feat: leetcode contest 310

Author: upupming

acwing/算法基础课/第五讲 动态规划/线性DP/896. 最长上升子序列 II_RMQ解法.cpp

@@ -1,6 +1,7 @@
 /*
 AcWing 3662 的简化版本
 时间复杂度 O(n log n)，不受数据大小影响
+https://www.acwing.com/solution/content/53104/
 */
 #include <algorithm>
 #include <cstring>

leetcode/contest-310/2406. Divide Intervals Into Minimum Number of Groups.cpp

@@ -0,0 +1,20 @@
+class Solution {
+   public:
+    int minGroups(vector<vector<int>>& ivs) {
+        sort(ivs.begin(), ivs.end(), [](auto& a, auto& b) {
+            return a[0] < b[0];
+        });
+        multiset<int> h;
+        for (auto& iv : ivs) {
+            int s = iv[0], e = iv[1];
+            auto it = (h.lower_bound(s));
+            if (!h.size() || it == h.begin()) {
+                h.insert(e);
+            } else {
+                h.erase(--it);
+                h.insert(e);
+            }
+        }
+        return h.size();
+    }
+};

leetcode/contest-310/6176. Most Frequent Even Element.js

@@ -0,0 +1,16 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const mostFrequentEven = function (nums) {
+  const cnt = {}
+  for (const x of nums) {
+    cnt[x] ??= 0
+    cnt[x]++
+  }
+  const ans = Object.entries(cnt).filter(x => x[0] % 2 === 0).sort((a, b) => {
+    if (a[1] === b[1]) return Number(a[0]) - Number(b[0])
+    return b[1] - a[1]
+  })
+  return ans[0]?.[0] ?? -1
+}

leetcode/contest-310/6177. Optimal Partition of String.js

@@ -0,0 +1,16 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+const partitionString = function (s) {
+  const n = s.length
+  let ans = 0
+  for (let i = 0; i < n;) {
+    ans++
+    const c = new Set()
+    let j = i
+    while (j < n && !c.has(s[j])) c.add(s[j++])
+    i = j
+  }
+  return ans
+}

leetcode/contest-310/6206. Longest Increasing Subsequence II.cpp

@@ -0,0 +1,53 @@
+const int N = 1e5 + 10;
+typedef long long LL;
+class Solution {
+    int h[N] = {0}, a[N] = {0}, n;
+    int lowbit(int x) {
+        return x & (-x);
+    }
+    void updata(int x, int k) {
+        while (x < N) {
+            h[x] = k;
+            int low = lowbit(x);
+            for (int i = 1; i < low; i <<= 1)
+                h[x] = max(h[x], h[x - i]);
+            x += lowbit(x);
+        }
+    }
+    int query(int x, int y) {
+        int ans = 0;
+        while (y >= x) {
+            ans = max(a[y], ans), y -= 1;
+            for (; y - lowbit(y) >= x; y -= lowbit(y))
+                ans = max(h[y], ans);
+        }
+        return ans;
+    }
+
+   public:
+    int lengthOfLIS(vector<int>& nums, int k) {
+        n = nums.size();
+        // vector<int> dp(N, 0);
+
+        for (int i = 1; i <= n; i++) {
+            int x = nums[i - 1];
+            // int len = 1;
+            // for (int j = max(1, a-k); j < a; j++) {
+            //     len = max(len, dp[j] + 1);
+            // }
+            int len = max(1, int(query(max(1, x - k), x - 1)) + 1);
+            // dp[a] = max(dp[a], len);
+            // int oldVal = ask(a, a);
+
+            // cout << x << " " << max(1, x-k) << " " << query(max(1, x-k), x-1) << " "<< len << endl;
+            a[x] = len;
+            updata(x, len);
+        }
+        int ans = 1;
+        for (int i = 1; i <= n; i++) {
+            int a = nums[i - 1];
+            ans = max(ans, int(query(a, a)));
+        }
+        return ans;
+    }
+};

template.md

@@ -646,6 +646,34 @@ void add(int x, int y) {
 }
 ```
 
+```cpp
+// https://www.cnblogs.com/qdscwyy/p/9759220.html
+
+int lowbit(int x) {
+    return x & (-x);
+}
+void updata(int x, int k) {
+    while (x <= n) {
+        h[x] = k;
+        int low = lowbit(x);
+        for (int i = 1; i < low; i <<= 1)
+            h[x] = max(h[x], h[x - i]);
+        x += lowbit(x);
+    }
+}
+// 区间查询 [x, y] 的 max
+int query(int x, int y) {
+    int ans = 0;
+    while (y >= x)
+    {
+        ans = max(a[y], ans), y -= 1;
+        for (; y-lowbit(y) >= x; y -= lowbit(y))
+            ans = max(h[y], ans);
+    }
+    return ans;
+}
+```
+
 ## 线段树
 
 RMQ 常用算法，树状数组基于区间划分，线段树则是基于分治。