From 832ef0d4a359eb6eff9c4a5378924c76489f3258 Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Sun, 7 Apr 2024 17:29:03 +0800
Subject: [PATCH] feat: add solutions to lc problem: No.3107
No.3107.Minimum Operations to Make Median of Array Equal to K
---
.../README.md | 109 +++++++++++++++++-
.../README_EN.md | 109 +++++++++++++++++-
.../Solution.cpp | 19 +++
.../Solution.go | 23 ++++
.../Solution.java | 18 +++
.../Solution.py | 17 +++
.../Solution.ts | 16 +++
7 files changed, 303 insertions(+), 8 deletions(-)
create mode 100644 solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.cpp
create mode 100644 solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.go
create mode 100644 solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.java
create mode 100644 solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.py
create mode 100644 solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.ts
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README.md b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README.md
index 3835b453045c4..a61515c7658d2 100644
--- a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README.md
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README.md
@@ -60,24 +60,125 @@
## 解法
-### 方法一
+### 方法一：贪心 + 排序
+
+我们首先对数组 $nums$ 进行排序，然后找到中位数的位置 $m$，那么初始时我们需要的操作次数就是 $|nums[m] - k|$。
+
+接下来，我们分情况讨论：
+
+- 如果 $nums[m] \gt k$，那么 $m$ 右侧的元素都大于等于 $k$，我们只需要将 $m$ 左侧的元素中，大于 $k$ 的元素减小到 $k$ 即可。
+- 如果 $nums[m] \le k$，那么 $m$ 左侧的元素都小于等于 $k$，我们只需要将 $m$ 右侧的元素中，小于 $k$ 的元素增大到 $k$ 即可。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 $nums$ 的长度。
<!-- tabs:start -->
```python
-
+class Solution:
+ def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
+ nums.sort()
+ n = len(nums)
+ m = n >> 1
+ ans = abs(nums[m] - k)
+ if nums[m] > k:
+ for i in range(m - 1, -1, -1):
+ if nums[i] <= k:
+ break
+ ans += nums[i] - k
+ else:
+ for i in range(m + 1, n):
+ if nums[i] >= k:
+ break
+ ans += k - nums[i]
+ return ans
```
```java
-
+class Solution {
+ public long minOperationsToMakeMedianK(int[] nums, int k) {
+ Arrays.sort(nums);
+ int n = nums.length;
+ int m = n >> 1;
+ long ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+}
```
```cpp
-
+class Solution {
+public:
+ long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
+ sort(nums.begin(), nums.end());
+ int n = nums.size();
+ int m = n >> 1;
+ long long ans = abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+};
```
```go
+func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
+ sort.Ints(nums)
+ n := len(nums)
+ m := n >> 1
+ ans = int64(abs(nums[m] - k))
+ if nums[m] > k {
+ for i := m - 1; i >= 0 && nums[i] > k; i-- {
+ ans += int64(nums[i] - k)
+ }
+ } else {
+ for i := m + 1; i < n && nums[i] < k; i++ {
+ ans += int64(k - nums[i])
+ }
+ }
+ return
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
+```
+```ts
+function minOperationsToMakeMedianK(nums: number[], k: number): number {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const m = n >> 1;
+ let ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (let i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+}
```
<!-- tabs:end -->
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README_EN.md b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README_EN.md
index 121032df5ebd3..ca4e8ddda30ca 100644
--- a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README_EN.md
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README_EN.md
@@ -58,24 +58,125 @@
## Solutions
-### Solution 1
+### Solution 1: Greedy + Sorting
+
+First, we sort the array $nums$ and find the position $m$ of the median. The initial number of operations we need is $|nums[m] - k|$.
+
+Next, we discuss in two cases:
+
+- If $nums[m] > k$, then all elements to the right of $m$ are greater than or equal to $k$. We only need to reduce the elements greater than $k$ on the left of $m$ to $k$.
+- If $nums[m] \le k$, then all elements to the left of $m$ are less than or equal to $k$. We only need to increase the elements less than $k$ on the right of $m$ to $k$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.
