2171

leetcode/2171. Removing Minimum Number of Magic Beans/README.md

@@ -0,0 +1,100 @@
+# [2171. Removing Minimum Number of Magic Beans (Medium)](https://leetcode.com/problems/removing-minimum-number-of-magic-beans/)
+
+<p>You are given an array of <strong>positive</strong> integers <code>beans</code>, where each integer represents the number of magic beans found in a particular magic bag.</p>
+
+<p><strong>Remove</strong> any number of beans (<strong>possibly none</strong>) from each bag such that the number of beans in each remaining <strong>non-empty</strong> bag (still containing <strong>at least one</strong> bean) is <strong>equal</strong>. Once a bean has been removed from a bag, you are <strong>not</strong> allowed to return it to any of the bags.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of magic beans that you have to remove</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> beans = [4,<u><strong>1</strong></u>,6,5]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+- We remove 1 bean from the bag with only 1 bean.
+  This results in the remaining bags: [4,<b><u>0</u></b>,6,5]
+- Then we remove 2 beans from the bag with 6 beans.
+  This results in the remaining bags: [4,0,<strong><u>4</u></strong>,5]
+- Then we remove 1 bean from the bag with 5 beans.
+  This results in the remaining bags: [4,0,4,<b><u>4</u></b>]
+We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that remove 4 beans or fewer.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> beans = [<strong><u>2</u></strong>,10,<u><strong>3</strong></u>,<strong><u>2</u></strong>]
+<strong>Output:</strong> 7
+<strong>Explanation:</strong>
+- We remove 2 beans from one of the bags with 2 beans.
+  This results in the remaining bags: [<u><strong>0</strong></u>,10,3,2]
+- Then we remove 2 beans from the other bag with 2 beans.
+  This results in the remaining bags: [0,10,3,<u><strong>0</strong></u>]
+- Then we remove 3 beans from the bag with 3 beans. 
+  This results in the remaining bags: [0,10,<u><strong>0</strong></u>,0]
+We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans.
+There are no other solutions that removes 7 beans or fewer.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= beans.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= beans[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+
+**Similar Questions**:
+* [Minimum Moves to Equal Array Elements II (Medium)](https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/)
+* [Minimum Operations to Reduce X to Zero (Medium)](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/)
+
+## Solution 1. Sorting
+
+Sort the original array `A`.
+
+If we select `A[i]` as the number of beans in a non-empty bag, the number of removals needed is `sum(A) - (N - i) * A[i]`.
+
+**Meaning of equation**
+
+For all `A[j] (j < i)`, they are completely removed, contributing `A[0] + .. + A[i-1]` removals.
+
+For all `A[j] (j >= i)`, they become all `A[i]`s, contributing `A[i] + .. + A[N-1] - (N-i) * A[i]` removals.
+
+Summing these two up, we get `sum(A) - (N - i) * A[i]`.
+
+Another way to think this is to remove every thing and recover `(N - i) * A[i]` beans that shouldn't be removed.
+
+**Why we should pick the number from `A`**
+
+Assume `A  = [1,5,10]`. If we pick a number that is not in `A`, say `3`, `A` becomes `[0,3,3]`. This is definitely not better than picking `A[i] = 5` resulting in `[0,5,5]`. So, a solution picking a non-existent number is always dominated by another solution picking an existing number.
+
+**Example:**
+
+`A = [1,4,5,6]`, `sum(A) = 16`
+
+* If we pick `A[0] = 1`, the result array is `[1,1,1,1]`, # of removals is `16 - (4 - 0) * 1 = 12`.
+* If we pick `A[1] = 4`, the result array is `[0,4,4,4]`, # of removals is `16 - (4 - 1) * 4 = 4`.
+* If we pick `A[2] = 5`, the result array is `[0,0,5,5]`, # of removals is `16 - (4 - 2) * 5 = 6`.
+* If we pick `A[3] = 6`, the result array is `[0,0,0,6]`, # of removals is `16 - (4 - 3) * 6 = 10`.
+
+```cpp
+// OJ: https://leetcode.com/contest/weekly-contest-280/problems/removing-minimum-number-of-magic-beans/
+// Author: github.com/lzl124631x
+// Time: O(NlogN)
+// Space: O(1)
+class Solution {
+public:
+    long long minimumRemoval(vector<int>& A) {
+        long N = A.size(), ans = LLONG_MAX, sum = accumulate(begin(A), end(A), 0L);
+        sort(begin(A), end(A));
+        for (int i = 0; i < N; ++i) ans = min(ans, sum - (N - i) * A[i]);
+        return ans;
+    }
+};
+```
+
+## Discuss
+
+https://leetcode.com/problems/removing-minimum-number-of-magic-beans/discuss/1766764