrereqTime, timeUntilComplete[prereq - 1])
+
+            if maxPrereqTime > -1:
+                timeUntilComplete[nextCourse - 1] = maxPrereqTime + time[nextCourse - 1]
+                maxOverallTimeFound = max(maxOverallTimeFound, timeUntilComplete[nextCourse - 1])
+                coursesCalculated += 1
+                toResolve.update(leadsToMap[nextCourse - 1])
+        
+
+
+        return maxOverallTimeFound

2050-parallel-courses-iii/README.md

@@ -0,0 +1,54 @@
+<h2><a href="https://leetcode.com/problems/parallel-courses-iii">2050. Parallel Courses III</a></h2><h3>Hard</h3><hr><p>You are given an integer <code>n</code>, which indicates that there are <code>n</code> courses labeled from <code>1</code> to <code>n</code>. You are also given a 2D integer array <code>relations</code> where <code>relations[j] = [prevCourse<sub>j</sub>, nextCourse<sub>j</sub>]</code> denotes that course <code>prevCourse<sub>j</sub></code> has to be completed <strong>before</strong> course <code>nextCourse<sub>j</sub></code> (prerequisite relationship). Furthermore, you are given a <strong>0-indexed</strong> integer array <code>time</code> where <code>time[i]</code> denotes how many <strong>months</strong> it takes to complete the <code>(i+1)<sup>th</sup></code> course.</p>
+
+<p>You must find the <strong>minimum</strong> number of months needed to complete all the courses following these rules:</p>
+
+<ul>
+	<li>You may start taking a course at <strong>any time</strong> if the prerequisites are met.</li>
+	<li><strong>Any number of courses</strong> can be taken at the <strong>same time</strong>.</li>
+</ul>
+
+<p>Return <em>the <strong>minimum</strong> number of months needed to complete all the courses</em>.</p>
+
+<p><strong>Note:</strong> The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex1.png" style="width: 392px; height: 232px;" /></strong>
+
+<pre>
+<strong>Input:</strong> n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
+<strong>Output:</strong> 8
+<strong>Explanation:</strong> The figure above represents the given graph and the time required to complete each course. 
+We start course 1 and course 2 simultaneously at month 0.
+Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
+Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex2.png" style="width: 500px; height: 365px;" /></strong>
+
+<pre>
+<strong>Input:</strong> n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
+<strong>Output:</strong> 12
+<strong>Explanation:</strong> The figure above represents the given graph and the time required to complete each course.
+You can start courses 1, 2, and 3 at month 0.
+You can complete them after 1, 2, and 3 months respectively.
+Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
+Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
+Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= relations.length &lt;= min(n * (n - 1) / 2, 5 * 10<sup>4</sup>)</code></li>
+	<li><code>relations[j].length == 2</code></li>
+	<li><code>1 &lt;= prevCourse<sub>j</sub>, nextCourse<sub>j</sub> &lt;= n</code></li>
+	<li><code>prevCourse<sub>j</sub> != nextCourse<sub>j</sub></code></li>
+	<li>All the pairs <code>[prevCourse<sub>j</sub>, nextCourse<sub>j</sub>]</code> are <strong>unique</strong>.</li>
+	<li><code>time.length == n</code></li>
+	<li><code>1 &lt;= time[i] &lt;= 10<sup>4</sup></code></li>
+	<li>The given graph is a directed acyclic graph.</li>
+</ul>

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

@@ -0,0 +1,17 @@
+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

@@ -0,0 +1,55 @@
+<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>