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feat: add solutions to lc problem: No.2098
No.2098.Subsequence of Size K With the Largest Even Sum
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‎solution/2000-2099/2098.Subsequence of Size K With the Largest Even Sum/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 排序**
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我们注意到，题目选取的是子序列，因此我们可以考虑先对数组进行排序。
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接下来，我们先贪心地选取最大的 $k$ 个数，如果这些数的和为偶数，则直接返回这个和 $ans$。
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否则，我们有两种贪心策略：
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1. 在最大的 $k$ 个数中，找到一个最小的偶数 $mi1$，然后在剩下的 $n - k$ 个数中，找到一个最大的奇数 $mx1$，将 $mi1$ 替换为 $mx1$，如果存在这样的替换，那么替换后的和 $ans - mi1 + mx1$ 一定是偶数；
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1. 在最大的 $k$ 个数中，找到一个最小的奇数 $mi2$，然后在剩下的 $n - k$ 个数中，找到一个最大的偶数 $mx2$，将 $mi2$ 替换为 $mx2$，如果存在这样的替换，那么替换后的和 $ans - mi2 + mx2$ 一定是偶数。
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我们取最大的偶数和作为答案。如果不存在偶数和，则返回 $-1$。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组长度。
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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 largestEvenSum(self, nums: List[int], k: int) -> int:
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nums.sort()
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ans = sum(nums[-k:])
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if ans % 2 == 0:
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return ans
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n = len(nums)
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mx1 = mx2 = -inf
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for x in nums[: n - k]:
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if x & 1:
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mx1 = x
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else:
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mx2 = x
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mi1 = mi2 = inf
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for x in nums[-k:][::-1]:
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if x & 1:
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mi2 = x
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else:
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mi1 = x
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ans = max(ans - mi1 + mx1, ans - mi2 + mx2, -1)
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return -1 if ans % 2 else ans
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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 largestEvenSum(int[] nums, int k) {
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Arrays.sort(nums);
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long ans = 0;
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int n = nums.length;
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for (int i = 0; i < k; ++i) {
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ans += nums[n - i - 1];
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}
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if (ans % 2 == 0) {
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return ans;
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}
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final int inf = 1 << 29;
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int mx1 = -inf, mx2 = -inf;
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for (int i = 0; i < n - k; ++i) {
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if (nums[i] % 2 == 1) {
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mx1 = nums[i];
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} else {
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mx2 = nums[i];
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}
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}
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int mi1 = inf, mi2 = inf;
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for (int i = n - 1; i >= n - k; --i) {
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if (nums[i] % 2 == 1) {
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mi2 = nums[i];
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} else {
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mi1 = nums[i];
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}
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}
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ans = Math.max(-1, Math.max(ans - mi1 + mx1, ans - mi2 + mx2));
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return ans % 2 != 0 ? -1 : ans;
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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 largestEvenSum(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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long long ans = 0;
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int n = nums.size();
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for (int i = 0; i < k; ++i) {
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ans += nums[n - i - 1];
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}
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if (ans % 2 == 0) {
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return ans;
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}
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const int inf = 1 << 29;
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int mx1 = -inf, mx2 = -inf;
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for (int i = 0; i < n - k; ++i) {
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if (nums[i] % 2) {
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mx1 = nums[i];
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} else {
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mx2 = nums[i];
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}
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}
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int mi1 = inf, mi2 = inf;
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for (int i = n - 1; i >= n - k; --i) {
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if (nums[i] % 2) {
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mi2 = nums[i];
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} else {
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mi1 = nums[i];
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}
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}
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ans = max(ans - mi1 + mx1, ans - mi2 + mx2);
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return ans % 2 || ans < 0 ? -1 : ans;
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}
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};
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```
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### **TypeScript**
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### **Go**
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```go
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func largestEvenSum(nums []int, k int) int64 {
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sort.Ints(nums)
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ans := 0
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n := len(nums)
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for i := 0; i < k; i++ {
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ans += nums[n-1-i]
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}
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if ans%2 == 0 {
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return int64(ans)
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}
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const inf = 1 << 29
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mx1, mx2 := -inf, -inf
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for _, x := range nums[:n-k] {
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if x%2 == 1 {
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mx1 = x
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} else {
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mx2 = x
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}
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}
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mi1, mi2 := inf, inf
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for i := n - 1; i >= n-k; i-- {
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if nums[i]%2 == 1 {
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mi2 = nums[i]
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} else {
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mi1 = nums[i]
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}
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}
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ans = max(-1, max(ans-mi1+mx1, ans-mi2+mx2))
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if ans%2 != 0 {
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return -1
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}
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return int64(ans)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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‎solution/2000-2099/2098.Subsequence of Size K With the Largest Even Sum/README_EN.md

