Problem Description

You start with a pointer at index 0 in an array of size arrLen. At each step, you can perform one of three actions:

• Move the pointer 1 position to the left
• Move the pointer 1 position to the right
• Stay in the same position

The pointer must always remain within the array bounds (cannot go to negative indices or indices >= arrLen).

Given two integers:

• steps: the exact number of steps you must take
• arrLen: the size of the array

Your task is to find the number of different ways to move the pointer such that after exactly steps steps, the pointer returns to index 0.

Since the answer can be very large, return it modulo 10^9 + 7.

For example, if steps = 3 and arrLen = 2, you could:

1. Stay at 0, stay at 0, stay at 0 (one way)
2. Go right to 1, go left to 0, stay at 0 (another way)
3. Stay at 0, go right to 1, go left to 0 (another way)
4. Go right to 1, stay at 1, go left to 0 (another way)

The solution uses dynamic programming with memoization. The function dfs(i, j) represents the number of ways to reach index 0 when currently at position i with j steps remaining. For each position and remaining steps, we consider all three possible moves (left, right, stay) and sum up the ways from those subsequent states.

Intuition

The key insight is that this is a path counting problem where we need to return to the starting position after a fixed number of steps. This naturally suggests a dynamic programming approach.

First, let's think about the constraints. With steps limited to 500, even if we move right at every step, we can only reach position 500. This means we don't need to consider positions beyond min(arrLen, steps) - any position further than steps away from 0 is unreachable since we wouldn't have enough steps to get there and return.

The problem has optimal substructure: the number of ways to reach position 0 in j steps from position i depends on the number of ways to reach position 0 in j-1 steps from positions i-1, i, and i+1 (corresponding to moving right, staying, or moving left in the current step).

We can think of this recursively: if we're at position i with j steps remaining, we can:

• Move left to position i-1 and solve for j-1 remaining steps
• Stay at position i and solve for j-1 remaining steps
• Move right to position i+1 and solve for j-1 remaining steps

The base cases are:

• If we're at position 0 with 0 steps remaining, we've found one valid way (return 1)
• If we go out of bounds or run out of steps at a non-zero position, this path is invalid (return 0)
• If the position is farther from 0 than the remaining steps, it's impossible to return (return 0)

Since the same state (position, remaining_steps) can be reached through different paths, we use memoization to avoid recalculating the same subproblems repeatedly. This transforms what would be an exponential time solution into a polynomial one.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization using Python's @cache decorator.

Function Definition: The core function dfs(i, j) represents the number of ways to return to position 0 when currently at position i with j steps remaining.

Base Cases:

1. Invalid states: If i > j (position is farther than remaining steps can reach), or i >= arrLen (out of right boundary), or i < 0 (out of left boundary), or j < 0 (negative steps), return 0.
2. Success state: If i == 0 and j == 0, we've successfully returned to position 0 with exactly the required steps, return 1.

Recursive Transition: For each valid state (i, j), we calculate the sum of ways from three possible moves:

ans = 0
for k in range(-1, 2):  # k can be -1, 0, or 1
    ans += dfs(i + k, j - 1)
    ans %= mod

• k = -1: Move left to position i - 1
• k = 0: Stay at position i
• k = 1: Move right to position i + 1

Each move consumes one step, so we recursively call with j - 1 remaining steps.

Modulo Operation: Since the answer can be large, we apply modulo 10^9 + 7 after each addition to prevent integer overflow.

Memoization: The @cache decorator automatically memoizes the results, storing computed values for each unique (i, j) pair. This prevents redundant calculations when the same state is reached through different paths.

Initial Call: The solution starts with dfs(0, steps), meaning we begin at position 0 with all steps available.

Time Complexity: O(steps × min(arrLen, steps)) as we compute each unique state at most once. Space Complexity: O(steps × min(arrLen, steps)) for the memoization cache.

Example Walkthrough

Let's walk through a small example with steps = 4 and arrLen = 3.

We need to find the number of ways to start at index 0, take exactly 4 steps, and return to index 0.

Starting with dfs(0, 4) (at position 0 with 4 steps remaining):

Step 1: From dfs(0, 4) We can make three moves:

• Move left: dfs(-1, 3) → returns 0 (out of bounds)
• Stay: dfs(0, 3) → need to compute
• Move right: dfs(1, 3) → need to compute

Step 2: Computing dfs(0, 3) From position 0 with 3 steps:

• Move left: dfs(-1, 2) → returns 0 (out of bounds)
• Stay: dfs(0, 2) → need to compute
• Move right: dfs(1, 2) → need to compute

Step 3: Computing dfs(1, 3) From position 1 with 3 steps:

• Move left: dfs(0, 2) → need to compute (or use memoized value)
• Stay: dfs(1, 2) → need to compute
• Move right: dfs(2, 2) → need to compute

Step 4: Computing dfs(0, 2) From position 0 with 2 steps:

• Move left: dfs(-1, 1) → returns 0 (out of bounds)
• Stay: dfs(0, 1) → need to compute
• Move right: dfs(1, 1) → need to compute

Step 5: Computing dfs(1, 2) From position 1 with 2 steps:

• Move left: dfs(0, 1) → need to compute (or use memoized value)
• Stay: dfs(1, 1) → need to compute
• Move right: dfs(2, 1) → returns 0 (position 2 > remaining steps 1, impossible to return)

