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 is 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 <= 500
• 1 <= arrLen <= 106

Code

Top Down DP with Memoization

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

class Solution {
    long mod = 1e9 + 7;
public:
    int numWays(int steps, int arrLen) {
        vector<vector<long>> dp(steps / 2 + 1, vector<long>(steps + 1, -1));
        return dfs(steps, 0, arrLen, dp);
    }
 
    int dfs(int step, int i, int n, vector<vector<long>>& dp) {
        if(i >= n || i < 0 || i > step) return 0;
        if(step == 0) {
            return i == 0;
        }
        
        if(dp[i][step] != -1) return dp[i][step];
 
        dp[i][step] = (((dfs(step - 1, i + 1, n, dp) % mod) + 
        dfs(step - 1, i - 1, n, dp) % mod) + dfs(step - 1, i, n, dp)) % mod;
 
        return (int) dp[i][step];
    }
};

Source

• Number of Ways to Stay in the Same Place After Some Steps - LeetCode