1269 - Number of Ways to Stay in the Same Place After Some Steps (Hard)

Problem Link

https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/

Problem Statement

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 <= 500
• 1 <= arrLen <= 10^6

Approach 1: Dynamic Programming

The first observation is that the computational complexity does not depend on $arrLen$. Instead, it is all about steps. If we have $n$ steps, we can only walk at most $n / 2$ steps to the left or the right. Therefore, we can use DFS with memoization to find out the answer.

Written by @wingkwong

class Solution {
public:
    int numWays(int steps, int arrLen) {
        int M = 1e9 + 7;
        // dp[i][j]: how many ways to reach i-th pos using j steps
        vector<vector<int>> dp(steps / 2 + 1, vector<int>(steps + 1, -1));
        function<long long(int,int)> dfs = [&](int pos, int steps) -> long long {
            // if we walk outside the array or use all the steps
            // then return 0
            if (pos < 0 || pos > arrLen - 1 || pos > steps) return 0;
            // if we use all the steps, return 1 only if pos is 0
            if (steps == 0) return pos == 0;
            // if it has been calculated, return directly
            if (dp[pos][steps] != -1) return dp[pos][steps];
            // memoize it
            return dp[pos][steps] = (
                // move to the left
                dfs(pos - 1, steps - 1) % M + 
                // stay at current position
                dfs(pos, steps - 1) % M + 
                // move to the right
                dfs(pos + 1, steps - 1) % M
            ) % M;
        };
        return dfs(0, steps);
    }
};

Written by @wingkwong

class Solution:
    def numWays(self, steps: int, arrLen: int) -> int:
        M = 10 ** 9 + 7
        @lru_cache(None)
        def dfs(pos, steps):
            # if we walk outside the array or use all the steps
            # then return 0
            if pos < 0 or pos > steps or pos > arrLen - 1: return 0
            # if we use all the steps, return 1 only if pos is 0
            if steps == 0: return pos == 0
            return (
                # move to the left
                dfs(pos - 1, steps - 1) +
                # stay at current position
                dfs(pos, steps - 1) +
                # move to the right
                dfs(pos + 1, steps - 1) 
            ) % M
        return dfs(0, steps)