Merge pull request #662 from 0xff-dev/1269

Add solution and test-cases for problem 1269

leetcode/1201-1300/1269.Number-of-Ways-to-Stay-in-the-Same-Place-After-Some-Steps/README.md

@@ -1,28 +1,38 @@
 # [1269.Number of Ways to Stay in the Same Place After Some Steps][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 have a pointer at index `0` in an array of size `arrLen`. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).
+
+Given two integers `steps` and `arrLen`, return the number of ways such that your pointer is still at index `0` after **exactly** steps steps. Since the answer may be too large, return it **modulo** `10^9 + 7`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: steps = 3, arrLen = 2
+Output: 4
+Explanation: There are 4 differents ways to stay at index 0 after 3 steps.
+Right, Left, Stay
+Stay, Right, Left
+Right, Stay, Left
+Stay, Stay, Stay
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Number of Ways to Stay in the Same Place After Some Steps
-```go
+```
+Input: steps = 2, arrLen = 4
+Output: 2
+Explanation: There are 2 differents ways to stay at index 0 after 2 steps
+Right, Left
+Stay, Stay
 ```
 
+**Example 3:**
+
+```
+Input: steps = 4, arrLen = 2
+Output: 8
+```
 
 ## 结语
 

leetcode/1201-1300/1269.Number-of-Ways-to-Stay-in-the-Same-Place-After-Some-Steps/Solution.go

@@ -1,5 +1,43 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+const mod1269 = 1000000007
+
+func Solution(steps int, arrLen int) int {
+	dp := make([][]int, arrLen)
+	if steps <= arrLen {
+		arrLen = steps
+	}
+	for i := 0; i < arrLen; i++ {
+		dp[i] = make([]int, steps+1)
+	}
+	for i := 0; i < arrLen; i++ {
+		for j := 0; j <= steps; j++ {
+			dp[i][j] = -1
+		}
+	}
+	var dirs = []int{-1, 0, 1}
+	var dfs func(int, int) int
+	dfs = func(index, useSteps int) int {
+		if useSteps == 0 {
+			if index == 0 {
+				return 1
+			}
+			return 0
+		}
+		if index >= arrLen {
+			return 0
+		}
+		if dp[index][useSteps] != -1 {
+			return dp[index][useSteps]
+		}
+		ans := 0
+		for _, dir := range dirs {
+			if ni := index + dir; ni >= 0 {
+				ans = (ans + dfs(ni, useSteps-1)) % mod1269
+			}
+		}
+		dp[index][useSteps] = ans
+		return ans
+	}
+	return dfs(0, steps)
 }

leetcode/1201-1300/1269.Number-of-Ways-to-Stay-in-the-Same-Place-After-Some-Steps/Solution_test.go

@@ -9,31 +9,32 @@ import (
 func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
-		name   string
-		inputs bool
-		expect bool
+		name          string
+		steps, arrLen int
+		expect        int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", 3, 2, 4},
+		{"TestCase2", 2, 4, 2},
+		{"TestCase3", 4, 2, 8},
+		{"TestCase4", 430, 148488, 525833932},
 	}
 
 	//	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.steps, c.arrLen)
 			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.steps, c.arrLen)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }