Solution Explanation: Number of Ways to Stay in the Same Place After Some Steps

This problem asks to find the number of ways to stay at index 0 after a given number of steps, moving left, right, or staying in place within an array. A recursive approach with memoization (dynamic programming) is efficient for this.

Approach: Memoized Depth-First Search (DFS)

The core idea is to use a recursive function dfs(i, j) that represents the number of ways to reach index i with j steps remaining.

• Base Cases:

  • If i is out of bounds (less than 0 or greater than or equal to arrLen), or j is negative, it's impossible to reach index i with j steps, so return 0.
  • If i is 0 and j is 0, we've reached the goal (index 0 after all steps), so return 1 (one successful way).

• Recursive Step: For a given dfs(i, j):

  • We can reach i from three positions in the previous step: i-1, i, and i+1.
  • Recursively calculate the number of ways to reach each of these positions (dfs(i-1, j-1), dfs(i, j-1), dfs(i+1, j-1)).
  • Sum these counts, take the modulo with 10^9 + 7 (to handle large numbers), and return the result.

• Memoization: To avoid redundant calculations, we store the results of dfs(i, j) calls in a memoization table (a 2D array f in the code). If dfs(i, j) is already calculated, we return the stored value directly, significantly improving performance.

Time and Space Complexity Analysis

• Time Complexity: O(steps * arrLen) in the worst case. The recursive function might explore all possible combinations of steps and positions, within the constraints of steps and arrLen. However, because of memoization, many subproblems are solved only once. In practice, it's often much less than O(steps * arrLen) because of the overlap in recursive calls.

• Space Complexity: O(steps * arrLen) due to the memoization table. This space is used to store the results of solved subproblems. The recursive call stack also contributes to space complexity, but its depth is at most the number of steps.

Code Examples (Multiple Languages)

The code examples below demonstrate the memoized DFS approach in various programming languages. They all follow the same algorithmic logic described above.

Python:

class Solution:
    def numWays(self, steps: int, arrLen: int) -> int:
        mod = 10**9 + 7
        dp = {}  # Memoization dictionary
 
        def dfs(i, steps_left):
            if (i, steps_left) in dp:
                return dp[(i, steps_left)]
            if steps_left == 0:
                return 1 if i == 0 else 0
            if i < 0 or i >= arrLen:
                return 0
            res = (dfs(i - 1, steps_left - 1) + dfs(i, steps_left - 1) + dfs(i + 1, steps_left - 1)) % mod
            dp[(i, steps_left)] = res
            return res
 
        return dfs(0, steps)
 

Java:

class Solution {
    private Integer[][] memo;
    private int arrLen;
    private int mod = (int) 1e9 + 7;
 
    public int numWays(int steps, int arrLen) {
        this.arrLen = arrLen;
        memo = new Integer[steps + 1][arrLen];
        return dfs(0, steps);
    }
 
    private int dfs(int pos, int stepsLeft) {
        if (stepsLeft == 0) return pos == 0 ? 1 : 0;
        if (pos < 0 || pos >= arrLen) return 0;
        if (memo[stepsLeft][pos] != null) return memo[stepsLeft][pos];
        int ans = (dfs(pos - 1, stepsLeft - 1) + dfs(pos, stepsLeft - 1) + dfs(pos + 1, stepsLeft - 1)) % mod;
        memo[stepsLeft][pos] = ans;
        return ans;
    }
}

C++:

class Solution {
public:
    int numWays(int steps, int arrLen) {
        int mod = 1e9 + 7;
        vector<vector<int>> memo(steps + 1, vector<int>(arrLen, -1));
 
        function<int(int, int)> dfs = [&](int pos, int stepsLeft) {
            if (stepsLeft == 0) return pos == 0 ? 1 : 0;
            if (pos < 0 || pos >= arrLen) return 0;
            if (memo[stepsLeft][pos] != -1) return memo[stepsLeft][pos];
            int ans = (dfs(pos - 1, stepsLeft - 1) + dfs(pos, stepsLeft - 1) + dfs(pos + 1, stepsLeft - 1)) % mod;
            memo[stepsLeft][pos] = ans;
            return ans;
        };
 
        return dfs(0, steps);
    }
};

(Other languages like JavaScript, Go, etc., would have similar structures, utilizing a memoization technique to store and reuse the results of subproblems.)

This detailed explanation, along with the diverse code samples, should provide a comprehensive understanding of how to solve the "Number of Ways to Stay in the Same Place After Some Steps" problem efficiently.