Merge pull request #1071 from 0xff-dev/2940

Add solution and test-cases for problem 2940

Author: 6boris

leetcode/2901-3000/2940.Find-Building-Where-Alice-and-Bob-Can-Meet/README.md

@@ -1,28 +1,41 @@
 # [2940.Find Building Where Alice and Bob Can Meet][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+You are given a **0-indexed** array `heights` of positive integers, where `heights[i]` represents the height of the `ith` building.
+
+If a person is in building `i`, they can move to any other building `j` if and only if `i < j` and `heights[i] < heights[j]`.
+
+You are also given another array `queries` where `queries[i] = [ai, bi]`. On the `ith` query, Alice is in building `ai` while Bob is in building `bi`.
+
+Return an array `ans` where `ans[i]` is **the index of the leftmost building** where Alice and Bob can meet on the `ith` query. If Alice and Bob cannot move to a common building on query `i`, set `ans[i]` to `-1`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
+Output: [2,5,-1,5,2]
+Explanation: In the first query, Alice and Bob can move to building 2 since heights[0] < heights[2] and heights[1] < heights[2]. 
+In the second query, Alice and Bob can move to building 5 since heights[0] < heights[5] and heights[3] < heights[5]. 
+In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
+In the fourth query, Alice and Bob can move to building 5 since heights[3] < heights[5] and heights[4] < heights[5].
+In the fifth query, Alice and Bob are already in the same building.  
+For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
+For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Find Building Where Alice and Bob Can Meet
-```go
 ```
-
+Input: heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
+Output: [7,6,-1,4,6]
+Explanation: In the first query, Alice can directly move to Bob's building since heights[0] < heights[7].
+In the second query, Alice and Bob can move to building 6 since heights[3] < heights[6] and heights[5] < heights[6].
+In the third query, Alice cannot meet Bob since Bob cannot move to any other building.
+In the fourth query, Alice and Bob can move to building 4 since heights[3] < heights[4] and heights[0] < heights[4].
+In the fifth query, Alice can directly move to Bob's building since heights[1] < heights[6].
+For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
+For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
+```
 
 ## 结语
 

leetcode/2901-3000/2940.Find-Building-Where-Alice-and-Bob-Can-Meet/Solution.go

@@ -1,5 +1,54 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import "sort"
+
+func Solution(heights []int, queries [][]int) []int {
+	// monotonic stack + sort + binarysearch
+	ll := len(queries)
+	indies := make([]int, ll)
+	for i := range ll {
+		if queries[i][0] > queries[i][1] {
+			queries[i][0], queries[i][1] = queries[i][1], queries[i][0]
+		}
+		indies[i] = i
+	}
+	sort.Slice(indies, func(i, j int) bool {
+		return queries[indies[i]][1] > queries[indies[j]][1]
+	})
+
+	ans := make([]int, ll)
+	for i := range ll {
+		ans[i] = -1
+	}
+
+	l := len(heights)
+	stack := make([]int, l)
+	stackIndex := l - 1
+	stack[stackIndex] = l - 1
+	cur := l - 2
+
+	for _, i := range indies {
+		a, b := queries[i][0], queries[i][1]
+		if a == b || heights[a] < heights[b] {
+			ans[i] = b
+			continue
+		}
+		if b == l-1 {
+			continue
+		}
+		target := max(heights[a], heights[b])
+		for ; cur > b; cur-- {
+			for ; stackIndex != l && heights[cur] >= heights[stack[stackIndex]]; stackIndex++ {
+			}
+			stackIndex--
+			stack[stackIndex] = cur
+		}
+		if idx := sort.Search(l-stackIndex, func(i int) bool {
+			return heights[stack[i+stackIndex]] > target
+		}); idx != l-stackIndex {
+			ans[i] = stack[idx+stackIndex]
+		}
+	}
+
+	return ans
 }

leetcode/2901-3000/2940.Find-Building-Where-Alice-and-Bob-Can-Meet/Solution_test.go

@@ -9,31 +9,31 @@ import (
 func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
-		name   string
-		inputs bool
-		expect bool
+		name    string
+		heights []int
+		queries [][]int
+		expect  []int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{6, 4, 8, 5, 2, 7}, [][]int{{0, 1}, {0, 3}, {2, 4}, {3, 4}, {2, 2}}, []int{2, 5, -1, 5, 2}},
+		{"TestCase2", []int{5, 3, 8, 2, 6, 1, 4, 6}, [][]int{{0, 7}, {3, 5}, {5, 2}, {3, 0}, {1, 6}}, []int{7, 6, -1, 4, 6}},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.heights, c.queries)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.heights, c.queries)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }