feat: finished problems for next week

Author: upupming

leetcode/残酷刷题/1649. Create Sorted Array through Instructions.js

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+/**
+ * @param {number[]} instructions
+ * @return {number}
+ */
+const createSortedArray = function (instructions) {
+  const n = instructions.length
+  instructions = instructions.reverse().map((a, idx) => [a, idx])
+  const a = Array(n).fill(0); const b = Array(n).fill(0)
+  const count = (i, j) => {
+    if (i === j) return
+    const t = i + j >> 1
+    count(i, t)
+    count(t + 1, j)
+    for (let x = i; x <= t; x++) {
+      const p = instructions[x][0]; const q = instructions[x][0]
+      // >p 的第一个数
+      let l = t + 1; let r = j + 1
+      while (l < r) {
+        const mid = l + r >> 1
+        if (mid === j + 1 || instructions[mid][0] > p) r = mid
+        else l = mid + 1
+      }
+      const p1 = l
+      // < q 的最后一个数
+      l = t, r = j
+      while (l < r) {
+        const mid = l + r + 1 >> 1
+        if (mid === t || instructions[mid][0] < q) l = mid
+        else r = mid - 1
+      }
+      const p2 = l
+      const idx = instructions[x][1]
+      a[idx] += j - p1 + 1
+      b[idx] += p2 - t
+    }
+
+    let x = i; let y = t + 1; const tmp = []
+    while (x <= t && y <= j) {
+      if (instructions[x][0] <= instructions[y][0]) tmp.push(instructions[x++])
+      else tmp.push(instructions[y++])
+    }
+    while (x <= t) tmp.push(instructions[x++])
+    while (y <= j) tmp.push(instructions[y++])
+    for (let k = i; k <= j; k++) instructions[k] = tmp[k - i]
+  }
+  count(0, n - 1)
+
+  let ans = 0
+  for (let i = 0; i < n; i++) {
+    ans += Math.min(a[i], b[i])
+    ans %= 1e9 + 7
+  }
+
+  return ans
+}

leetcode/残酷刷题/1697. Checking Existence of Edge Length Limited Paths.js

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+/**
+ * @param {number} n
+ * @param {number[][]} edgeList
+ * @param {number[][]} queries
+ * @return {boolean[]}
+ */
+const distanceLimitedPathsExist = function (n, edgeList, queries) {
+  const fa = Array(n).fill(0).map((_, idx) => idx)
+  edgeList.sort((a, b) => a[2] - b[2])
+  const ans = []; let i = 0; const m = edgeList.length
+  const get = (x) => {
+    if (x === fa[x]) return x
+    return fa[x] = get(fa[x])
+  }
+  const merge = (x, y) => {
+    const [fx, fy] = [get(x), get(y)]
+    fa[fx] = fy
+  }
+  queries.map((q, idx) => [...q, idx]).sort((a, b) => a[2] - b[2])
+    .forEach(([x, y, d, idx]) => {
+      while (i < m && edgeList[i][2] < d) {
+        merge(edgeList[i][0], edgeList[i][1])
+        i++
+      }
+      ans[idx] = get(x) === get(y)
+    })
+  return ans
+}

leetcode/残酷刷题/327. Count of Range Sum.js

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+/**
+ * @param {number[]} nums
+ * @param {number} lower
+ * @param {number} upper
+ * @return {number}
+ */
+ var countRangeSum = function(nums, lower, upper) {
+  const sum = [0], n = nums.length
+  for (let i = 0; i < n; i++) sum[i+1] = sum[i]+nums[i]
+  let ans = 0
+
+  const count = (i, j) => {
+      if (i === j) return
+      const t = i+j>>1
+      count(i, t)
+      count(t+1, j)
+      for (let x = i; x <= t; x++) {
+          let p = sum[x]+lower, q = sum[x]+upper
+          // >=p 的第一个数
+          let l = t+1, r = j+1
+          while (l < r) {
+              const mid = l+r>>1
+              if (mid === j+1 || sum[mid] >= p) r = mid
+              else l = mid+1
+          }
+          let p1 = l
+          // <= q 的最后一个数
+          l = t, r = j
+          while (l < r) {
+              const mid = l+r+1>>1
+              if (mid === t || sum[mid] <= q) l = mid
+              else r = mid-1
+          }
+          let p2 = l
+          // console.log(sum, i, j, x, p1, p2)
+          ans += p2-p1+1
+      }
+
+      let x = i, y = t+1, tmp = []
+      while (x <= t && y <= j) {
+          if (sum[x] <= sum[y]) tmp.push(sum[x++])
+          else tmp.push(sum[y++])
+      }
+      while (x <= t) tmp.push(sum[x++])
+      while (y <= j) tmp.push(sum[y++])
+      for (let k = i; k <= j; k++) sum[k] = tmp[k-i]
+  }
+  count(0, n)
+
+  return ans
+};

leetcode/残酷刷题/493. Reverse Pairs.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const reversePairs = function (nums) {
+  const n = nums.length
+  let ans = 0
+  const count = (i, j) => {
+    if (i === j) return
+    const t = i + j >> 1
+    count(i, t)
+    count(t + 1, j)
+    for (let x = i; x <= t; x++) {
+      const q = nums[x] / 2
+      // < q 的最后一个数
+      l = t, r = j
+      while (l < r) {
+        const mid = l + r + 1 >> 1
+        if (mid === t || nums[mid] < q) l = mid
+        else r = mid - 1
+      }
+      ans += l - t
+    }
+
+    let x = i; let y = t + 1; const tmp = []
+    while (x <= t && y <= j) {
+      if (nums[x] <= nums[y]) tmp.push(nums[x++])
+      else tmp.push(nums[y++])
+    }
+    while (x <= t) tmp.push(nums[x++])
+    while (y <= j) tmp.push(nums[y++])
+    for (let k = i; k <= j; k++) nums[k] = tmp[k - i]
+  }
+  count(0, n - 1)
+  return ans
+}