Add LeetCode Weekly Contest 259

Author: EndlessCheng

leetcode/weekly/259/a/a.go

@@ -0,0 +1,13 @@
+package main
+
+// github.com/EndlessCheng/codeforces-go
+func finalValueAfterOperations(operations []string) (ans int) {
+	for _, op := range operations {
+		if op[1] == '+' {
+			ans++
+		} else {
+			ans--
+		}
+	}
+	return
+}

leetcode/weekly/259/a/a_test.go

@@ -0,0 +1,33 @@
+package main
+
+// 前缀最大值+后缀最小值
+
+// github.com/EndlessCheng/codeforces-go
+func sumOfBeauties(a []int) (ans int) {
+	n := len(a)
+	sufMin := make([]int, n) // 后缀最小值
+	sufMin[n-1] = a[n-1]
+	for i := n - 2; i > 1; i-- {
+		sufMin[i] = min(sufMin[i+1], a[i])
+	}
+	preMax := a[0] // 前缀最大值
+	for i := 1; i < n-1; i++ {
+		v := a[i]
+		if preMax < v && v < sufMin[i+1] {
+			ans += 2
+		} else if a[i-1] < v && v < a[i+1] {
+			ans++
+		}
+		if v > preMax {
+			preMax = v
+		}
+	}
+	return
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}

leetcode/weekly/259/b/b.go

@@ -0,0 +1,40 @@
+package main
+
+// github.com/EndlessCheng/codeforces-go
+type DetectSquares struct{}
+
+var row [1001][]int
+var col = [1001]map[int]int{}
+
+func Constructor() (_ DetectSquares) {
+	row = [1001][]int{}
+	for i := range col {
+		col[i] = map[int]int{}
+	}
+	return
+}
+
+func (DetectSquares) Add(point []int) {
+	x, y := point[0], point[1]
+	row[y] = append(row[y], x)
+	col[x][y]++
+}
+
+func (DetectSquares) Count(point []int) (ans int) {
+	x, y := point[0], point[1]
+	for _, x2 := range row[y] {
+		if x2 != x {
+			d := abs(x2 - x) // 横向距离
+			ans += col[x][y-d] * col[x2][y-d] // 将 x-x2 当作正方形顶边两个点，计算正方形底边两点的组合方案
+			ans += col[x][y+d] * col[x2][y+d] // 将 x-x2 当作正方形底边两个点，计算正方形顶边两点的组合方案
+		}
+	}
+	return
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}

leetcode/weekly/259/b/b_test.go

@@ -0,0 +1,99 @@
+package main
+
+/*
+
+根据题意，答案长度不会超过 $\lceil\dfrac{n}{k}\rceil$，而 $n<8k$，故答案长度不超过 $7$。
+
+我们可以统计出 $s$ 中个数不低于 $k$ 的字符，答案只能由这些字符注册，而这些字符的个数不会超过 $\lceil\dfrac{n}{k}\rceil$，由于 $n<8k$，故字符个数不超过 $7$。
+
+这些然后暴力枚举这些字符的某个排列，然后贪心地匹配子序列，从中找到符合题目要求的子序列。
+
+*/
+
+// github.com/EndlessCheng/codeforces-go
+func longestSubsequenceRepeatedK(s string, k int) (ans string) {
+	n := len(s)
+	right := [26]int{}
+	for i := range right {
+		right[i] = n
+	}
+	nxt := make([][26]int, n)
+	cnt := [26]int{}
+	for i := n - 1; i >= 0; i-- {
+		nxt[i] = right
+		right[s[i]-'a'] = i
+		cnt[s[i]-'a']++
+	}
+
+	// 计算所有可能出现在 ans 中的字符，包括重复的
+	// 倒着统计，这样下面计算排列时的第一个合法方案就是答案，从而提前退出
+	a := []byte{}
+	for i := 25; i >= 0; i-- {
+		for c := cnt[i]; c >= k; c -= k {
+			a = append(a, 'a'+byte(i))
+		}
+	}
+
+	for m := len(a); m > 0 && ans == ""; m-- {
+		permutations(len(a), m, func(ids []int) bool {
+			t := make([]byte, m)
+			for i, id := range ids {
+				t[i] = a[id]
+			}
+			i, j := 0, 0
+			if t[0] == s[0] {
+				j = 1
+			}
+			for {
+				i = nxt[i][t[j%m]-'a']
+				if i == n {
+					break
+				}
+				j++
+			}
+			if j >= k*len(t) {
+				ans = string(t)
+				return true
+			}
+			return false
+		})
+	}
+	return
+}
+
+// 生成 n 选 r 的排列
+func permutations(n, r int, do func(ids []int) bool) {
+	ids := make([]int, n)
+	for i := range ids {
+		ids[i] = i
+	}
+	if do(ids[:r]) {
+		return
+	}
+	cycles := make([]int, r)
+	for i := range cycles {
+		cycles[i] = n - i
+	}
+	for {
+		i := r - 1
+		for ; i >= 0; i-- {
+			cycles[i]--
+			if cycles[i] == 0 {
+				tmp := ids[i]
+				copy(ids[i:], ids[i+1:])
+				ids[n-1] = tmp
+				cycles[i] = n - i
+			} else {
+				j := cycles[i]
+				ids[i], ids[n-j] = ids[n-j], ids[i]
+				if do(ids[:r]) {
+					return
+				}
+				break
+			}
+		}
+		if i == -1 {
+			return
+		}
+	}
+}