1269. Number of Ways to Stay in the Same Place After Some Steps

LeetCode 1269. Number of Ways to Stay in the Same Place After Some Steps

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 still at index 0 after exactlysteps steps.

Since the answer may be too large, return it modulo 10^9 + 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

Constraints:

• 1 <= steps <= 500

• 1 <= arrLen <= 10^6

Tags

Dynamic Programming

Solution

We define dp[steps][idx] as the number of ways to back to index 0 with exactly steps moves. The number of rows is steps, yet that of columns is min(arrLen, steps/2+1). If a pointer from index 0 wants to go back with steps moves, it can only reach as far as index steps/2. We initialize dp[0][*] = 1. For each step i from 1 to steps, we have the transition function dp[i][j] = dp[i-1][j-1] + dp[i-1][j]+dp[i-1][j+1]. Finally, we return dp[steps][0]. We notice that the state of the current step is only related to its pervious step, thus, we can compress the 2D array into 1D with the length of min(arrLen, steps/2+1).

Complexity

• Time complexity: $O(steps×min(arrLen,steps))$

• Space complexity: $O(min(arrLen,steps))$

Code

func numWays(steps int, arrLen int) int {
	const MOD = 1e9 + 7
	minCols := arrLen
  // If a pointer from index 0 wants to go back with `steps` steps,
	// it can only reach as far as index steps/2.
	if c := steps/2 + 1; c < minCols {
		minCols = c
	}
	dp := make([]int, minCols)
	dp[0] = 1
	for i := 1; i <= steps; i++ {
		dpNext := make([]int, minCols)
		for j := 0; j < minCols; j++ {
			dpNext[j] = dp[j]
			if j-1 >= 0 {
				dpNext[j] = (dpNext[j] + dp[j-1]) % MOD
			}
			if j+1 < minCols {
				dpNext[j] = (dpNext[j] + dp[j+1]) % MOD
			}
		}
		dp = dpNext
	}
	return dp[0]
}

Reference

1. 停在原地的方案数