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Commit c12f418

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feat: add solutions to lc problem: No.2919 (#1906)
No.2919.Minimum Increment Operations to Make Array Beautiful
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7 files changed

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‎solution/2900-2999/2919.Minimum Increment Operations to Make Array Beautiful/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:动态规划**
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我们定义 $f,ドル $g,ドル $h$ 表示前 $i$ 项中,分别以最后三项作为子数组的最大值所需要的最小增量运算数,初始时 $f = 0,ドル $g = 0,ドル $h = 0$。
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接下来,我们遍历数组 $nums,ドル对于每个 $x,ドル我们需要更新 $f,ドル $g,ドル $h$ 的值,使其满足题目要求,即:
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$$
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\begin{aligned}
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f' &= g \\
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g' &= h \\
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h' &= \min(f, g, h) + \max(k - x, 0)
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\end{aligned}
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$$
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最后,我们只需要返回 $f,ドル $g,ドル $h$ 中的最小值即可。
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时间复杂度 $O(n),ドル其中 $n$ 为数组长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minIncrementOperations(self, nums: List[int], k: int) -> int:
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f = g = h = 0
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for x in nums:
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f, g, h = g, h, min(f, g, h) + max(k - x, 0)
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return min(f, g, h)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long minIncrementOperations(int[] nums, int k) {
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long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return Math.min(Math.min(f, g), h);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long minIncrementOperations(vector<int>& nums, int k) {
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long long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long long hh = min({f, g, h}) + max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return min({f, g, h});
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}
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};
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```
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### **Go**
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```go
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func minIncrementOperations(nums []int, k int) int64 {
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var f, g, h int
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for _, x := range nums {
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f, g, h = g, h, min(min(f, g), h)+max(k-x, 0)
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}
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return int64(min(min(f, g), h))
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}
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```
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### **TypeScript**
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```ts
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function minIncrementOperations(nums: number[], k: number): number {
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let [f, g, h] = [0, 0, 0];
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for (const x of nums) {
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[f, g, h] = [g, h, Math.min(f, g, h) + Math.max(k - x, 0)];
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}
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return Math.min(f, g, h);
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}
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```
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### **...**

‎solution/2900-2999/2919.Minimum Increment Operations to Make Array Beautiful/README_EN.md‎

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## Solutions
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**Solution 1: Dynamic Programming**
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We define $f,ドル $g,ドル and $h$ as the minimum number of increment operations needed to get the maximum value from the last three items in the first $i$ items, initially $f = 0,ドル $g = 0,ドル $h = 0$.
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Next, we traverse the array $nums$. For each $x,ドル we need to update the values of $f,ドル $g,ドル and $h$ to meet the requirements of the problem, that is:
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$$
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\begin{aligned}
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f' &= g \\
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g' &= h \\
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h' &= \min(f, g, h) + \max(k - x, 0)
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\end{aligned}
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$$
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Finally, we only need to return the minimum value among $f,ドル $g,ドル and $h$.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def minIncrementOperations(self, nums: List[int], k: int) -> int:
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f = g = h = 0
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for x in nums:
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f, g, h = g, h, min(f, g, h) + max(k - x, 0)
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return min(f, g, h)
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```
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### **Java**
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```java
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class Solution {
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public long minIncrementOperations(int[] nums, int k) {
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long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return Math.min(Math.min(f, g), h);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long minIncrementOperations(vector<int>& nums, int k) {
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long long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long long hh = min({f, g, h}) + max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return min({f, g, h});
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}
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};
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```
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### **Go**
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```go
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func minIncrementOperations(nums []int, k int) int64 {
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var f, g, h int
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for _, x := range nums {
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f, g, h = g, h, min(min(f, g), h)+max(k-x, 0)
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}
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return int64(min(min(f, g), h))
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}
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```
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### **TypeScript**
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```ts
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function minIncrementOperations(nums: number[], k: number): number {
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let [f, g, h] = [0, 0, 0];
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for (const x of nums) {
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[f, g, h] = [g, h, Math.min(f, g, h) + Math.max(k - x, 0)];
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}
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return Math.min(f, g, h);
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}
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```
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### **...**
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class Solution {
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public:
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long long minIncrementOperations(vector<int>& nums, int k) {
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long long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long long hh = min({f, g, h}) + max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return min({f, g, h});
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}
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};
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func minIncrementOperations(nums []int, k int) int64 {
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var f, g, h int
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for _, x := range nums {
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f, g, h = g, h, min(min(f, g), h)+max(k-x, 0)
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}
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return int64(min(min(f, g), h))
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}
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class Solution {
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public long minIncrementOperations(int[] nums, int k) {
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long f = 0, g = 0, h = 0;
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for (int x : nums) {
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long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
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f = g;
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g = h;
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h = hh;
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}
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return Math.min(Math.min(f, g), h);
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}
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}
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class Solution:
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def minIncrementOperations(self, nums: List[int], k: int) -> int:
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f = g = h = 0
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for x in nums:
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f, g, h = g, h, min(f, g, h) + max(k - x, 0)
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return min(f, g, h)
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function minIncrementOperations(nums: number[], k: number): number {
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let [f, g, h] = [0, 0, 0];
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for (const x of nums) {
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[f, g, h] = [g, h, Math.min(f, g, h) + Math.max(k - x, 0)];
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}
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return Math.min(f, g, h);
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}

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