feat: problems of next week

Author: upupming

leetcode/残酷刷题/2398. Maximum Number of Robots Within Budget.js

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+/**
+ * @param {number[]} chargeTimes
+ * @param {number[]} runningCosts
+ * @param {number} budget
+ * @return {number}
+ */
+const maximumRobots = function (c, r, b) {
+  const n = c.length; const q = []
+  let ans = 0
+  for (let i = 0, j = 0, sum = 0; i < n; i++) {
+    // [i, j)
+    while (j < n) {
+      if (j >= i) {
+        // 如果加上 j 会出问题，就停止
+        const max = Math.max(c[j], q.length ? c[q[0]] : 0)
+        if (max + (j - i + 1) * (sum + r[j]) > b) break
+
+        while (q.length && c[q[q.length - 1]] <= c[j]) q.pop()
+        q.push(j)
+      }
+      sum += r[j]
+
+      j++
+    }
+    // console.log(i, j, sum, q[0], q.length ? c[q[0]] : 0)
+    ans = Math.max(ans, j - i)
+
+    while (q.length && q[0] <= i) q.shift()
+    sum -= r[i]
+  }
+  return ans
+}

leetcode/残酷刷题/862. Shortest Subarray with Sum at Least K.js

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+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+const shortestSubarray = function (nums, k) {
+  const n = nums.length
+  const sum = [0]; const q = [0]; let ans = 1e10
+  for (let i = 1; i <= n; i++) {
+    sum[i] = sum[i - 1] + nums[i - 1]
+    // 尽量短
+    while (q.length && sum[i] - sum[q[1]] >= k) q.shift()
+    if (sum[i] - sum[q[0]] >= k) ans = Math.min(i - q[0], ans)
+    while (q.length && sum[q[q.length - 1]] >= sum[i]) q.pop()
+    q.push(i)
+  }
+  return ans === 1e10 ? -1 : ans
+}