LeetCode in Kotlin
1269. Number of Ways to Stay in the Same Place After Some Steps
Hard
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 still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
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
Example 2:
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
Constraints:
1 <= steps <= 5001 <= arrLen <= 106
Solution
class Solution {
private var n = 0
private lateinit var dp: Array<IntArray>
private fun dfs(i: Int, st: Int): Int {
if (i < 0 || i >= n) {
return 0
}
if (st == 0) {
return if (i == 0) 1 else 0
}
if (dp[i][st] == -1) {
val mod = 1000000007
dp[i][st] = ((dfs(i + 1, st - 1) + dfs(i, st - 1)) % mod + dfs(i - 1, st - 1)) % mod
}
return dp[i][st]
}
fun numWays(steps: Int, arrLen: Int): Int {
n = Math.min(steps, arrLen)
dp = Array(n) { IntArray(steps + 1) }
for (i in 0 until n) {
dp[i].fill(-1)
}
dfs(0, steps)
return dp[0][steps]
}
}