[pull] master from wisdompeak:master

Binary_Search/2528.Maximize-the-Minimum-Powered-City/2528.Maximize-the-Minimum-Powered-City.cpp

@@ -0,0 +1,44 @@
+using LL = long long;
+class Solution {
+public:
+    long long maxPower(vector<int>& stations, int r, int k) 
+    {
+        LL left = 0,  right = LLONG_MAX;
+        while (left < right)
+        {
+            LL mid = right-(right-left)/2;
+            if (isOK(stations, r, k, mid))
+                left = mid;
+            else
+                right = mid-1;
+        }
+        return left;
+    }
+
+    bool isOK(vector<int>stations, int r, LL k, LL m)
+    {
+        int n = stations.size();
+        LL sum = 0;
+        for (int i=0; i<=min(n-1, r-1); i++)
+            sum += stations[i];
+
+        for (int i=0; i<n; i++)
+        {
+            if (i+r < n)            
+                sum += stations[i+r];            
+
+            if (i-r-1 >= 0)
+                sum -= stations[i-r-1];
+
+            if (sum >= m) continue;
+
+            LL diff = m - sum;
+            if (diff > k) 
+                return false;
+            stations[min(n-1, i+r)] += diff;
+            sum = m;
+            k -= diff;
+        }
+        return true;
+    }
+};

Binary_Search/2528.Maximize-the-Minimum-Powered-City/Readme.md

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+### 2528.Maximize-the-Minimum-Powered-City
+
+我们假设The minimum power of a city是m，那么意味着每个城市都可以得到至少m的供应。我们就顺着走一遍每个城市，看它是否已经实现了周围[i-r,i+r]范围内的滑窗和大于等于m。如果不够m，那么我们必然会贪心地增加最优端的那个城市的供给（因为它能覆盖更多未来的城市）补齐至m。一路走下去，如果总共需要补建的电力站小于等于k，那么就说明m是可以实现的，于是可以尝试猜测更大的m。如果需要补建的电力站大于k，那么就说明m不可行，就尝试猜测更小的m。

Bit_Manipulation/2527.Find-Xor-Beauty-of-Array/2527.Find-Xor-Beauty-of-Array.cpp

@@ -0,0 +1,11 @@
+class Solution {
+public:
+    int xorBeauty(vector<int>& nums) 
+    {
+        int sum = 0;
+        for (int x: nums)
+            sum ^= x;
+        return sum;
+
+    }
+};

Readme.md

@@ -118,6 +118,7 @@
 [2439.Minimize-Maximum-of-Array](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2439.Minimize-Maximum-of-Array) (H-)    
 [2517.Maximum-Tastiness-of-Candy-Basket](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2517.Maximum-Tastiness-of-Candy-Basket) (M+)      
 [2513.Minimize-the-Maximum-of-Two-Arrays](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2513.Minimize-the-Maximum-of-Two-Arrays) (H)     
+[2528.Maximize-the-Minimum-Powered-City](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2528.Maximize-the-Minimum-Powered-City) (H-)      
     * ``Find K-th Element``       
 [215.Kth-Largest-Element-in-an-Array](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/215.Kth-Largest-Element-in-an-Array) (M)   
 [287.Find-the-Duplicate-Number](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/287.Find-the-Duplicate-Number) (H-)   
@@ -859,6 +860,7 @@
 [1734.Decode-XORed-Permutation](https://github.com/wisdompeak/LeetCode/tree/master/Bit_Manipulation/1734.Decode-XORed-Permutation) (M+)  
 [1738.Find-Kth-Largest-XOR-Coordinate-Value](https://github.com/wisdompeak/LeetCode/tree/master/Bit_Manipulation/1738.Find-Kth-Largest-XOR-Coordinate-Value) (M+)  
 [1835.Find-XOR-Sum-of-All-Pairs-Bitwise-AND](https://github.com/wisdompeak/LeetCode/tree/master/Bit_Manipulation/1835.Find-XOR-Sum-of-All-Pairs-Bitwise-AND) (M)  
+[2527.Find-Xor-Beauty-of-Array](https://github.com/wisdompeak/LeetCode/tree/master/Bit_Manipulation/2527.Find-Xor-Beauty-of-Array) (H)    
 * ``Bit Mask``   
 [320.Generalized-Abbreviation](https://github.com/wisdompeak/LeetCode/tree/master/String/320.Generalized-Abbreviation) (M)      
 [1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters](https://github.com/wisdompeak/LeetCode/tree/master/Bit_Manipulation/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters) (M+)