[LeetCode Sync] Runtime - 1476 ms (60.83%), Memory - 33.3 MB (41.24%)

Author: NinePiece2

Solutions/2618-maximize-the-minimum-powered-city/README.md

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+<p>You are given a <strong>0-indexed</strong> integer array <code>stations</code> of length <code>n</code>, where <code>stations[i]</code> represents the number of power stations in the <code>i<sup>th</sup></code> city.</p>
+
+<p>Each power station can provide power to every city in a fixed <strong>range</strong>. In other words, if the range is denoted by <code>r</code>, then a power station at city <code>i</code> can provide power to all cities <code>j</code> such that <code>|i - j| &lt;= r</code> and <code>0 &lt;= i, j &lt;= n - 1</code>.</p>
+
+<ul>
+	<li>Note that <code>|x|</code> denotes <strong>absolute</strong> value. For example, <code>|7 - 5| = 2</code> and <code>|3 - 10| = 7</code>.</li>
+</ul>
+
+<p>The <strong>power</strong> of a city is the total number of power stations it is being provided power from.</p>
+
+<p>The government has sanctioned building <code>k</code> more power stations, each of which can be built in any city, and have the same range as the pre-existing ones.</p>
+
+<p>Given the two integers <code>r</code> and <code>k</code>, return <em>the <strong>maximum possible minimum power</strong> of a city, if the additional power stations are built optimally.</em></p>
+
+<p><strong>Note</strong> that you can build the <code>k</code> power stations in multiple cities.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> 
+One of the optimal ways is to install both the power stations at city 1. 
+So stations will become [1,4,4,5,0].
+- City 0 is provided by 1 + 4 = 5 power stations.
+- City 1 is provided by 1 + 4 + 4 = 9 power stations.
+- City 2 is provided by 4 + 4 + 5 = 13 power stations.
+- City 3 is provided by 5 + 4 = 9 power stations.
+- City 4 is provided by 5 + 0 = 5 power stations.
+So the minimum power of a city is 5.
+Since it is not possible to obtain a larger power, we return 5.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+It can be proved that we cannot make the minimum power of a city greater than 4.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == stations.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= stations[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= r&nbsp;&lt;= n - 1</code></li>
+	<li><code>0 &lt;= k&nbsp;&lt;= 10<sup>9</sup></code></li>
+</ul>

Solutions/2618-maximize-the-minimum-powered-city/solution.py

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+class Solution:
+    def maxPower(self, stations: List[int], r: int, k: int) -> int:
+        n = len(stations)
+        f = [0] * (n + 1)
+        for i, val in enumerate(stations):
+            left, right = max(0, i - r), min(i + r, n - 1)
+            f[left] += val
+            f[right + 1] -= val
+        
+        s = list(accumulate(f))
+
+        def checker(x, k):
+            f = [0] * (n + 1)
+            t = 0
+            for i in range(n):
+                t += f[i]
+                dist = x - (s[i] + t)
+                if dist > 0:
+                    if k < dist:
+                        return False
+                    k -= dist
+                    j = min(i + r, n - 1)
+                    left, right = max(0, j - r), min(j + r, n - 1)
+                    f[left] += dist
+                    f[right + 1] -= dist
+                    t += dist
+            return True
+        
+        left, right = 0, 1 << 40
+        while right > left:
+            mid = (left + right + 1) // 2
+            if checker(mid, k):
+                left = mid
+            else:
+                right = mid - 1
+        return left