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feat: add solutions to lc problem: No.2195 #2790

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May 11, 2024
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feat: add solutions to lc problem: No.2195 #2790

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93 changes: 46 additions & 47 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -44,47 +44,42 @@ nums 最终元素和为 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36 ，这是所有情况

### 方法一：排序 + 贪心 + 数学

我们不妨向数组中加入两个哨兵节点，分别为 $0$ 和 $2 \times 10^9$。

然后我们对数组进行排序，那么对于数组中的任意两个相邻的元素 $a$ 和 $b$，区间 $[a+1, b-1]$ 中的整数都是未出现在数组中的，我们可以将这些整数加入到数组中。

因此，我们从小到大遍历数组中的相邻元素对 $(a, b)$，对于每个相邻元素对，我们计算区间 $[a+1, b-1]$ 中的整数个数 $m$，那么这 $m$ 个整数的和为 $\frac{m \times (a+1 + a+m)}{2}$，我们将这个和累加到答案中，并将 $k$ 减去 $m$。如果 $k$ 减到 $0$，我们就可以停止遍历了，返回答案。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组的长度。

<!-- tabs:start -->

```python
class Solution:
def minimalKSum(self, nums: List[int], k: int) -> int:
nums.append(0)
nums.append(2 * 10**9)
nums.extend([0, 2 * 10**9])
nums.sort()
ans = 0
for a, b in pairwise(nums):
n = min(k, b - a - 1)
if n <= 0:
continue
k -= n
ans += (a + 1 + a + n) * n // 2
if k == 0:
break
m = max(0, min(k, b - a - 1))
ans += (a + 1 + a + m) * m // 2
k -= m
return ans
```

```java
class Solution {
public long minimalKSum(int[] nums, int k) {
int[] arr = new int[nums.length + 2];
arr[arr.length - 1] = (int) 2e9;
for (int i = 0; i < nums.length; ++i) {
arr[i + 1] = nums[i];
}
int n = nums.length;
int[] arr = new int[n + 2];
arr[1] = 2 * 1000000000;
System.arraycopy(nums, 0, arr, 2, n);
Arrays.sort(arr);
long ans = 0;
for (int i = 1; i < arr.length; ++i) {
int a = arr[i - 1], b = arr[i];
int n = Math.min(k, b - a - 1);
if (n <= 0) {
continue;
}
k -= n;
ans += (long) (a + 1 + a + n) * n / 2;
if (k == 0) {
break;
}
for (int i = 0; i < n + 1 && k > 0; ++i) {
int m = Math.max(0, Math.min(k, arr[i + 1] - arr[i] - 1));
ans += (arr[i] + 1L + arr[i] + m) * m / 2;
k -= m;
}
return ans;
}
Expand All @@ -99,37 +94,41 @@ public:
nums.push_back(2e9);
sort(nums.begin(), nums.end());
long long ans = 0;
for (int i = 1; i < nums.size(); ++i) {
int a = nums[i - 1], b = nums[i];
int n = min(k, b - a - 1);
if (n <= 0) continue;
k -= n;
ans += 1ll * (a + 1 + a + n) * n / 2;
if (k == 0) break;
for (int i = 0; i < nums.size() - 1 && k > 0; ++i) {
int m = max(0, min(k, nums[i + 1] - nums[i] - 1));
ans += 1LL * (nums[i] + 1 + nums[i] + m) * m / 2;
k -= m;
}
return ans;
}
};
```

```go
func minimalKSum(nums []int, k int) int64 {
nums = append(nums, 0, 2e9)
func minimalKSum(nums []int, k int) (ans int64) {
nums = append(nums, []int{0, 2e9}...)
sort.Ints(nums)
ans := 0
for i := 1; i < len(nums); i++ {
a, b := nums[i-1], nums[i]
n := min(k, b-a-1)
if n <= 0 {
continue
}
k -= n
ans += (a + 1 + a + n) * n / 2
if k == 0 {
break
}
for i, b := range nums[1:] {
a := nums[i]
m := max(0, min(k, b-a-1))
ans += int64(a+1+a+m) * int64(m) / 2
k -= m
}
return int64(ans)
return ans
}
```

```ts
function minimalKSum(nums: number[], k: number): number {
nums.push(...[0, 2 * 10 ** 9]);
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 0; i < nums.length - 1; ++i) {
const m = Math.max(0, Math.min(k, nums[i + 1] - nums[i] - 1));
ans += Number((BigInt(nums[i] + 1 + nums[i] + m) * BigInt(m)) / BigInt(2));
k -= m;
}
return ans;
}
```

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95 changes: 47 additions & 48 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/README_EN.md
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Expand Up @@ -41,49 +41,44 @@ The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return

## Solutions

### Solution 1
### Solution 1: Sorting + Greedy + Mathematics

We can add two sentinel nodes to the array, which are $0$ and $2 \times 10^9$.

