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

Problem Description

In this problem, we are given an array of size arrLen, and we start with a pointer at index 0. We can take a total of steps number of moves, and at each move, we have three choices:

• Move the pointer to the left (if the pointer is not at the leftmost position),
• Move the pointer to the right (if not at the rightmost position),
• Or don't move the pointer at all.

The goal is to calculate the number of unique ways we can make these moves so that after exactly steps steps, our pointer is returned to the starting position at index 0. Because the answer can potentially be very large, we will return the result modulo 10^9 + 7, which is a common technique to prevent integer overflow in programming contests.

Intuition

The problem is a perfect candidate for dynamic programming because it deals with counting ways and optimizing over a range of choices at each step.

Intuitively, we can approach this using a depth-first search (DFS) strategy that explores all possible movements from a given position at a step. However, a naive DFS solution would be too slow because it involves a lot of repeated calculations. Therefore, we can apply memoization to cache the results of subproblems and avoid redundant computations.

The DFS function dfs(i, j) keeps track of the current index i and the remaining steps j. Here are the key points:

• If the pointer is outside the array bounds (i < 0 or i >= arrLen) or if the number of steps remaining is not sufficient to return to the start (i > j), there are no valid ways, so we return 0.
• If the pointer is at the starting position and no more steps remain (i == 0 and j == 0), we've found a valid way, so we return 1.
• Otherwise, we recursively call dfs for all potential moves: stay in place, move left, or move right. We do this by decreasing the number of remaining steps j by one and either keeping the index i the same or adjusting it by -1 or 1 depending on the move.

We sum the results from the three recursive calls, apply the modulo operation to keep the number within the bounds, and then return the sum as the answer from the current state. The @cache decorator in Python ensures that once a subproblem has been solved, its result is stored and can be reused in subsequent calls without recalculating.

Our main function numWays initializes the modulo value and calls the helper function dfs starting from the first position with all steps available, and returns the total count of unique ways.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution makes use of recursive depth-first search (DFS) combined with memoization, which is a form of dynamic programming where we remember the results of the subproblems we have already solved.

DFS and Recursion

The dfs(i, j) function is the heart of our solution. It recursively searches through all potential moves starting from a given index i with j steps remaining. The recursion allows the function to branch out into every possible move scenario. When we reach the base cases — either stepping out of the array bounds or reaching index 0 with no steps left — the function stops and returns either 0 (not a valid scenario) or 1 (a valid scenario), respectively.

Memoization with Caching

To optimize the recursive calls, we use the @cache decorator from the Python functools module. This automatically stores any results returned by dfs(i, j) so that if the function is called again with the same arguments i and j, it does not compute everything from scratch but instead returns the cached result. This greatly reduces the number of calculations and thus speeds up the execution.

Modulo Operation

The mod variable is set to 10**9 + 7, a large prime number, and it's used to perform modulo operations to keep the numbers we work with within the boundaries of integer limits. This is a common approach in problems involving counting, where the numbers can grow exponentially. In our code, after updating the ans (the number of ways to return to index 0), we take ans % mod to ensure that we only keep the remainder and avoid integer overflow.

Execution

The helper function dfs is called from the numWays method with parameters 0 and steps, which signify that we start at index 0 with all steps remaining. By exploring all possibilities and pruning unnecessary ones using memoization, we are able to calculate the exact number of ways to return to the start after the stipulated number of steps.

1class Solution:
2    def numWays(self, steps: int, arrLen: int) -> int:
3        @cache
4        def dfs(i, j):
5            if i > j or i >= arrLen or i < 0 or j < 0:
6                return 0
7            if i == 0 and j == 0:
8                return 1
9            ans = 0
10            for k in range(-1, 2):
11                ans += dfs(i + k, j - 1)
12                ans %= mod
13            return ans
14
15        mod = 10**9 + 7
16        return dfs(0, steps)

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have arrLen = 4 and steps = 3. This means we have an array [0, 1, 2, 3] (indices of the array) and we can make three moves. We want to find out in how many unique ways we can return to index 0.

1. We start at (index = 0, steps = 3) and call dfs(0, 3).
2. From here, we can move to index 1, stay at index 0, or try to move to the left, which isn't possible since we're at the leftmost position.
3. When we call dfs(1, 2) (move to the right):
   • We can move back to index 0, go further to index 2 (which isn't wise because we don't have enough steps to come back), or stay at index 1.
   • The recursive call dfs(0, 1) from this position will lead us back to the starting index with one step remaining. Here, we can only stay in place for the last step.
   • Hence, from the move to index 1 and back, we get one unique way to return to the starting point.
4. When we call dfs(0, 2) (stay in place):
   • We are again presented with the choice to stay, move right to index 1, or move left (not possible).
   • If we stay and call dfs(0, 1), our next step can only be to stay again at index 0, which gives us another unique way.
   • If we move to index 1 and call dfs(1, 1), the only option is to move back to index 0, which we can do.
   • Thus, staying in place on the first move gives us two more unique ways.
5. We wouldn't consider moving left from the starting position since it is not possible.
6. Each call to dfs saves its result, so any overlapping subproblems don't require re-computation.
7. After exploring all possibilities, we find that there are 1 + 2 = 3 unique ways to return to index 0 in 3 steps for an arrLen of 4.

Putting this into the context of our algorithm, the recursive calls and the memoization ensure that we efficiently count all unique ways with the DFS approach, considering the implications of each possible move at every step. The modulo operation is applied to each update to the number of ways, which keeps our numbers within the bounds of typical integer values.

Solution Implementation

Time and Space Complexity

The given code defines a dynamic programming function dfs(i, j) where i is the current position and j is the remaining number of steps. The function is memoized using Python's cache decorator, meaning previously computed results for certain (i, j) pairs will be stored and reused.

Time Complexity

The time complexity depends on the number of unique recursive calls to dfs(i, j). Let min(steps, arrLen) be m which represents the maximum possible unique positions we can be on the path.

Given that from each position i, we can move to i-1, i, or i+1 for the next step, the recursion creates a ternary tree of calls. However, the use of memoization cuts down the number of unique calls to the actual number of different states the problem can be in.

There are at most m different positions and at each position, we can have at most steps different j values (number of steps remaining). Therefore, the number of unique states is O(m * steps).

For each state (i, j), we consider 3 possible next states by iterating over k = -1, 0, 1, each constant time operations. Hence, the total time complexity is:

O(m * steps * 3) which simplifies to O(m * steps).

Since m = min(steps, arrLen), the time complexity becomes O(min(steps, arrLen) * steps).

Space Complexity

The space complexity consists of the space used by the call stack during recursion and the space used to store the memoized results.

1. Call Stack: In the worst-case scenario, the maximum depth of the recursive call stack is equal to the number of steps steps, because we can only move steps times before we run out of moves.

2. Memoization Dictionary: There are O(m * steps) unique states being stored in the dictionary, where m is min(steps, arrLen) as discussed earlier.

Therefore, the overall space complexity, accounting for both the call stack and the memoization storage, is O(m * steps + steps) which simplifies to O(m * steps) where m = min(steps, arrLen).

Putting it all together, the space complexity is also O(min(steps, arrLen) * steps).

Learn more about how to find time and space complexity quickly using problem constraints.