2400. Number of Ways to Reach a Position After Exactly k Steps

Problem Description

You start at a startPos on an infinite number line and want to reach an endPos. You can take steps to the left or right, and you have to take exactly k steps to do this. The goal is to find how many different ways you can do this. A way is different from another if the sequence of steps is different. Since you are moving on an infinite number line, negative positions are possible. The position reached after k steps must be exactly endPos. Since the number of ways could be very large, you need to return the result modulo 10^9 + 7 to keep the number manageable.

Intuition

Given the need to move a specific number of steps and end at a specific position, this is a classic problem that can be addressed using Dynamic Programming or memoization to avoid redundant calculations.

First, consider the absolute difference between the startPos and endPos: this is the minimum number of moves required to get from start to end without any extra steps. From there, any additional steps must eventually be counteracted by an equal number of steps in the opposite direction.

To calculate the number of ways, you can use recursion. With each recursive call, you decrement k because you're making a step. If k reaches 0, you check if you've arrived at endPos: if so, that's one valid way; if not, you return 0 as it's not a valid sequence of steps.

The solution uses a depth-first search (DFS) approach where each 'node' is a position after a certain number of steps. The DFS explores all possible sequences of left and right steps until k steps are performed. Memoization (@cache decorator) is key here, as it stores the result of previous computations. This is crucial since there are many overlapping subproblems, especially as k grows larger.

The dfs(i, j) function checks whether you can reach the relative position i (startPos - endPos) in j steps. It evaluates two scenarios at each step: moving right and moving left. The modulus operation is used on the sum of possibilities to keep the numbers within the constraints of the modulo.

Since movement on the number line is symmetrical, taking an absolute of the i when moving left is a significant optimization. Movements X steps to the right and Y steps to the left result in the same position as Y steps to the right and X steps to the left if X and Y are flipped. So you can always consider moves to the right and moves toward zero, enabling the dfs function to memoize more effectively as it encounters fewer unique (i, j) pairs.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

The solution employs depth-first search (DFS) with memoization to systematically explore all possible step sequences that amount to exactly k steps and arrive at endPos. The following patterns and algorithms are used:

1. Recursion: The dfs(i, j) function calls itself to explore each possibility. It increments or decrements the current position, respectively, for a right or left step, and decreases the remaining steps j by one.

2. Base Cases: The function has two critical base cases:

   • If j == 0 (no more steps left), check if the current position i is 0 (which means startPos == endPos). If so, return 1, representing one valid way to reach the end position; otherwise return 0, as it’s not possible to reach endPos without steps left.
   • If i > j (the current position is farther from 0 than the number of steps remaining) or j < 0 (somehow the steps went negative, which should not happen in this approach), return 0 because reaching endPos is impossible.

3. Memoization: The @cache decorator is used in Python to store the results of dfs calls, which prevents redundant calculations of the same (i, j) state. Since many positions will be visited multiple times with the same number of remaining steps, storing these results significantly speeds up the computation.

4. Modulo Arithmetic: The modulo operation % mod ensures that intermediate sums do not exceed the problem's specified limit (10**9 + 7). This is done to avoid integer overflow and is a common practice in problems dealing with large numbers.

5. Symmetry and Absolute Value: When a step is taken to the left, the function calls itself with abs(i - 1) instead of -(i - 1). This is because taking a step to the left from position i is equivalent to taking a step to the right from position -i due to the line's symmetry. Using the absolute value maximizes the effectiveness of memoization because it reduces the number of unique states.

The code does not explicitly create a data structure for memoization; it relies on the cache mechanism provided by the Python functools module, which internally creates a mapping of arguments to the dfs function to their results.

Finally, the result of the dfs function call for abs(startPos - endPos), k gives us the number of ways to reach from startPos to endPos in k steps. It respects the symmetry of the problem and allows for an efficient computation of the result.

Example Walkthrough

Let's go through a small example to illustrate the solution approach.

Suppose you start at startPos = 1 and want to reach endPos = 3. You have k = 3 steps to do it. How many different ways can you do this?

