feat: add solutions to lc problem: No.3422 (#3956)

No.3422.Minimum Operations to Make Subarray Elements Equal

Author: yanglbme

solution/3400-3499/3422.Minimum Operations to Make Subarray Elements Equal/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3422.Minimum%20Operations%20to%20Make%20Subarray%20Elements%20Equal/README.md
+---
+
+<!-- problem:start -->
+
+# [3422. Minimum Operations to Make Subarray Elements Equal 🔒](https://leetcode.cn/problems/minimum-operations-to-make-subarray-elements-equal)
+
+[English Version](/solution/3400-3499/3422.Minimum%20Operations%20to%20Make%20Subarray%20Elements%20Equal/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can perform the following operation any number of times:</p>
+
+<ul>
+	<li>Increase or decrease any element of <code>nums</code> by 1.</li>
+</ul>
+
+<p>Return the <strong>minimum</strong> number of operations required to ensure that <strong>at least</strong> one <span data-keyword="subarray">subarray</span> of size <code>k</code> in <code>nums</code> has all elements equal.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [4,-3,2,1,-4,6], k = 3</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Use 4 operations to add 4 to <code>nums[1]</code>. The resulting array is <span class="example-io"><code>[4, 1, 2, 1, -4, 6]</code>.</span></li>
+	<li><span class="example-io">Use 1 operation to subtract 1 from <code>nums[2]</code>. The resulting array is <code>[4, 1, 1, 1, -4, 6]</code>.</span></li>
+	<li><span class="example-io">The array now contains a subarray <code>[1, 1, 1]</code> of size <code>k = 3</code> with all elements equal. Hence, the answer is 5.</span></li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [-2,-2,3,1,4], k = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>
+	<p>The subarray <code>[-2, -2]</code> of size <code>k = 2</code> already contains all equal elements, so no operations are needed. Hence, the answer is 0.</p>
+	</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>6</sup> &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
+	<li><code>2 &lt;= k &lt;= nums.length</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：有序集合
+
+根据题目描述，我们需要找到一个长度为 $k$ 的子数组，通过最少的操作使得子数组中的所有元素相等，即我们需要找到一个长度为 $k$ 的子数组，使得子数组中所有元素变成这 $k$ 个元素的中位数所需的最少操作次数最小。
+
+我们可以使用两个有序集合 $l$ 和 $r$ 分别维护 $k$ 个元素的左右两部分，其中 $l$ 用于存储 $k$ 个元素中较小的一部分，$r$ 用于存储 $k$ 个元素中较大的一部分，并且 $l$ 的元素个数要么等于 $r$ 的元素个数，要么比 $r$ 的元素个数少一个，这样 $r$ 的最小值就是 $k$ 个元素中的中位数。
+
+时间复杂度 $O(n \times \log k)$，空间复杂度 $O(k)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minOperations(self, nums: List[int], k: int) -> int:
+        l = SortedList()
+        r = SortedList()
+        s1 = s2 = 0
+        ans = inf
+        for i, x in enumerate(nums):
+            l.add(x)
+            s1 += x
+            y = l.pop()
+            s1 -= y
+            r.add(y)
+            s2 += y
+            if len(r) - len(l) > 1:
+                y = r.pop(0)
+                s2 -= y
+                l.add(y)
+                s1 += y
+            if i >= k - 1:
+                ans = min(ans, s2 - r[0] * len(r) + r[0] * len(l) - s1)
+                j = i - k + 1
+                if nums[j] in r:
+                    r.remove(nums[j])
+                    s2 -= nums[j]
+                else:
+                    l.remove(nums[j])
+                    s1 -= nums[j]
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public long minOperations(int[] nums, int k) {
+        TreeMap<Integer, Integer> l = new TreeMap<>();
+        TreeMap<Integer, Integer> r = new TreeMap<>();
+        long s1 = 0, s2 = 0;
+        int sz1 = 0, sz2 = 0;
+        long ans = Long.MAX_VALUE;
+        for (int i = 0; i < nums.length; ++i) {
+            l.merge(nums[i], 1, Integer::sum);
+            s1 += nums[i];
+            ++sz1;
+            int y = l.lastKey();
+            if (l.merge(y, -1, Integer::sum) == 0) {
+                l.remove(y);
+            }
+            s1 -= y;
+            --sz1;
+            r.merge(y, 1, Integer::sum);
+            s2 += y;
+            ++sz2;
+            if (sz2 - sz1 > 1) {
+                y = r.firstKey();
+                if (r.merge(y, -1, Integer::sum) == 0) {
+                    r.remove(y);
+                }
+                s2 -= y;
+                --sz2;
+                l.merge(y, 1, Integer::sum);
+                s1 += y;
+                ++sz1;
+            }
+            if (i >= k - 1) {
+                ans = Math.min(ans, s2 - r.firstKey() * sz2 + r.firstKey() * sz1 - s1);
+                int j = i - k + 1;
+                if (r.containsKey(nums[j])) {
+                    if (r.merge(nums[j], -1, Integer::sum) == 0) {
+                        r.remove(nums[j]);
+                    }
+                    s2 -= nums[j];
+                    --sz2;
+                } else {
+                    if (l.merge(nums[j], -1, Integer::sum) == 0) {
+                        l.remove(nums[j]);
+                    }
+                    s1 -= nums[j];
+                    --sz1;
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    long long minOperations(vector<int>& nums, int k) {
+        multiset<int> l, r;
+        long long s1 = 0, s2 = 0, ans = 1e18;
+        for (int i = 0; i < nums.size(); ++i) {
+            l.insert(nums[i]);
+            s1 += nums[i];
+            int y = *l.rbegin();
+            l.erase(l.find(y));
+            s1 -= y;
+            r.insert(y);
+            s2 += y;
+            if (r.size() - l.size() > 1) {
+                y = *r.begin();
+                r.erase(r.find(y));
+                s2 -= y;
+                l.insert(y);
+                s1 += y;
+            }
+            if (i >= k - 1) {
+                long long x = *r.begin();
+                ans = min(ans, s2 - x * (int) r.size() + x * (int) l.size() - s1);
+                int j = i - k + 1;
+                if (r.contains(nums[j])) {
+                    r.erase(r.find(nums[j]));
+                    s2 -= nums[j];
+                } else {
+                    l.erase(l.find(nums[j]));
+                    s1 -= nums[j];
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->