Then we sort the array. For any two adjacent elements $a$ and $b$ in the array, the integers in the interval $[a+1, b-1]$ do not appear in the array, and we can add these integers to the array.

Therefore, we traverse the adjacent element pairs $(a, b)$ in the array from small to large. For each adjacent element pair, we calculate the number of integers $m$ in the interval $[a+1, b-1]$. The sum of these $m$ integers is $\frac{m \times (a+1 + a+m)}{2}$. We add this sum to the answer and subtract $m$ from $k$. If $k$ is reduced to $0$, we can stop the traversal and return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.

<!-- tabs:start -->

```python
class Solution:
def minimalKSum(self, nums: List[int], k: int) -> int:
nums.append(0)
nums.append(2 * 10**9)
nums.extend([0, 2 * 10**9])
nums.sort()
ans = 0
for a, b in pairwise(nums):
n = min(k, b - a - 1)
if n <= 0:
continue
k -= n
ans += (a + 1 + a + n) * n // 2
if k == 0:
break
m = max(0, min(k, b - a - 1))
ans += (a + 1 + a + m) * m // 2
k -= m
return ans
```

```java
class Solution {
public long minimalKSum(int[] nums, int k) {
int[] arr = new int[nums.length + 2];
arr[arr.length - 1] = (int) 2e9;
for (int i = 0; i < nums.length; ++i) {
arr[i + 1] = nums[i];
}
int n = nums.length;
int[] arr = new int[n + 2];
arr[1] = 2 * 1000000000;
System.arraycopy(nums, 0, arr, 2, n);
Arrays.sort(arr);
long ans = 0;
for (int i = 1; i < arr.length; ++i) {
int a = arr[i - 1], b = arr[i];
int n = Math.min(k, b - a - 1);
if (n <= 0) {
continue;
}
k -= n;
ans += (long) (a + 1 + a + n) * n / 2;
if (k == 0) {
break;
}
for (int i = 0; i < n + 1 && k > 0; ++i) {
int m = Math.max(0, Math.min(k, arr[i + 1] - arr[i] - 1));
ans += (arr[i] + 1L + arr[i] + m) * m / 2;
k -= m;
}
return ans;
}
Expand All @@ -98,37 +93,41 @@ public:
nums.push_back(2e9);
sort(nums.begin(), nums.end());
long long ans = 0;
for (int i = 1; i < nums.size(); ++i) {
int a = nums[i - 1], b = nums[i];
int n = min(k, b - a - 1);
if (n <= 0) continue;
k -= n;
ans += 1ll * (a + 1 + a + n) * n / 2;
if (k == 0) break;
for (int i = 0; i < nums.size() - 1 && k > 0; ++i) {
int m = max(0, min(k, nums[i + 1] - nums[i] - 1));
ans += 1LL * (nums[i] + 1 + nums[i] + m) * m / 2;
k -= m;
}
return ans;
}
};
```

```go
func minimalKSum(nums []int, k int) int64 {
nums = append(nums, 0, 2e9)
func minimalKSum(nums []int, k int) (ans int64) {
nums = append(nums, []int{0, 2e9}...)
sort.Ints(nums)
ans := 0
for i := 1; i < len(nums); i++ {
a, b := nums[i-1], nums[i]
n := min(k, b-a-1)
if n <= 0 {
continue
}
k -= n
ans += (a + 1 + a + n) * n / 2
if k == 0 {
break
}
for i, b := range nums[1:] {
a := nums[i]
m := max(0, min(k, b-a-1))
ans += int64(a+1+a+m) * int64(m) / 2
k -= m
}
return int64(ans)
return ans
}
```

```ts
function minimalKSum(nums: number[], k: number): number {
nums.push(...[0, 2 * 10 ** 9]);
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 0; i < nums.length - 1; ++i) {
const m = Math.max(0, Math.min(k, nums[i + 1] - nums[i] - 1));
ans += Number((BigInt(nums[i] + 1 + nums[i] + m) * BigInt(m)) / BigInt(2));
k -= m;
}
return ans;
}
```

