feat: problems for the next week

Author: upupming

leetcode/contest-292/6059. Check if There Is a Valid Parentheses String Path.js

@@ -4,28 +4,25 @@
  */
 const hasValidPath = function (grid) {
   const m = grid.length; const n = grid[0].length
-  grid = grid.map(g => g.map((a) => a === '(' ? 1 : -1))
-  if ((m + n) % 2 === 0) return false
-  const max = Array(m).fill(0).map(() => Array(n).fill(-1e5))
-  const min = Array(m).fill(0).map(() => Array(n).fill(1e5))
-  if (grid[0][0] >= 0) min[0][0] = grid[0][0], max[0][0] = grid[0][0]
-  // console.log(grid)
+  const dp = Array(m).fill(0).map(() => Array(n).fill(0).map(() => new Set()))
+
   for (let i = 0; i < m; i++) {
     for (let j = 0; j < n; j++) {
-      // 从 [i-1, j] 过来
-      if (i >= 1) {
-        // tmpMin < 0 但是 tmpMax >= 0 的情况，我们 i, j 仍然合法，min 值为 0
-        const tmpMin = min[i - 1][j] + grid[i][j]; const tmpMax = max[i - 1][j] + grid[i][j]
-        if (tmpMax >= 0) min[i][j] = Math.min(min[i][j], Math.max(tmpMin, 0))
-        if (tmpMax >= 0) max[i][j] = Math.max(max[i][j], tmpMax)
+      if (i === 0 && j === 0) {
+        if (grid[0][0] === '(') dp[0][0].add(1)
+        continue
       }
-      // 从 [i, j-1] 过来
-      if (j >= 1) {
-        const tmpMin = min[i][j - 1] + grid[i][j]; const tmpMax = max[i][j - 1] + grid[i][j]
-        if (tmpMax >= 0) min[i][j] = Math.min(min[i][j], Math.max(tmpMin, 0))
-        if (tmpMax >= 0) max[i][j] = Math.max(max[i][j], tmpMax)
+
+      const s = new Set()
+      if (i - 1 >= 0) [...dp[i - 1][j]].forEach(x => s.add(x))
+      if (j - 1 >= 0) [...dp[i][j - 1]].forEach(x => s.add(x))
+
+      for (const c of s) {
+        const x = c + (grid[i][j] === '(' ? 1 : -1)
+        if (x >= 0) dp[i][j].add(x)
+        // console.log(i, j, dp[i][j])
       }
     }
   }
-  return min[m - 1][n - 1] === 0 && max[m - 1][n - 1] >= 0
+  return dp[m - 1][n - 1].has(0)
 }

leetcode/残酷刷题/1388. Pizza With 3n Slices.js

@@ -0,0 +1,27 @@
+/**
+ * @param {number[]} slices
+ * @return {number}
+ */
+const maxSizeSlices = function (s) {
+  const n = s.length
+  return Math.max(
+    // 第 0 个不选
+    solve(1, n - 1, n / 3),
+    // 最后一个不选
+    solve(0, n - 2, n / 3)
+  )
+  function solve (l, r, cnt) {
+    // dp[i][j][k] 到 i 为止，选了 j 个的最大价值,k=0表示最后一个不选
+    const dp = Array(r + 1).fill(0).map(() => Array(cnt + 1).fill(0).map(() => Array(2).fill(0)))
+    dp[l][1][1] = s[l]
+
+    for (let i = l + 1; i <= r; i++) {
+      for (let j = 1; j <= Math.min(cnt, i - l + 1); j++) {
+        dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1])
+        dp[i][j][1] = dp[i - 1][j - 1][0] + s[i]
+      }
+    }
+
+    return Math.max(dp[r][cnt][0], dp[r][cnt][1])
+  }
+}

leetcode/残酷刷题/198. House Robber.js

@@ -0,0 +1,14 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+ var rob = function(nums) {
+  const n = nums.length
+  const dp = Array(n).fill(0).map(() => [])
+  dp[0][0] = 0, dp[0][1] = nums[0]
+  for (let i = 1; i < n; i++) {
+      dp[i][0] = Math.max(dp[i-1][1], dp[i-1][0])
+      dp[i][1] = dp[i-1][0] + nums[i]
+  }
+  return Math.max(dp[n-1][0], dp[n-1][1])
+};

leetcode/残酷刷题/213. House Robber II.js

@@ -0,0 +1,21 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+ var rob = function(nums) {
+  if (nums.length === 1) return nums[0]
+  const n = nums.length
+  // dp[i][j][k] j 表示第 i 个有没有被 rob
+  // k 表示第 1 个有没有被 rob
+  const dp = Array(n).fill(0).map(() => Array(2).fill(0).map(() => Array(2).fill(0)))
+  dp[0][0][0] = 0, dp[0][0][1] = -1
+  dp[0][1][0] = -1, dp[0][1][1] = nums[0]
+  for (let i = 1; i < n; i++) {
+      dp[i][0][0] = Math.max(dp[i-1][1][0], dp[i-1][0][0])
+      dp[i][0][1] = Math.max(dp[i-1][1][1], dp[i-1][0][1])
+
+      dp[i][1][0] = Math.max(dp[i-1][0][0], 0) + nums[i]
+      dp[i][1][1] = Math.max(dp[i-1][0][1], 0) + nums[i]
+  }
+  return Math.max(dp[n-1][0][0], dp[n-1][0][1], dp[n-1][1][0])
+};

leetcode/残酷刷题/2318. Number of Distinct Roll Sequences.js

@@ -0,0 +1,34 @@
+/**
+ * @param {number} n
+ * @return {number}
+ */
+const distinctSequences = function (n) {
+  if (n === 1) return 6
+  // dp[i][j][k] 长为 i, 最后两个元素是 j, k
+  // dp[1][.][.] = 1
+  const dp = Array(n).fill(0).map(() => Array(7).fill(0).map(() => Array(7).fill(0)))
+  const gcd = (a, b) => b ? gcd(b, a % b) : a
+  for (let j = 1; j <= 6; j++) {
+    for (let k = 1; k <= 6; k++) {
+      if (j !== k && gcd(j, k) === 1) dp[1][j][k] = 1
+    }
+  }
+  let ans = 0; const P = 1e9 + 7
+  for (let i = 2; i < n; i++) {
+    for (let j = 1; j <= 6; j++) {
+      for (let k = 1; k <= 6; k++) {
+        for (let l = 1; l <= 6; l++) {
+          if (gcd(l, k) > 1) continue
+          if (l === k || l === j) continue
+          dp[i][k][l] = (dp[i][k][l] + dp[i - 1][j][k]) % P
+        }
+      }
+    }
+  }
+  for (let j = 1; j <= 6; j++) {
+    for (let k = 1; k <= 6; k++) {
+      ans = (ans + dp[n - 1][j][k]) % P
+    }
+  }
+  return ans
+}