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feat: add solutions to lc problem: No.1673
No.1673.Find the Most Competitive Subsequence
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‎.github/workflows/sync.yml

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This file was deleted.

‎solution/1600-1699/1671.Minimum Number of Removals to Make Mountain Array/README.md

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那么最终答案就是 $n - \max(left[i] + right[i] - 1)$，其中 $1 \leq i \leq n$，并且 $left[i] \gt 1$ 且 $right[i] \gt 1$。
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时间复杂度 $O(n^2)$。
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<!-- tabs:start -->
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### **Python3**

‎solution/1600-1699/1672.Richest Customer Wealth/README.md

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@@ -53,6 +53,12 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：求和**
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遍历 `accounts`，求出每一行的和，然后求出最大值。
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时间复杂度 $O(m\times n)$。
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<!-- tabs:start -->
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### **Python3**
@@ -62,7 +68,7 @@
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```python
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class Solution:
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def maximumWealth(self, accounts: List[List[int]]) -> int:
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return max(sum(account) for account in accounts)
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return max(sum(v) for v in accounts)
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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 int maximumWealth(int[][] accounts) {
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int res = 0;
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for (int[] account : accounts) {
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int t = 0;
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for (int money : account) {
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t += money;
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int ans = 0;
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for (var e : accounts) {
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int s = 0;
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for (int v : e) {
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s += v;
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}
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res = Math.max(res, t);
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ans = Math.max(ans, s);
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}
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return res;
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return ans;
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}
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}
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```
@@ -91,10 +97,11 @@ class Solution {
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class Solution {
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public:
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int maximumWealth(vector<vector<int>>& accounts) {
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int res = 0;
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for (auto& account : accounts)
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res = max(res, accumulate(account.begin(), account.end(), 0));
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return res;
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int ans = 0;
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for (auto& v : accounts) {
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ans = max(ans, accumulate(v.begin(), v.end(), 0));
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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
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func maximumWealth(accounts [][]int) int {
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res := 0
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for _, account := range accounts {
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t := 0
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for _, money := range account {
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t += money
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}
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if t > res {
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res = t
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}
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}
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return res
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ans := 0
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for _, e := range accounts {
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s := 0
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for _, v := range e {
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s += v
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}
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if ans < s {
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ans = s
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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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‎solution/1600-1699/1672.Richest Customer Wealth/README_EN.md

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@@ -57,23 +57,23 @@ The 2nd customer is the richest with a wealth of 10.</pre>
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```python
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class Solution:
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def maximumWealth(self, accounts: List[List[int]]) -> int:
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return max(sum(account) for account in accounts)
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return max(sum(v) for v in accounts)
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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 int maximumWealth(int[][] accounts) {
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int res = 0;
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for (int[] account : accounts) {
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int t = 0;
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for (int money : account) {
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t += money;
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int ans = 0;
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for (var e : accounts) {
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int s = 0;
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for (int v : e) {
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s += v;
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}
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res = Math.max(res, t);
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ans = Math.max(ans, s);
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}
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return res;
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return ans;
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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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int maximumWealth(vector<vector<int>>& accounts) {
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int res = 0;
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for (auto& account : accounts)
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res = max(res, accumulate(account.begin(), account.end(), 0));
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return res;
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int ans = 0;
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for (auto& v : accounts) {
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ans = max(ans, accumulate(v.begin(), v.end(), 0));
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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
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func maximumWealth(accounts [][]int) int {
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res := 0
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for _, account := range accounts {
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t := 0
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for _, money := range account {
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t += money
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}
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if t > res {
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res = t
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}
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}
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return res
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ans := 0
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for _, e := range accounts {
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s := 0
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for _, v := range e {
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s += v
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}
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if ans < s {
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ans = s
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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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‎solution/1600-1699/1672.Richest Customer Wealth/Solution.cpp

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class Solution {
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public:
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int maximumWealth(vector<vector<int>>& accounts) {
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int res = 0;
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for (auto& account : accounts)
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res = max(res, accumulate(account.begin(), account.end(), 0));
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return res;
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int ans = 0;
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for (auto& v : accounts) {
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ans = max(ans, accumulate(v.begin(), v.end(), 0));
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}
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return ans;
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}
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};

