Add LeetCode Weekly Contest 258

Author: EndlessCheng

leetcode/258/a/a.go

@@ -0,0 +1,19 @@
+package main
+
+import "strings"
+
+// 模拟
+
+// github.com/EndlessCheng/codeforces-go
+func reversePrefix(s string, ch byte) string {
+	j := strings.IndexByte(s, ch)
+	if j < 0 {
+		return s
+	}
+	t := []byte(s)
+	for i := 0; i < j; i++ {
+		t[i], t[j] = t[j], t[i]
+		j--
+	}
+	return string(t)
+}

leetcode/258/a/a_test.go

@@ -0,0 +1,30 @@
+package main
+
+/*
+哈希表统计比值的最简分数
+
+将宽高比 w/h 化简为最简分数，用哈希表统计最简分数的个数。若有 $m$ 个宽高比相同的矩形，则可互换的矩形对有 $C_m2=\dfrac{m\cdot(m-1}{2}$ 个。遍历哈希表，累加所有可互换的矩形对的个数即为答案。
+
+*/
+
+// github.com/EndlessCheng/codeforces-go
+func interchangeableRectangles(rectangles [][]int) (ans int64) {
+	cnt := map[[2]int]int64{}
+	for _, p := range rectangles {
+		// 计算 w/h 的最简分数，计入哈希表
+		w, h := p[0], p[1]
+		g := gcd(w, h)
+		cnt[[2]int{w / g, h / g}]++
+	}
+	for _, m := range cnt {
+		ans += m * (m - 1) / 2
+	}
+	return
+}
+
+func gcd(a, b int) int {
+	for a != 0 {
+		a, b = b%a, a
+	}
+	return b
+}

leetcode/258/b/b.go

@@ -0,0 +1,35 @@
+package main
+
+// Go 爆搜
+
+// github.com/EndlessCheng/codeforces-go
+func maxProduct(s string) (ans int) {
+	var a, b []byte
+	var f func(int)
+	f = func(i int) {
+		if i == len(s) {
+			if len(a)*len(b) > ans && isPalindromic(a) && isPalindromic(b) {
+				ans = len(a) * len(b)
+			}
+			return
+		}
+		f(i + 1)
+		a = append(a, s[i])
+		f(i + 1)
+		a = a[:len(a)-1]
+		b = append(b, s[i])
+		f(i + 1)
+		b = b[:len(b)-1]
+	}
+	f(0)
+	return
+}
+
+func isPalindromic(a []byte) bool {
+	for i, n := 0, len(a); i < n/2; i++ {
+		if a[i] != a[n-1-i] {
+			return false
+		}
+	}
+	return true
+}

leetcode/258/b/b_test.go

@@ -0,0 +1,36 @@
+package main
+
+// github.com/EndlessCheng/codeforces-go
+func smallestMissingValueSubtree(parents []int, a []int) []int {
+	n := len(parents)
+	g := make([][]int, n)
+	for w := 1; w < n; w++ {
+		v := parents[w]
+		g[v] = append(g[v], w)
+	}
+	mex := make([]int, n)
+	var f func(int) map[int]bool
+	f = func(v int) map[int]bool {
+		set := map[int]bool{}
+		mex[v] = 1
+		for _, w := range g[v] {
+			s := f(w)
+			if len(s) > len(set) {
+				set, s = s, set
+			}
+			for x := range s {
+				set[x] = true
+			}
+			if mex[w] > mex[v] {
+				mex[v] = mex[w]
+			}
+		}
+		set[a[v]] = true
+		for set[mex[v]] {
+			mex[v]++
+		}
+		return set
+	}
+	f(0)
+	return mex
+}