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### **Python3**
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```python
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class Solution:
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def largestEvenSum(self, nums: List[int], k: int) -> int:
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nums.sort()
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ans = sum(nums[-k:])
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if ans % 2 == 0:
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return ans
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n = len(nums)
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mx1 = mx2 = -inf
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for x in nums[: n - k]:
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if x & 1:
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mx1 = x
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else:
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mx2 = x
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mi1 = mi2 = inf
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for x in nums[-k:][::-1]:
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if x & 1:
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mi2 = x
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else:
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mi1 = x
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ans = max(ans - mi1 + mx1, ans - mi2 + mx2, -1)
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return -1 if ans % 2 else ans
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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 largestEvenSum(int[] nums, int k) {
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Arrays.sort(nums);
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long ans = 0;
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int n = nums.length;
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for (int i = 0; i < k; ++i) {
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ans += nums[n - i - 1];
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}
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if (ans % 2 == 0) {
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return ans;
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}
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final int inf = 1 << 29;
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int mx1 = -inf, mx2 = -inf;
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for (int i = 0; i < n - k; ++i) {
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if (nums[i] % 2 == 1) {
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mx1 = nums[i];
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} else {
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mx2 = nums[i];
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}
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}
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int mi1 = inf, mi2 = inf;
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for (int i = n - 1; i >= n - k; --i) {
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if (nums[i] % 2 == 1) {
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mi2 = nums[i];
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} else {
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mi1 = nums[i];
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}
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}
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ans = Math.max(-1, Math.max(ans - mi1 + mx1, ans - mi2 + mx2));
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return ans % 2 != 0 ? -1 : ans;
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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 largestEvenSum(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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long long ans = 0;
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int n = nums.size();
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for (int i = 0; i < k; ++i) {
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ans += nums[n - i - 1];
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}
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if (ans % 2 == 0) {
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return ans;
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}
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const int inf = 1 << 29;
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int mx1 = -inf, mx2 = -inf;
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for (int i = 0; i < n - k; ++i) {
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if (nums[i] % 2) {
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mx1 = nums[i];
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} else {
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mx2 = nums[i];
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}
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}
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int mi1 = inf, mi2 = inf;
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for (int i = n - 1; i >= n - k; --i) {
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if (nums[i] % 2) {
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mi2 = nums[i];
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} else {
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mi1 = nums[i];
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}
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}
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ans = max(ans - mi1 + mx1, ans - mi2 + mx2);
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return ans % 2 || ans < 0 ? -1 : ans;
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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 largestEvenSum(nums []int, k int) int64 {
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sort.Ints(nums)
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ans := 0
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n := len(nums)
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for i := 0; i < k; i++ {
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ans += nums[n-1-i]
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}
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if ans%2 == 0 {
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return int64(ans)
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}
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const inf = 1 << 29
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mx1, mx2 := -inf, -inf
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for _, x := range nums[:n-k] {
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if x%2 == 1 {
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mx1 = x
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} else {
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mx2 = x
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}
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}
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mi1, mi2 := inf, inf
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for i := n - 1; i >= n-k; i-- {
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if nums[i]%2 == 1 {
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mi2 = nums[i]
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} else {
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mi1 = nums[i]
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}
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}
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ans = max(-1, max(ans-mi1+mx1, ans-mi2+mx2))
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if ans%2 != 0 {
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return -1
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}
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return int64(ans)
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**

‎solution/2000-2099/2098.Subsequence of Size K With the Largest Even Sum/Solution.cpp

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class Solution {
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public:
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long long largestEvenSum(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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long long ans = 0;
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int n = nums.size();
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for (int i = 0; i < k; ++i) {
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ans += nums[n - i - 1];
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}
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if (ans % 2 == 0) {
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return ans;
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}
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const int inf = 1 << 29;
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int mx1 = -inf, mx2 = -inf;
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for (int i = 0; i < n - k; ++i) {
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if (nums[i] % 2) {
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mx1 = nums[i];
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} else {
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mx2 = nums[i];
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}
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}
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int mi1 = inf, mi2 = inf;
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for (int i = n - 1; i >= n - k; --i) {
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if (nums[i] % 2) {
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mi2 = nums[i];
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} else {
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mi1 = nums[i];
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}
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}
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ans = max(ans - mi1 + mx1, ans - mi2 + mx2);
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return ans % 2 || ans < 0 ? -1 : ans;
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}
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};

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