diff --git a/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/2387.Median-of-a-Row-Wise-Sorted-Matrix.cpp b/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/2387.Median-of-a-Row-Wise-Sorted-Matrix.cpp
new file mode 100644
index 000000000..211ed8d22
--- /dev/null
+++ b/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/2387.Median-of-a-Row-Wise-Sorted-Matrix.cpp
@@ -0,0 +1,25 @@
+class Solution {
+public:
+ int matrixMedian(vector<vector<int>>& grid) 
+ {
+ int m = grid.size();
+ int n = grid[0].size(); 
+ int k = (m*n+1)/2;
+ 
+ int left = 0, right = INT_MAX;
+ while (left < right) + { + int mid = left+(right-left)/2; + int count = 0; + for (int i=0; i<m; i++) + count += upper_bound(grid[i].begin(), grid[i].end(), mid) - grid[i].begin(); + if (count < k) + left = mid + 1; + else + right = mid; + } + + return left; + + } +}; diff --git a/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/Readme.md b/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/Readme.md new file mode 100644 index 000000000..4c42ebded --- /dev/null +++ b/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix/Readme.md @@ -0,0 +1,9 @@ +### 2387.Median-of-a-Row-Wise-Sorted-Matrix + +令`k=(m*n+1)/2`,本题就是求矩阵里的从小到大的第k个元素。本质和`215.Kth-Largest-Element-in-an-Array`相同,只不过需要将各行`smallerOrEqual(mid)`的结果累加起来得到count。 +```cpp +if (count < k) + left = mid+1; +else + right = mid; +``` diff --git a/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/2361.Minimum-Costs-Using-the-Train-Line.cpp b/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/2361.Minimum-Costs-Using-the-Train-Line.cpp new file mode 100644 index 000000000..e29f45d2d --- /dev/null +++ b/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/2361.Minimum-Costs-Using-the-Train-Line.cpp @@ -0,0 +1,26 @@ +using LL = long long; +class Solution { + LL dp[100005][2]; +public: + vector<long long> minimumCosts(vector<int>& regular, vector<int>& express, int expressCost) 
+ {
+ int n = regular.size();
+ regular.insert(regular.begin(), 0);
+ express.insert(express.begin(), 0);
+ 
+ dp[0][0] = 0;
+ dp[0][1] = expressCost;
+ 
+ vector<ll>rets; 
+ 
+ for (int i=1; i<=n; i++) + { + dp[i][0] = min(dp[i-1][0] + regular[i], dp[i-1][1] + regular[i]); + dp[i][1] = min(dp[i-1][1] + express[i], dp[i-1][0] + expressCost + express[i]); + + rets.push_back(min(dp[i][0], dp[i][1])); + } + + return rets; + } +}; diff --git a/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/Readme.md b/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/Readme.md new file mode 100644 index 000000000..de0a1a977 --- /dev/null +++ b/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line/Readme.md @@ -0,0 +1,8 @@ +### 2361.Minimum-Costs-Using-the-Train-Line + +很明显,状态变量dp[i][0]表示到达第i个车站的regular所需要的最小代价,dp[i][1]表示到达第i个车站的express所需要的最小代价。于是有转移方程: +```cpp +dp[i][0] = min(dp[i-1][0] + regular[i], dp[i-1][1] + regular[i]); +dp[i][1] = min(dp[i-1][1] + express[i], dp[i-1][0] + expressCost + express[i]); +``` +注意我们不需要考虑dp[i][0]与dp[i][1]之间的转移。这是因为,我们如果想要从dp[i][0]转移到dp[i][1],其目的一定只是为了后续得到dp[i+1][1]。单独从第i站的角度来看,只要到了regular或express都算达成了任务,两者间的跳转对于第i站而言没有意义。 diff --git a/Readme.md b/Readme.md index dfb0414aa..21ae6ac95 100644 --- a/Readme.md +++ b/Readme.md @@ -53,6 +53,7 @@ [159.Longest-Substring-with-At-Most-Two-Distinct-Characters](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/159.Longest-Substring-with-At-Most-Two-Distinct-Characters)(H-) [340.Longest-Substring-with-At-Most-K-Distinct-Characters](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/340.Longest-Substring-with-At-Most-K-Distinct-Characters) (H) [992.Subarrays-with-K-Different-Integers](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/992.Subarrays-with-K-Different-Integers) (H-) +[2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K) (M) * ``Two pointers for two seuqences`` [986.Interval-List-Intersections](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/986.Interval-List-Intersections) (M) [1229.Meeting-Scheduler](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/1229.Meeting-Scheduler) (M+) @@ -128,6 +129,7 @@ [793.Preimage-Size-of-Factorial-Zeroes-Function](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/793.Preimage-Size-of-Factorial-Zeroes-Function) (H-) [1201.Ugly-Number-III](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/1201.Ugly-Number-III) (H-) [1539.Kth-Missing-Positive-Number](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/1539.Kth-Missing-Positive-Number) (H-) +[2387.Median-of-a-Row-Wise-Sorted-Matrix](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2387.Median-of-a-Row-Wise-Sorted-Matrix) (H-) #### [Hash Map](https://github.com/wisdompeak/LeetCode/tree/master/Hash) [049.Group-Anagrams](https://github.com/wisdompeak/LeetCode/tree/master/Hash/049.Group-Anagrams) (M+) @@ -678,6 +680,7 @@ [2036.Maximum-Alternating-Subarray-Sum](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2036.Maximum-Alternating-Subarray-Sum) (M+) [2143.Choose-Numbers-From-Two-Arrays-in-Range](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2143.Choose-Numbers-From-Two-Arrays-in-Range) (H) [2318.Number-of-Distinct-Roll-Sequences](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2318.Number-of-Distinct-Roll-Sequences) (H-) +[2361.Minimum-Costs-Using-the-Train-Line](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2361.Minimum-Costs-Using-the-Train-Line) (M+) * ``基本型 II`` [368.Largest-Divisible-Subset](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/368.Largest-Divisible-Subset) (M+) [300.Longest-Increasing-Subsequence](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/300.Longest-Increasing-Subsequence) (M+) @@ -952,6 +955,7 @@ [133.Clone-Graph](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/133.Clone-Graph) (M+) [213.House-Robber-II](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/213.House-Robber-II) (H-) [337.House-Robber-III](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/337.House-Robber-III) (M+) +[2378.Choose-Edges-to-Maximize-Score-in-a-Tree](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree) (H-) [390.Elimination-Game](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/390.Elimination-Game) (H) [395.Longest-Substring-with-At-Least-K-Repeating-Characters](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/395.Longest-Substring-with-At-Least-K-Repeating-Characters) (H) [397.Integer-Replacement](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/397.Integer-Replacement) (M+) diff --git a/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/2378.Choose-Edges-to-Maximize-Score-in-a-Tree.cpp b/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/2378.Choose-Edges-to-Maximize-Score-in-a-Tree.cpp new file mode 100644 index 000000000..447ca3eff --- /dev/null +++ b/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/2378.Choose-Edges-to-Maximize-Score-in-a-Tree.cpp @@ -0,0 +1,52 @@ +using LL = long long; +class Solution { + vector<pair<int,ll>>children[100005];
+ LL memo[100005][2];
+public:
+ long long maxScore(vector<vector<int>>& edges) 
+ {
+ int n = edges.size();
+ int root = -1;
+ for (int i=0; i<n; i++) + { + if (edges[i][0]==-1) + { + root = i; + continue; + } + int par = edges[i][0]; + int weight = edges[i][1]; + children[par].push_back({i, weight}); + } + + return dfs(root, 0); + } + + LL dfs(int node, int status) + { + if (memo[node][status]!=0) + return memo[node][status]; + + if (status == 0) + { + LL sum = 0; + for (auto& [child, weight]: children[node]) + sum += dfs(child, 0); + + LL maxSum = sum; + for (auto& [child, weight]: children[node]) + maxSum = max(maxSum, sum - dfs(child, 0) + dfs(child, 1) + weight); + + memo[node][0] = maxSum; + return maxSum; + } + else + { + LL sum = 0; + for (auto& [child, weight]: children[node]) + sum += dfs(child, 0); + memo[node][1] = sum; + return sum; + } + } +}; diff --git a/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/Readme.md b/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/Readme.md new file mode 100644 index 000000000..457b2c703 --- /dev/null +++ b/Recursion/2378.Choose-Edges-to-Maximize-Score-in-a-Tree/Readme.md @@ -0,0 +1,11 @@ +### 2378.Choose-Edges-to-Maximize-Score-in-a-Tree + +我们不难发现,如果从parent到node的edge被选中的话,那么node到它所有children的edge都不能选中。如果从parent到node的edge不被选中的话,那么node到它所有children的edge里只能最多选一条。 + +所以我们定义`dfs(node, 1)`表示从parent到node的edge被选中,那么以node为根的子树的最大收益。于是有 +```cpp +dfs(node, 1) = sum{dfs(child, 0)}; +``` +同理,定义`dfs(node, 0)`表示从parent到node的edge不被选中,那么以node为根的子树的最大收益。因为我们允许有一条node的子边被选中,所以需要遍历这个选择。为了便于计算,我们先求出`Sum = sum{dfs(child, 0)}`,那么对于任何一个child,如果它的边被选中的话,则整棵树的最大收益就是`Sum - dfs(child, 0) + dfs(child, 1) + weight`. 我们遍历child,再返回最大的作为`dfs(node,1)`. + +显然,此题需要记忆化来避免重复计算。 diff --git a/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K.cpp b/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K.cpp new file mode 100644 index 000000000..296089a64 --- /dev/null +++ b/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K.cpp @@ -0,0 +1,31 @@ +using LL = long long; +class Solution { +public: + long long maximumSubarraySum(vector<int>& nums, int k) 
+ {
+ LL ret = 0;
+ unordered_map<int,int>Map;
+ int count = 0;
+ LL sum = 0;
+ for (int i=0; i<nums.size(); i++) + { + Map[nums[i]]++; + if (Map[nums[i]]==1) + count++; + sum += nums[i]; + + if (i>=k-1)
+ {
+ if (count == k)
+ ret = max(ret, sum);
+ 
+ Map[nums[i-k+1]]--;
+ sum -= nums[i-k+1];
+ if (Map[nums[i-k+1]]==0)
+ count--; 
+ }
+ }
+ 
+ return ret;
+ }
+};
diff --git a/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/Readme.md b/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/Readme.md
new file mode 100644
index 000000000..01afc5d50
--- /dev/null
+++ b/Two_Pointers/2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K/Readme.md
@@ -0,0 +1,17 @@
+### 2461.Maximum-Sum-of-Distinct-Subarrays-With-Length-K
+
+非常普通的固定长度的滑窗。
+
+用一个HashMap来记录每个number出现的次数。用count来表示滑窗内的distinct number的数量。count的变动依据如下:
+1. 当新加入一个数字时
+```cpp
+Map[num]++;
+if (Map[num]==1) 
+ count++;
+```
+2. 当移出一个数字时
+```cpp
+Map[num]--;
+if (Map[num]==0) 
+ count--;
+```
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