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feat: add solutions to lc/lcp problems
* Add solutions to lcp problem: No.40 * Add solution to lc problem: No.1921.Eliminate Maximum Number of Monsters
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‎lcp/LCP 40. 心算挑战/README.md

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@@ -32,22 +32,149 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 贪心**
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排序先取最大的 $cnt$ 个数，如果和为偶数则直接返回答案。
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否则，找一个已取的最小奇数换成剩余未取的最大偶数，或者找一个已取的最小偶数换成剩下未取的最大奇数，取两者中较大的。
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时间复杂度 $O(nlogn)$。
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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 maxmiumScore(self, cards: List[int], cnt: int) -> int:
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cards.sort(reverse=True)
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t = cards[:cnt]
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ans = sum(t)
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if ans % 2 == 0:
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return ans
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a = min([v for v in t if v & 1], default=inf)
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b = min([v for v in t if v % 2 == 0], default=inf)
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c = max([v for v in cards[cnt:] if v % 2 == 0], default=-inf)
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d = max([v for v in cards[cnt:] if v & 1], default=-inf)
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return max(ans - a + c, ans - b + d, 0)
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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 int maxmiumScore(int[] cards, int cnt) {
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Arrays.sort(cards);
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int ans = 0;
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int n = cards.length;
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for (int i = 0; i < cnt; ++i) {
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ans += cards[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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int inf = 0x3f3f3f3f;
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int a = inf, b = inf;
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for (int i = 0; i < cnt; ++i) {
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int v = cards[n - i - 1];
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if (v % 2 == 1) {
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a = Math.min(a, v);
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} else {
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b = Math.min(b, v);
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}
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}
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int c = -inf, d = -inf;
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for (int i = cnt; i < n; ++i) {
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int v = cards[n - i - 1];
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if (v % 2 == 0) {
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c = Math.max(c, v);
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} else {
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d = Math.max(d, v);
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}
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}
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return Math.max(0, Math.max(ans - a + c, ans - b + d));
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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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int maxmiumScore(vector<int>& cards, int cnt) {
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sort(cards.begin(), cards.end());
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reverse(cards.begin(), cards.end());
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int ans = 0, n = cards.size();
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for (int i = 0; i < cnt; ++i) ans += cards[i];
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if (ans % 2 == 0) return ans;
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int inf = 0x3f3f3f3f;
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int a = inf, b = inf, c = -inf, d = -inf;
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for (int i = 0; i < cnt; ++i)
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{
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int v = cards[i];
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if (v % 2 == 1) a = min(a, v);
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else b = min(b, v);
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}
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for (int i = cnt; i < n; ++i)
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{
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int v = cards[i];
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if (v % 2 == 0) c = max(c, v);
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else d = max(d, v);
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}
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return max(0, max(ans - a + c, ans - b + d));
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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 maxmiumScore(cards []int, cnt int) int {
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ans := 0
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sort.Slice(cards, func(i, j int) bool { return cards[i] > cards[j] })
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for _, v := range cards[:cnt] {
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ans += v
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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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inf := 0x3f3f3f3f
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a, b, c, d := inf, inf, -inf, -inf
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for _, v := range cards[:cnt] {
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if v%2 == 1 {
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a = min(a, v)
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} else {
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b = min(b, v)
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}
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}
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for _, v := range cards[cnt:] {
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if v%2 == 0 {
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c = max(c, v)
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} else {
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d = max(d, v)
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}
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}
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return max(0, max(ans-a+c, ans-b+d))
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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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func min(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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### **...**

‎lcp/LCP 40. 心算挑战/Solution.cpp

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class Solution {
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public:
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int maxmiumScore(vector<int>& cards, int cnt) {
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sort(cards.begin(), cards.end());
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reverse(cards.begin(), cards.end());
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int ans = 0, n = cards.size();
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for (int i = 0; i < cnt; ++i) ans += cards[i];
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if (ans % 2 == 0) return ans;
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int inf = 0x3f3f3f3f;
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int a = inf, b = inf, c = -inf, d = -inf;
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for (int i = 0; i < cnt; ++i)
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{
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int v = cards[i];
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if (v % 2 == 1) a = min(a, v);
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else b = min(b, v);
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}
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for (int i = cnt; i < n; ++i)
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{
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int v = cards[i];
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if (v % 2 == 0) c = max(c, v);
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else d = max(d, v);
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}
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return max(0, max(ans - a + c, ans - b + d));
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}
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};

