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feat: add solutions to lc problem: No.3107 (#2551)
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‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README.md

+105-4
Original file line numberDiff line numberDiff line change
@@ -60,24 +60,125 @@
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## 解法
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### 方法一
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### 方法一：贪心 + 排序
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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 $|nums[m] - k|$。
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接下来，我们分情况讨论：
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- 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。
70+
- 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。
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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。
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6574
<!-- tabs:start -->
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```python
68-
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class Solution:
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def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
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nums.sort()
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n = len(nums)
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m = n >> 1
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ans = abs(nums[m] - k)
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if nums[m] > k:
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for i in range(m - 1, -1, -1):
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if nums[i] <= k:
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break
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ans += nums[i] - k
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else:
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for i in range(m + 1, n):
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if nums[i] >= k:
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break
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ans += k - nums[i]
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return ans
6994
```
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7196
```java
72-
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class Solution {
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public long minOperationsToMakeMedianK(int[] nums, int k) {
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Arrays.sort(nums);
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int n = nums.length;
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int m = n >> 1;
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long ans = Math.abs(nums[m] - k);
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if (nums[m] > k) {
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for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
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ans += nums[i] - k;
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}
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} else {
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for (int i = m + 1; i < n && nums[i] < k; ++i) {
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ans += k - nums[i];
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}
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}
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return ans;
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}
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}
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```
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75117
```cpp
76-
118+
class Solution {
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public:
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long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int n = nums.size();
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int m = n >> 1;
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long long ans = abs(nums[m] - k);
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if (nums[m] > k) {
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for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
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ans += nums[i] - k;
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}
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} else {
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for (int i = m + 1; i < n && nums[i] < k; ++i) {
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ans += k - nums[i];
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}
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}
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return ans;
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}
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};
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```
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```go
140+
func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
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sort.Ints(nums)
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n := len(nums)
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m := n >> 1
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ans = int64(abs(nums[m] - k))
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if nums[m] > k {
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for i := m - 1; i >= 0 && nums[i] > k; i-- {
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ans += int64(nums[i] - k)
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}
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} else {
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for i := m + 1; i < n && nums[i] < k; i++ {
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ans += int64(k - nums[i])
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}
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}
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return
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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```ts
166+
function minOperationsToMakeMedianK(nums: number[], k: number): number {
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nums.sort((a, b) => a - b);
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const n = nums.length;
169+
const m = n >> 1;
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let ans = Math.abs(nums[m] - k);
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if (nums[m] > k) {
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for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
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ans += nums[i] - k;
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}
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} else {
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for (let i = m + 1; i < n && nums[i] < k; ++i) {
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ans += k - nums[i];
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/README_EN.md

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@@ -58,24 +58,125 @@
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## Solutions
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61-
### Solution 1
61+
### Solution 1: Greedy + Sorting
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First, we sort the array $nums$ and find the position $m$ of the median. The initial number of operations we need is $|nums[m] - k|$.
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Next, we discuss in two cases:
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- 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 $k$.
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- 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 $k$.
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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 $nums$.
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<!-- tabs:start -->
6473

6574
```python
66-
75+
class Solution:
76+
def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
77+
nums.sort()
78+
n = len(nums)
79+
m = n >> 1
80+
ans = abs(nums[m] - k)
81+
if nums[m] > k:
82+
for i in range(m - 1, -1, -1):
83+
if nums[i] <= k:
84+
break
85+
ans += nums[i] - k
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else:
87+
for i in range(m + 1, n):
88+
if nums[i] >= k:
89+
break
90+
ans += k - nums[i]
91+
return ans
6792
```
6893

6994
```java
70-
95+
class Solution {
96+
public long minOperationsToMakeMedianK(int[] nums, int k) {
97+
Arrays.sort(nums);
98+
int n = nums.length;
99+
int m = n >> 1;
100+
long ans = Math.abs(nums[m] - k);
101+
if (nums[m] > k) {
102+
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
103+
ans += nums[i] - k;
104+
}
105+
} else {
106+
for (int i = m + 1; i < n && nums[i] < k; ++i) {
107+
ans += k - nums[i];
108+
}
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}
110+
return ans;
111+
}
112+
}
71113
```
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73115
```cpp
74-
116+
class Solution {
117+
public:
118+
long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
119+
sort(nums.begin(), nums.end());
120+
int n = nums.size();
121+
int m = n >> 1;
122+
long long ans = abs(nums[m] - k);
123+
if (nums[m] > k) {
124+
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
125+
ans += nums[i] - k;
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}
127+
} else {
128+
for (int i = m + 1; i < n && nums[i] < k; ++i) {
129+
ans += k - nums[i];
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}
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}
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return ans;
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}
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};
75135
```
76136
77137
```go
138+
func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
139+
sort.Ints(nums)
140+
n := len(nums)
141+
m := n >> 1
142+
ans = int64(abs(nums[m] - k))
143+
if nums[m] > k {
144+
for i := m - 1; i >= 0 && nums[i] > k; i-- {
145+
ans += int64(nums[i] - k)
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}
147+
} else {
148+
for i := m + 1; i < n && nums[i] < k; i++ {
149+
ans += int64(k - nums[i])
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}
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}
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return
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}
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func abs(x int) int {
156+
if x < 0 {
157+
return -x
158+
}
159+
return x
160+
}
161+
```
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163+
```ts
164+
function minOperationsToMakeMedianK(nums: number[], k: number): number {
165+
nums.sort((a, b) => a - b);
166+
const n = nums.length;
167+
const m = n >> 1;
168+
let ans = Math.abs(nums[m] - k);
169+
if (nums[m] > k) {
170+
for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
171+
ans += nums[i] - k;
172+
}
173+
} else {
174+
for (let i = m + 1; i < n && nums[i] < k; ++i) {
175+
ans += k - nums[i];
176+
}
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}
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return ans;
179+
}
79180
```
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81182
<!-- tabs:end -->

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.cpp

+19
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
1+
class Solution {
2+
public:
3+
long long minOperationsToMakeMedianK(vector<int>& nums, int k) {
4+
sort(nums.begin(), nums.end());
5+
int n = nums.size();
6+
int m = n >> 1;
7+
long long ans = abs(nums[m] - k);
8+
if (nums[m] > k) {
9+
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
10+
ans += nums[i] - k;
11+
}
12+
} else {
13+
for (int i = m + 1; i < n && nums[i] < k; ++i) {
14+
ans += k - nums[i];
15+
}
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}
17+
return ans;
18+
}
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};

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.go

+23
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
1+
func minOperationsToMakeMedianK(nums []int, k int) (ans int64) {
2+
sort.Ints(nums)
3+
n := len(nums)
4+
m := n >> 1
5+
ans = int64(abs(nums[m] - k))
6+
if nums[m] > k {
7+
for i := m - 1; i >= 0 && nums[i] > k; i-- {
8+
ans += int64(nums[i] - k)
9+
}
10+
} else {
11+
for i := m + 1; i < n && nums[i] < k; i++ {
12+
ans += int64(k - nums[i])
13+
}
14+
}
15+
return
16+
}
17+
18+
func abs(x int) int {
19+
if x < 0 {
20+
return -x
21+
}
22+
return x
23+
}

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.java

+18
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
class Solution {
2+
public long minOperationsToMakeMedianK(int[] nums, int k) {
3+
Arrays.sort(nums);
4+
int n = nums.length;
5+
int m = n >> 1;
6+
long ans = Math.abs(nums[m] - k);
7+
if (nums[m] > k) {
8+
for (int i = m - 1; i >= 0 && nums[i] > k; --i) {
9+
ans += nums[i] - k;
10+
}
11+
} else {
12+
for (int i = m + 1; i < n && nums[i] < k; ++i) {
13+
ans += k - nums[i];
14+
}
15+
}
16+
return ans;
17+
}
18+
}

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.py

+17
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
class Solution:
2+
def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
3+
nums.sort()
4+
n = len(nums)
5+
m = n >> 1
6+
ans = abs(nums[m] - k)
7+
if nums[m] > k:
8+
for i in range(m - 1, -1, -1):
9+
if nums[i] <= k:
10+
break
11+
ans += nums[i] - k
12+
else:
13+
for i in range(m + 1, n):
14+
if nums[i] >= k:
15+
break
16+
ans += k - nums[i]
17+
return ans

‎solution/3100-3199/3107.Minimum Operations to Make Median of Array Equal to K/Solution.ts

+16
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1+
function minOperationsToMakeMedianK(nums: number[], k: number): number {
2+
nums.sort((a, b) => a - b);
3+
const n = nums.length;
4+
const m = n >> 1;
5+
let ans = Math.abs(nums[m] - k);
6+
if (nums[m] > k) {
7+
for (let i = m - 1; i >= 0 && nums[i] > k; --i) {
8+
ans += nums[i] - k;
9+
}
10+
} else {
11+
for (let i = m + 1; i < n && nums[i] < k; ++i) {
12+
ans += k - nums[i];
13+
}
14+
}
15+
return ans;
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}

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