1. First, check the absolute difference between startPos and endPos, which is |3 - 1| = 2. This is the minimum number of steps to reach from start to end without any extra steps.

2. To account for the exact k steps, you can take 1 additional step to the left and then move back to the right, resulting in 3 total steps (right -> left -> right).

Let's use the dfs(i, j) recursion with memoization to explore the possibilities:

• The initial call is dfs(abs(1-3), 3) which simplifies to dfs(2, 3).

• The dfs function will now calculate this in two ways: consider a step to the left and consider a step to the right.

• When considering a step to the right (dfs(2+1, 2)), the state becomes dfs(3, 2).

  • Here, since i = 3 is greater than j = 2, we return 0 because you can't reach position 0 in just 2 steps.

• When considering a step to the left (dfs(abs(2-1), 2)), the state becomes dfs(1, 2).

  • Now, for dfs(1, 2), again we consider steps left and right:
    • Step to the right: dfs(1+1, 1) simplifies to dfs(2, 1), which returns 0 because i > j.
    • Step to the left: dfs(abs(1-1), 1) simplifies to dfs(0, 1).
      • At this point we have dfs(0, 1). Considering further steps:
        • Step to the right: dfs(0+1, 0) simplifies to dfs(1, 0), which returns 0 because i is not equal to 0.
        • Step to the left: dfs(abs(0-1), 0) simplifies to dfs(1, 0), which also returns 0.

However, from the initial dfs(1, 2), if we make one more DFS call with one step to the left (dfs(abs(1-1), 1)), the result is dfs(0, 1). This state has already been evaluated when it was encountered via the dfs(1, 2) from the first step to the right. Hence, it will not compute again but fetch the result from the memoization cache, which should be zero since we learned no way exists to end at 0 starting from 1 with 1 step.

To summarize, from startPos = 1 to endPos = 3 with k = 3 steps, there are no valid ways to achieve the end position.

The use of memoization is crucial in this example because it avoids the re-computation of states that have already been visited, such as dfs(1, 2) being re-computed after dfs(1, 0). By caching the results, we speed up the computation and also maintain the count of possible ways within the required modulo.

Solution Implementation

Time and Space Complexity

The given Python code defines a Solution class with a method numberOfWays that calculates the number of ways to reach the endPos from the startPos in exactly k steps. The solution uses a depth-first search (DFS) approach with memoization.

Time Complexity

The time complexity of this algorithm largely depends on the parameters startPos, endPos, and k, and how many unique states the DFS function generates.

There are at most k + 1 levels to explore because each level corresponds to a step, and we do not go beyond k steps. At each step, the number of positions that we could be at increases, however due to the absolute value used in the second call to dfs, two calls may result in the same next state (e.g., dfs(1, j-1) and dfs(-1, j-1) both lead to dfs(1, j-1)), effectively reducing the number of unique states. Therefore, for each step, there can be up to 2 * (j - 1) + 1 unique positions to consider because we could either move left or right from each position or stay at the same position.

Considering memoization, which ensures that each unique state is only computed once, the time complexity can be estimated by the number of unique (i, j) pairs for which the dfs function is called. Hence, the total number of unique calls to dfs is O(k^2), leading to a time complexity of T(n) = O(k^2).

Space Complexity

The space complexity is determined by the size of the cache (to memoize the DFS calls) and the maximum depth of the recursion stack (which occurs when the DFS explores from j = k down to j = 0).

The cache potentially stores all unique states that the DFS visits, which we've established above is O(k^2).

The maximum depth of the recursive call stack occurs when the function continually recurses without finding any cached results, which would mean a maximum depth of k. This is a very rare case, though, because memoization ensures that even deep recursive chains would quickly find cached results and not recurse to their maximum possible depth.

However, for the sake of computation, we consider the worst-case scenario without memoization's benefits, which is O(k).

Therefore, the space complexity is determined by the larger of the cache's size or the recursion stack's depth. In this case, the cache's size is larger; hence the space complexity is S(n) = O(k^2).

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