feat: leetcode contest 301

Author: upupming

leetcode/contest-301/6112. Minimum Amount of Time to Fill Cups.js

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+/**
+ * @param {number[]} amount
+ * @return {number}
+ */
+ var fillCups = function(a) {
+  a = a.sort((x, y) => x-y).filter(x => x!= 0)
+  let ans = 0
+  while (a.length) {
+      // console.log(a)
+      if (a.length >= 2) a[0]--, a[a.length-1]--
+      else a[0]--
+      a = a.sort((x, y) => x-y).filter(x => x!= 0)
+      ans++
+  }
+  return ans
+};

leetcode/contest-301/6113. Smallest Number in Infinite Set.cpp

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+class SmallestInfiniteSet {
+    set<int> s;
+
+   public:
+    SmallestInfiniteSet() {
+        for (int i = 1; i <= 1e3; i++) s.insert(i);
+    }
+
+    int popSmallest() {
+        int x = *s.begin();
+        s.erase(s.begin());
+        return x;
+    }
+
+    void addBack(int num) {
+        s.insert(num);
+    }
+};
+
+/**
+ * Your SmallestInfiniteSet object will be instantiated and called as such:
+ * SmallestInfiniteSet* obj = new SmallestInfiniteSet();
+ * int param_1 = obj->popSmallest();
+ * obj->addBack(num);
+ */

leetcode/contest-301/6114. Move Pieces to Obtain a String.js

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+/**
+ * @param {string} start
+ * @param {string} target
+ * @return {boolean}
+ */
+const canChange = function (s, t) {
+  const ca = s.split('').filter(x => x != '_').join('') === t.split('').filter(x => x != '_').join('')
+  if (!ca) return false
+  let a = []; let b = []
+  for (let i = 0; i < s.length; i++) {
+    if (s[i] === 'L') a.push(i)
+    if (t[i] === 'L') b.push(i)
+  }
+  if (a.length !== b.length) return false
+  for (let i = 0; i < a.length; i++) if (!(a[i] >= b[i])) return false
+
+  a = [], b = []
+  for (let i = 0; i < s.length; i++) {
+    if (s[i] === 'R') a.push(i)
+    if (t[i] === 'R') b.push(i)
+  }
+  if (a.length !== b.length) return false
+  for (let i = 0; i < a.length; i++) if (!(a[i] <= b[i])) return false
+
+  return true
+}

leetcode/contest-301/6115. Count the Number of Ideal Arrays.ts

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+function idealArrays (n: number, maxValue: number): number {
+  // f[i][j] 表示 arr[i] 与 arr[i - 1] 不能相等 的情况下，以 i 为最后一个元素，长度为 j 的方案数
+  // 因为 maxValue <= 1e4, j <= log_2 maxValue <= 13.2 < 14
+  const MAX_LEN = 14; const P = 1e9 + 7
+
+  // maxValue log maxValue 时间复杂度求因子
+  const frac = Array(maxValue + 1).fill(0).map<number[]>(() => [])
+  for (let i = 1; i <= maxValue; i++) {
+    for (let j = 2 * i; j <= maxValue; j += i) {
+      if (j % i === 0) frac[j].push(i)
+    }
+  }
+
+  // maxValue log maxValue log maxValue
+  const f = Array(maxValue + 1).fill(0).map(() => Array<number>(MAX_LEN + 1).fill(0))
+  for (let i = 1; i <= maxValue; i++) {
+    f[i][1] = 1
+    for (let j = 2; j <= MAX_LEN; j++) {
+      for (const k of frac[i]) f[i][j] = (f[i][j] + f[k][j - 1]) % P
+    }
+  }
+
+  // n log maxValue
+  const c = Array(n + 1).fill(0).map(() => Array<number>(MAX_LEN + 1).fill(0))
+  c[0][0] = 1
+  for (let i = 1; i <= n; i++) {
+    c[i][0] = 1
+    for (let j = 1; j <= MAX_LEN && j <= i; j++) {
+      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % P
+    }
+  }
+
+  // maxValue log maxValue
+  let ans = 0
+  // 真正的长度为 n，arr[i] 与 arr[i - 1] 不能相等的情况下的长度为 t，在 n-1 个位置选 t-1 个位置作为值改变的起始点
+  for (let i = 1; i <= maxValue; i++) {
+    for (let j = 1; j <= MAX_LEN; j++) {
+      ans = (ans + Number(BigInt(c[n - 1][j - 1]) * BigInt(f[i][j]) % BigInt(P))) % P
+    }
+  }
+  return ans
+};
+
+export declare const a: 1