feat: leetcode contest 313

Author: upupming

leetcode/contest-313/6192. Number of Common Factors.js

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+/**
+ * @param {number} a
+ * @param {number} b
+ * @return {number}
+ */
+const commonFactors = function (a, b) {
+  let ans = 0
+  for (let i = 1; i <= Math.floor(Math.sqrt(a)); i++) {
+    if (a % i === 0) {
+      if (b % i === 0)ans++
+      if (a / i !== i && b % (a / i) === 0)ans++
+    }
+  }
+  return ans
+}

leetcode/contest-313/6193. Maximum Sum of an Hourglass.js

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+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+const maxSum = function (g) {
+  const m = g.length; const n = g[0].length
+  let ans = 0
+  for (let i = 0; i + 2 < m; i++) {
+    for (let j = 0; j + 2 < n; j++) {
+      let t = 0
+      for (let l = i; l <= i + 2; l++) {
+        for (let s = j; s <= j + 2; s++) {
+          t += g[l][s]
+        }
+      }
+      t -= g[i + 1][j] + g[i + 1][j + 2]
+
+      ans = Math.max(ans, t)
+    }
+  }
+  return ans
+}

leetcode/contest-313/6194. Minimize XOR.js

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+/**
+ * @param {number} num1
+ * @param {number} num2
+ * @return {number}
+ */
+const minimizeXor = function (num1, num2) {
+  let s = num1.toString(2)
+  let t = num2.toString(2)
+  const n = Math.max(s.length, t.length)
+  // console.log('s', s, typeof s, 't', t)
+  s = s.padStart(n, '0').split('').map(Number)
+  t = t.padStart(n, '0').split('').map(Number)
+  const cnt = t.reduce((p, c) => p + c, 0)
+  const ans = Array(n).fill(-1)
+  let used = 0
+
+  for (let i = 0; i < n && used < cnt; i++) {
+    if (s[i] === 1) ans[i] = 1, used++
+  }
+  for (let i = n - 1; i >= 0 && used < cnt; i--) {
+    if (ans[i] === -1) ans[i] = 1, used++
+  }
+  for (let i = 0; i < n; i++) {
+    if (ans[i] === -1) ans[i] = 0
+  }
+  return parseInt(ans.join(''), 2)
+}

leetcode/contest-313/6195. Maximum Deletions on a String-double-hash.js

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+/**
+ * @param {string} s
+ * @return {number}
+ */
+const deleteString = function (s) {
+  // 最好以 1 为开始下标，比较好处理 p 数组
+  const n = s.length
+  s = ' ' + s
+
+  const P = 131
+  // 三哈希（双哈希的拓展）避免冲突
+  const Q = 1e4 + 7; const Q1 = 1e5 + 7; const Q2 = 1e6 + 7
+  const f = [0]; const p = [1]
+  const f1 = [0]; const p1 = [1]
+  const f2 = [0]; const p2 = [1]
+  for (let i = 1; i <= n; i++) {
+    f[i] = (f[i - 1] * P % Q + (s[i].charCodeAt(0) - 'a'.charCodeAt(0))) % Q
+    p[i] = p[i - 1] * P % Q
+
+    f1[i] = (f1[i - 1] * P % Q1 + (s[i].charCodeAt(0) - 'a'.charCodeAt(0))) % Q1
+    p1[i] = p1[i - 1] * P % Q1
+
+    f2[i] = (f2[i - 1] * P % Q2 + (s[i].charCodeAt(0) - 'a'.charCodeAt(0))) % Q2
+    p2[i] = p2[i - 1] * P % Q2
+  }
+  const hash = (i, j) => {
+    return ((f[j] - f[i - 1] * p[j - i + 1] % Q + Q) % Q) &&
+           ((f1[j] - f1[i - 1] * p1[j - i + 1] % Q1 + Q1) % Q1) &&
+           ((f2[j] - f2[i - 1] * p2[j - i + 1] % Q2 + Q2) % Q2)
+  }
+
+  const dp = Array(n + 1).fill(1)
+
+  for (let i = n; i >= 1; i--) {
+    for (let len = 1; i + (len * 2) - 1 <= n; len++) {
+      // [i, j]; [j+1, k]
+      const j = i + len - 1; const k = i + (len * 2) - 1
+      if (hash(i, j) === hash(j + 1, k)) {
+        dp[i] = Math.max(dp[i], dp[j + 1] + 1)
+      }
+    }
+  }
+  return dp[1]
+}

leetcode/contest-313/6195. Maximum Deletions on a String.js

@@ -0,0 +1,34 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+const deleteString = function (s) {
+  // 最好以 1 为开始下标，比较好处理 p 数组
+  const n = s.length
+  s = ' ' + s
+
+  const P = 131n
+  const Q = 1n << 64n
+  const f = [0n]; const p = [1n]
+  for (let i = 1; i <= n; i++) {
+    f[i] = (f[i - 1] * P % Q + (BigInt(s[i].charCodeAt(0) - 'a'.charCodeAt(0)))) % Q
+    p[i] = p[i - 1] * P % Q
+  }
+  const hash = (i, j) => {
+    // 乘完之后的取模非常重要，漏掉就会 WA（出现较小负数导致 + 一个 Q 不够）
+    return (f[j] - f[i - 1] * p[j - i + 1] % Q + Q) % Q
+  }
+
+  const dp = Array(n + 1).fill(1)
+
+  for (let i = n; i >= 1; i--) {
+    for (let len = 1; i + (len * 2) - 1 <= n; len++) {
+      // [i, j]; [j+1, k]
+      const j = i + len - 1; const k = i + (len * 2) - 1
+      if (hash(i, j) === hash(j + 1, k)) {
+        dp[i] = Math.max(dp[i], dp[j + 1] + 1)
+      }
+    }
+  }
+  return dp[1]
+}