<!-- tabs:start -->
```python
-
+class Solution:
+ def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
+ nums.sort()
+ n = len(nums)
+ m = n >> 1
+ ans = abs(nums[m] - k)
+ if nums[m] > k:
+ for i in range(m - 1, -1, -1):
+ if nums[i] <= k:
+ break
+ ans += nums[i] - k
+ else:
+ for i in range(m + 1, n):
+ if nums[i] >= k:
+ break
+ ans += k - nums[i]
+ return ans
```
```java
-
+class Solution {
+ public long minOperationsToMakeMedianK(int[] nums, int k) {
+ Arrays.sort(nums);
+ int n = nums.length;
+ int m = n >> 1;
+ long ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+}
```
```cpp
-
+class Solution {
+public:
+ long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
+ sort(nums.begin(), nums.end());
+ int n = nums.size();
+ int m = n >> 1;
+ long long ans = abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+};
```
```go
+func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
+ sort.Ints(nums)
+ n := len(nums)
+ m := n >> 1
+ ans = int64(abs(nums[m] - k))
+ if nums[m] > k {
+ for i := m - 1; i >= 0 && nums[i] > k; i-- {
+ ans += int64(nums[i] - k)
+ }
+ } else {
+ for i := m + 1; i < n && nums[i] < k; i++ {
+ ans += int64(k - nums[i])
+ }
+ }
+ return
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
+```
+```ts
+function minOperationsToMakeMedianK(nums: number[], k: number): number {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const m = n >> 1;
+ let ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (let i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+}
```
<!-- tabs:end -->
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.cpp b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.cpp
new file mode 100644
index 0000000000000..3a037af1c0727
--- /dev/null
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.cpp
@@ -0,0 +1,19 @@
+class Solution {
+public:
+ long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
+ sort(nums.begin(), nums.end());
+ int n = nums.size();
+ int m = n >> 1;
+ long long ans = abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+};
\ No newline at end of file
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.go b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.go
new file mode 100644
index 0000000000000..6185bf2b8e399
--- /dev/null
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.go
@@ -0,0 +1,23 @@
+func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
+ sort.Ints(nums)
+ n := len(nums)
+ m := n >> 1
+ ans = int64(abs(nums[m] - k))
+ if nums[m] > k {
+ for i := m - 1; i >= 0 && nums[i] > k; i-- {
+ ans += int64(nums[i] - k)
+ }
+ } else {
+ for i := m + 1; i < n && nums[i] < k; i++ {
+ ans += int64(k - nums[i])
+ }
+ }
+ return
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
\ No newline at end of file
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.java b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.java
new file mode 100644
index 0000000000000..ce1191d540b66
--- /dev/null
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.java
@@ -0,0 +1,18 @@
+class Solution {
+ public long minOperationsToMakeMedianK(int[] nums, int k) {
+ Arrays.sort(nums);
+ int n = nums.length;
+ int m = n >> 1;
+ long ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (int i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+ }
+}
\ No newline at end of file
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.py b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.py
new file mode 100644
index 0000000000000..0374b49e4590a
--- /dev/null
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.py
@@ -0,0 +1,17 @@
+class Solution:
+ def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
+ nums.sort()
+ n = len(nums)
+ m = n >> 1
+ ans = abs(nums[m] - k)
+ if nums[m] > k:
+ for i in range(m - 1, -1, -1):
+ if nums[i] <= k:
+ break
+ ans += nums[i] - k
+ else:
+ for i in range(m + 1, n):
+ if nums[i] >= k:
+ break
+ ans += k - nums[i]
+ return ans
diff --git a/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.ts b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.ts
new file mode 100644
index 0000000000000..e8605cac83f35
--- /dev/null
+++ b/solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.ts
@@ -0,0 +1,16 @@
+function minOperationsToMakeMedianK(nums: number[], k: number): number {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const m = n >> 1;
+ let ans = Math.abs(nums[m] - k);
+ if (nums[m] > k) {
+ for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
+ ans += nums[i] - k;
+ }
+ } else {
+ for (let i = m + 1; i < n && nums[i] < k; ++i) {
+ ans += k - nums[i];
+ }
+ }
+ return ans;
+}