‎lcp/LCP 40. 心算挑战/Solution.go

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func maxmiumScore(cards []int, cnt int) int {
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ans := 0
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sort.Slice(cards, func(i, j int) bool { return cards[i] > cards[j] })
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for _, v := range cards[:cnt] {
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ans += v
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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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inf := 0x3f3f3f3f
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a, b, c, d := inf, inf, -inf, -inf
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for _, v := range cards[:cnt] {
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if v%2 == 1 {
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a = min(a, v)
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} else {
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b = min(b, v)
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}
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}
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for _, v := range cards[cnt:] {
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if v%2 == 0 {
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c = max(c, v)
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} else {
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d = max(d, v)
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}
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}
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return max(0, max(ans-a+c, ans-b+d))
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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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func min(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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}

‎lcp/LCP 40. 心算挑战/Solution.java

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class Solution {
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public int maxmiumScore(int[] cards, int cnt) {
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Arrays.sort(cards);
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int ans = 0;
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int n = cards.length;
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for (int i = 0; i < cnt; ++i) {
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ans += cards[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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int inf = 0x3f3f3f3f;
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int a = inf, b = inf;
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for (int i = 0; i < cnt; ++i) {
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int v = cards[n - i - 1];
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if (v % 2 == 1) {
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a = Math.min(a, v);
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} else {
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b = Math.min(b, v);
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}
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}
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int c = -inf, d = -inf;
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for (int i = cnt; i < n; ++i) {
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int v = cards[n - i - 1];
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if (v % 2 == 0) {
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c = Math.max(c, v);
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} else {
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d = Math.max(d, v);
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}
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}
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return Math.max(0, Math.max(ans - a + c, ans - b + d));
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}
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}

‎lcp/LCP 40. 心算挑战/Solution.py

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class Solution:
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def maxmiumScore(self, cards: List[int], cnt: int) -> int:
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cards.sort(reverse=True)
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t = cards[:cnt]
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ans = sum(t)
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if ans % 2 == 0:
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return ans
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a = min([v for v in t if v & 1], default=inf)
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b = min([v for v in t if v % 2 == 0], default=inf)
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c = max([v for v in cards[cnt:] if v % 2 == 0], default=-inf)
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d = max([v for v in cards[cnt:] if v & 1], default=-inf)
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return max(ans - a + c, ans - b + d, 0)

‎solution/1900-1999/1921.Eliminate Maximum Number of Monsters/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 贪心**
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用 $times$ 数组记录每个怪物最晚可被消灭的时间。对于第 $i$ 个怪物，最晚可被消灭的时间满足：
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$$times[i] = \lfloor \frac{dist[i]-1}{speed[i]} \rfloor$$
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我们对 $times$ 数组升序排列，然后遍历 $times$ 数组。对于第 $i$ 个怪物，如果 $times[i] \geq i$，说明第 $i$ 个怪物可以被消灭，否则说明第 $i$ 个怪物无法被消灭，直接返回 $i$ 即可。
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若所有怪物都可以被消灭，则返回 $n$。
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时间复杂度 $O(nlogn)$。
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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 eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
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n = len(dist)
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times = [(dist[i] - 1) // speed[i] for i in range(n)]
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times = [(d - 1) // s for d, s in zip(dist, speed)]
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times.sort()
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for i in range(n):
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if times[i] < i:
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for i, t in enumerate(times):
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if t < i:
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return i
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return n
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return len(dist)
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```
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### **Java**
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func eliminateMaximum(dist []int, speed []int) int {
159170
n := len(dist)
160171
times := make([]int, n)
161-
for i := 0; i < n; i++ {
162-
times[i] = (dist[i] - 1) / speed[i]
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for i, d := range dist {
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times[i] = (d - 1) / speed[i]
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}
164175
sort.Ints(times)
165-
for i := 0; i < n; i++ {
166-
if times[i] < i {
176+
for i, t := range times {
177+
if t < i {
167178
return i
168179
}
169180
}
170181
return n
171182
}
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```
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### **TypeScript**
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```ts
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function eliminateMaximum(dist: number[], speed: number[]): number {
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const n = dist.length;
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const times = new Array(n).fill(0);
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for (let i = 0; i < n; ++i) {
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times[i] = Math.floor((dist[i] - 1) / speed[i]);
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}
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times.sort((a, b) => a - b);
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for (let i = 0; i < n; ++i) {
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if (times[i] < i) {
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return i;
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}
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}
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return n;
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}
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```
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### **...**
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```

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