‎solution/1600-1699/1672.Richest Customer Wealth/Solution.go

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@@ -1,13 +1,13 @@
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func maximumWealth(accounts [][]int) int {
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res := 0
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for _, account := range accounts {
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t := 0
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for _, money := range account {
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t += money
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ans := 0
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for _, e := range accounts {
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s := 0
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for _, v := range e {
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s += v
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}
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if t > res {
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res = t
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if ans < s {
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ans = s
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}
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}
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return res
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return ans
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}

‎solution/1600-1699/1672.Richest Customer Wealth/Solution.java

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@@ -1,13 +1,13 @@
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class Solution {
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public int maximumWealth(int[][] accounts) {
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int res = 0;
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for (int[] account : accounts) {
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int t = 0;
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for (int money : account) {
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t += money;
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int ans = 0;
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for (var e : accounts) {
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int s = 0;
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for (int v : e) {
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s += v;
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}
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res = Math.max(res, t);
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ans = Math.max(ans, s);
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}
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return res;
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return ans;
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}
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}

‎solution/1600-1699/1672.Richest Customer Wealth/Solution.py

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@@ -1,3 +1,3 @@
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class Solution:
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def maximumWealth(self, accounts: List[List[int]]) -> int:
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return max(sum(account) for account in accounts)
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return max(sum(v) for v in accounts)

‎solution/1600-1699/1673.Find the Most Competitive Subsequence/README.md

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@@ -43,22 +43,96 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：栈**
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我们从左到右遍历数组 `nums`，维护一个栈 `stk`，遍历过程中，如果当前元素 `nums[i]` 小于栈顶元素，且栈中元素个数加上 $n-i$ 大于 $k$，则将栈顶元素出栈，直到不满足上述条件为止。此时如果栈中元素个数小于 $k$，则将当前元素入栈。
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遍历结束后，栈中元素即为所求。
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时间复杂度 $O(n)$，空间复杂度 $O(k)$。其中 $n$ 为数组 `nums` 的长度。
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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:
62+
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
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stk = []
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n = len(nums)
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for i, v in enumerate(nums):
66+
while stk and stk[-1] > v and len(stk) + n - i > k:
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stk.pop()
68+
if len(stk) < k:
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stk.append(v)
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return stk
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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
78+
class Solution {
79+
public int[] mostCompetitive(int[] nums, int k) {
80+
Deque<Integer> stk = new ArrayDeque<>();
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int n = nums.length;
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for (int i = 0; i < nums.length; ++i) {
83+
while (!stk.isEmpty() && stk.peek() > nums[i] && stk.size() + n - i > k) {
84+
stk.pop();
85+
}
86+
if (stk.size() < k) {
87+
stk.push(nums[i]);
88+
}
89+
}
90+
int[] ans = new int[stk.size()];
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for (int i = ans.length - 1; i >= 0; --i) {
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ans[i] = stk.pop();
93+
}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
102+
class Solution {
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public:
104+
vector<int> mostCompetitive(vector<int>& nums, int k) {
105+
vector<int> stk;
106+
int n = nums.size();
107+
for (int i = 0; i < n; ++i) {
108+
while (stk.size() && stk.back() > nums[i] && stk.size() + n - i > k) {
109+
stk.pop_back();
110+
}
111+
if (stk.size() < k) {
112+
stk.push_back(nums[i]);
113+
}
114+
}
115+
return stk;
116+
}
117+
};
118+
```
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### **Go**
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```go
123+
func mostCompetitive(nums []int, k int) []int {
124+
stk := []int{}
125+
n := len(nums)
126+
for i, v := range nums {
127+
for len(stk) > 0 && stk[len(stk)-1] > v && len(stk)+n-i > k {
128+
stk = stk[:len(stk)-1]
129+
}
130+
if len(stk) < k {
131+
stk = append(stk, v)
132+
}
133+
}
134+
return stk
135+
}
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```
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### **...**

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