Step 6: Computing base cases with 1 step remaining

• dfs(0, 1): Can only stay → dfs(0, 0) = 1 ✓
• dfs(1, 1): Can only move left → dfs(0, 0) = 1 ✓

Step 7: Backtracking with computed values

• dfs(0, 2) = 0 + 1 + 1 = 2 (from steps: left=0, stay=1, right=1)
• dfs(1, 2) = 1 + 1 + 0 = 2 (from steps: left=1, stay=1, right=0)
• dfs(2, 2) = 1 + 0 + 0 = 1 (can only move left then left again)

Continuing this process:

• dfs(0, 3) = 0 + 2 + 2 = 4
• dfs(1, 3) = 2 + 2 + 1 = 5

Finally:

• dfs(0, 4) = 0 + 4 + 5 = 9

So there are 9 different ways to return to position 0 after exactly 4 steps.

Some example paths:

1. Stay → Stay → Stay → Stay
2. Right → Left → Stay → Stay
3. Right → Stay → Left → Stay
4. Right → Right → Left → Left
5. Stay → Right → Left → Stay (and 4 more...)

Solution Implementation

Time and Space Complexity

The time complexity is O(steps × min(steps, arrLen)), and the space complexity is O(steps × min(steps, arrLen)).

The analysis can be refined to O(steps²) for both time and space complexity when considering the practical constraints. Here's why:

For the recursive function dfs(i, j):

• Parameter j ranges from 0 to steps
• Parameter i represents the position in the array, ranging from 0 to arrLen - 1
• However, due to the nature of the problem, we can only move at most steps distance from the starting position 0, so the effective range of i is bounded by min(steps, arrLen - 1)

Since we start at position 0 and have steps moves, with each move allowing us to go left, right, or stay, the maximum distance we can reach from origin is steps. Therefore:

• The number of unique states is O(steps × min(steps, arrLen))
• When arrLen > steps, the effective positions we can reach are limited to [0, steps], making it O(steps²)
• When arrLen ≤ steps, we have O(steps × arrLen) states

The reference answer simplifies this to O(steps × steps) because in most practical scenarios, the positions we can actually reach are limited by the number of steps, effectively making the complexity O(steps²) for both time and space.

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

Common Pitfalls

1. Not Optimizing the Search Space

A critical optimization that's easy to miss is that we don't need to consider positions beyond min(arrLen, steps // 2 + 1). Since we start at position 0 and must return to 0, the maximum distance we can travel is steps // 2 (go out and come back). Many solutions fail to include this optimization, leading to unnecessary computation and potential timeout for large inputs.

Problem Code:

@cache
def dfs(position: int, remaining_steps: int) -> int:
    if position < 0 or position >= arrLen or remaining_steps < 0:
        return 0
    # ... rest of the code

Solution:

# Optimize the array length to avoid unnecessary states
max_position = min(arrLen, steps // 2 + 1)

@cache
def dfs(position: int, remaining_steps: int) -> int:
    if position < 0 or position >= max_position or remaining_steps < 0:
        return 0
    # ... rest of the code

2. Forgetting to Apply Modulo Consistently

Another common mistake is forgetting to apply the modulo operation after accumulating results, which can cause integer overflow in languages with fixed integer sizes or incorrect results.

Problem Code:

ways = 0
for move in [-1, 0, 1]:
    ways += dfs(position + move, remaining_steps - 1)
return ways  # Forgot to apply modulo!

Solution:

ways = 0
for move in [-1, 0, 1]:
    ways += dfs(position + move, remaining_steps - 1)
    ways %= MOD  # Apply modulo after each addition
return ways

3. Incorrect Base Case Ordering

Checking base cases in the wrong order can lead to incorrect results. The "too far to return" check (position > remaining_steps) should come after boundary checks to avoid index errors.

Problem Code:

def dfs(position: int, remaining_steps: int) -> int:
    # Checking distance before boundaries - could cause issues
    if position > remaining_steps:
        return 0
    if position < 0 or position >= arrLen:
        return 0

Solution:

def dfs(position: int, remaining_steps: int) -> int:
    # Check boundaries first
    if position < 0 or position >= arrLen or remaining_steps < 0:
        return 0
    # Then check if too far to return
    if position > remaining_steps:
        return 0

4. Memory Limit Issues with Large Inputs

For very large values of steps and arrLen, the memoization cache can grow too large. Using the optimization from pitfall #1 is crucial to prevent this.

Complete Optimized Solution:

class Solution:
    def numWays(self, steps: int, arrLen: int) -> int:
        from functools import cache
      
        MOD = 10**9 + 7
        # Key optimization: limit search space
        max_position = min(arrLen, steps // 2 + 1)
      
        @cache
        def dfs(position: int, remaining_steps: int) -> int:
            # Boundary and validity checks
            if position < 0 or position >= max_position or remaining_steps < 0:
                return 0
          
            # Optimization: impossible to return to 0
            if position > remaining_steps:
                return 0
          
            # Success case
            if position == 0 and remaining_steps == 0:
                return 1
          
            # Try all three moves
            ways = 0
            for move in [-1, 0, 1]:
                ways += dfs(position + move, remaining_steps - 1)
                ways %= MOD
          
            return ways
      
        return dfs(0, steps)