Expand Down
11 changes: 4 additions & 7 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -5,13 +5,10 @@ class Solution {
nums.push_back(2e9);
sort(nums.begin(), nums.end());
long long ans = 0;
for (int i = 1; i < nums.size(); ++i) {
int a = nums[i - 1], b = nums[i];
int n = min(k, b - a - 1);
if (n <= 0) continue;
k -= n;
ans += 1ll * (a + 1 + a + n) * n / 2;
if (k == 0) break;
for (int i = 0; i < nums.size() - 1 && k > 0; ++i) {
int m = max(0, min(k, nums[i + 1] - nums[i] - 1));
ans += 1LL * (nums[i] + 1 + nums[i] + m) * m / 2;
k -= m;
}
return ans;
}
Expand Down
23 changes: 8 additions & 15 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/Solution.go
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,11 @@
func minimalKSum(nums []int, k int) int64 {
nums = append(nums, 0, 2e9)
func minimalKSum(nums []int, k int) (ans int64) {
nums = append(nums, []int{0, 2e9}...)
sort.Ints(nums)
ans := 0
for i := 1; i < len(nums); i++ {
a, b := nums[i-1], nums[i]
n := min(k, b-a-1)
if n <= 0 {
continue
}
k -= n
ans += (a + 1 + a + n) * n / 2
if k == 0 {
break
}
for i, b := range nums[1:] {
a := nums[i]
m := max(0, min(k, b-a-1))
ans += int64(a+1+a+m) * int64(m) / 2
k -= m
}
return int64(ans)
return ans
}
24 changes: 8 additions & 16 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,23 +1,15 @@
class Solution {
public long minimalKSum(int[] nums, int k) {
int[] arr = new int[nums.length + 2];
arr[arr.length - 1] = (int) 2e9;
for (int i = 0; i < nums.length; ++i) {
arr[i + 1] = nums[i];
}
int n = nums.length;
int[] arr = new int[n + 2];
arr[1] = 2 * 1000000000;
System.arraycopy(nums, 0, arr, 2, n);
Arrays.sort(arr);
long ans = 0;
for (int i = 1; i < arr.length; ++i) {
int a = arr[i - 1], b = arr[i];
int n = Math.min(k, b - a - 1);
if (n <= 0) {
continue;
}
k -= n;
ans += (long) (a + 1 + a + n) * n / 2;
if (k == 0) {
break;
}
for (int i = 0; i < n + 1 && k > 0; ++i) {
int m = Math.max(0, Math.min(k, arr[i + 1] - arr[i] - 1));
ans += (arr[i] + 1L + arr[i] + m) * m / 2;
k -= m;
}
return ans;
}
Expand Down
13 changes: 4 additions & 9 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,15 +1,10 @@
class Solution:
def minimalKSum(self, nums: List[int], k: int) -> int:
nums.append(0)
nums.append(2 * 10**9)
nums.extend([0, 2 * 10**9])
nums.sort()
ans = 0
for a, b in pairwise(nums):
n = min(k, b - a - 1)
if n <= 0:
continue
k -= n
ans += (a + 1 + a + n) * n // 2
if k == 0:
break
m = max(0, min(k, b - a - 1))
ans += (a + 1 + a + m) * m // 2
k -= m
return ans
11 changes: 11 additions & 0 deletions solution/2100-2199/2195.Append K Integers With Minimal Sum/Solution.ts
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Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,11 @@
function minimalKSum(nums: number[], k: number): number {
nums.push(...[0, 2 * 10 ** 9]);
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 0; i < nums.length - 1; ++i) {
const m = Math.max(0, Math.min(k, nums[i + 1] - nums[i] - 1));
ans += Number((BigInt(nums[i] + 1 + nums[i] + m) * BigInt(m)) / BigInt(2));
k -= m;
}
return ans;
}
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