feat: problems of next week

Author: upupming

leetcode/残酷刷题/1425. Constrained Subsequence Sum.js

@@ -0,0 +1,21 @@
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+const constrainedSubsetSum = function (nums, k) {
+  const n = nums.length
+  const q = [[-1e10, -1]]; const dp = Array(n + 1).fill(0); let ans = -1e10
+  for (let i = 0; i < n; i++) {
+    // 区间长度超过 k，去掉
+    while (q.length && i - q[0][1] > k) q.shift()
+
+    dp[i] = Math.max(0, q[0][0]) + nums[i]
+    ans = Math.max(ans, dp[i])
+
+    // 远的，还小就要去掉
+    while (q.length && q[q.length - 1][0] <= dp[i]) q.pop()
+    q.push([dp[i], i])
+  }
+  return ans
+}

leetcode/残酷刷题/1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.js

@@ -0,0 +1,23 @@
+/**
+ * @param {number[]} nums
+ * @param {number} limit
+ * @return {number}
+ */
+const longestSubarray = function (nums, limit) {
+  const n = nums.length; let ans = 0; const qMin = [[1e10, -1]]; const qMax = [[-1e10, -1]]
+  for (let i = 0, j = 0; i < n; i++) {
+    while ((qMin[0][0] === 1e10 || qMax[0][0] === -1e10 || Math.abs(qMax[0][0] - qMin[0][0]) <= limit)) {
+      ans = Math.max(ans, j - i)
+      if (j === n) break
+      while (qMax.length && qMax[qMax.length - 1][0] <= nums[j]) qMax.pop()
+      while (qMin.length && qMin[qMin.length - 1][0] >= nums[j]) qMin.pop()
+      qMax.push([nums[j], j])
+      qMin.push([nums[j], j])
+      j++
+    }
+
+    while (qMin.length && qMin[0][1] <= i) qMin.shift()
+    while (qMin.length && qMax[0][1] <= i) qMax.shift()
+  }
+  return ans
+}

leetcode/残酷刷题/1499. Max Value of Equation.js

@@ -0,0 +1,21 @@
+/**
+ * @param {number[][]} points
+ * @param {number} k
+ * @return {number}
+ */
+const findMaxValueOfEquation = function (points, k) {
+  const n = points.length
+  const q = []; let ans = -1e10
+  // xj-xi+yi+yj
+  for (let i = 0, j = 0; i < n; i++) {
+    while (j < n && points[j][0] - points[i][0] <= k) {
+      while (q.length && q[q.length - 1][0] <= points[j][0] + points[j][1]) q.pop()
+      q.push([points[j][0] + points[j][1], j])
+      j++
+    }
+    while (q.length && q[0][1] <= i) q.shift()
+
+    if (q.length) { ans = Math.max(ans, q[0][0] - points[i][0] + points[i][1]) }
+  }
+  return ans
+}

leetcode/残酷刷题/1562. Find Latest Group of Size M.js

@@ -0,0 +1,26 @@
+/**
+ * @param {number[]} arr
+ * @param {number} m
+ * @return {number}
+ */
+const findLatestStep = function (arr, m) {
+  const day = []; const n = arr.length; const q = []
+  let ans = -1
+  for (let i = 0; i < arr.length; i++) {
+    day[arr[i]] = i + 1
+  }
+  // 区间 [i-m+1, i] 被设置为 1，求区间中所有 1 的最大 day 值
+  for (let i = 1; i <= n; i++) {
+    while (q.length && i - q[0][1] >= m) q.shift()
+    while (q.length && q[q.length - 1][0] <= day[i]) q.pop()
+    q.push([day[i], i])
+
+    if (i < m) continue
+    const maxDay = q[0][0]
+    const left = day[i - m] ?? 1e10; const right = day[i + 1] ?? 1e10
+    if (left > maxDay && right > maxDay) {
+      ans = Math.max(ans, Math.min(left, right, n + 1) - 1)
+    }
+  }
+  return ans
+}

leetcode/残酷刷题/239. Sliding Window Maximum.js

@@ -0,0 +1,17 @@
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+const maxSlidingWindow = function (nums, k) {
+  const q = []; const ans = []
+  for (let i = 0; i < nums.length; i++) {
+    // 越远，又越小的数不需要保留
+    while (q.length && q[q.length - 1][0] < nums[i]) q.pop()
+    q.push([nums[i], i])
+    // 长度不能超过 k
+    while (q.length && i - q[0][1] >= k) q.shift()
+    if (i >= k - 1) ans.push(q[0][0])
+  }
+  return ans
+}

leetcode/残酷刷题/903. Valid Permutations for DI Sequence.js

@@ -0,0 +1,29 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+const numPermsDISequence = function (s) {
+  const n = s.length; const P = 1e9 + 7
+  s = '#' + s
+  const dp = Array(n + 1).fill(0).map(() => Array(n + 1).fill(0))
+  dp[0][0] = 1
+  // dp[i][j] 表示 0~i 的满足题目条件的排列组合，且最后一位为 j 的方案数目
+  for (let i = 1; i <= n; i++) {
+    for (let j = 0; j <= i; j++) {
+      if (s[i] === 'I') {
+        for (let k = 0; k < j; k++) {
+          dp[i][j] += dp[i - 1][k]
+          dp[i][j] %= P
+        }
+      } else {
+        for (let k = j; k <= i - 1; k++) {
+          dp[i][j] += dp[i - 1][k]
+          dp[i][j] %= P
+        }
+      }
+    }
+  }
+  let ans = 0
+  for (let i = 0; i <= n; i++) ans = (ans + dp[n][i]) % P
+  return ans
+}