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

Input: You are given the integers `startPos`, `endPos`, and `k`. These represent the starting position, the destination, and the number of steps respectively.

Example: startPos = 3, endPos = 5, k = 4

Constraints:

• 1 <= startPos, endPos, k <= 1000

Output: Return the number of ways to reach `endPos` from `startPos` in exactly `k` steps, modulo 10^9 + 7.

Example: Output: 4

Constraints:

Goal: The goal is to find the number of valid ways to move from `startPos` to `endPos` in exactly `k` steps, adhering to the movement rules (left or right in each step).

Steps:

• 1. Calculate the absolute difference `d` between `startPos` and `endPos`.

• 2. Use dynamic programming or recursion with memoization to count the number of ways to reach the target by moving either left or right.

• 3. Return the count modulo 10^9 + 7.

Goal: The problem constraints ensure that the input values are within a manageable range, making it feasible to solve using dynamic programming or recursion with memoization.

Steps:

• The length of `startPos`, `endPos`, and `k` is limited to 1000, which ensures that brute force solutions are not feasible and dynamic programming is necessary.

Assumptions:

• The input values are valid and within the specified range.

• The number of ways can be very large, so modulo operation must be used.

• Input: startPos = 1, endPos = 2, k = 3

• Explanation: We can reach position 2 from 1 in exactly 3 steps in three distinct ways: 1 -> 2 -> 3 -> 2, 1 -> 2 -> 1 -> 2, and 1 -> 0 -> 1 -> 2. Therefore, the output is 3.

• Input: startPos = 2, endPos = 5, k = 10

• Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps. Hence, the output is 0.

Approach: The approach involves using a dynamic programming technique or recursion with memoization to count the number of valid ways to move from `startPos` to `endPos` in `k` steps.

Observations:

• The problem can be reduced to finding paths with exactly `k` steps on a number line, which can be modeled as a dynamic programming problem.

• We need to calculate the number of paths efficiently, considering both left and right moves at each step.

Steps:

• 1. Define a recursive function `dfs(k, d)` where `k` is the remaining number of steps and `d` is the current distance from `startPos` to `endPos`.

• 2. Implement memoization to store results of subproblems and avoid redundant calculations.

• 3. Return the number of ways modulo 10^9 + 7.

Empty Inputs:

• This problem does not have an empty input case since the input size is guaranteed to be within the range 1 to 1000.

Large Inputs:

• For large values of `k`, the recursive approach should be optimized using dynamic programming or memoization to avoid recalculating results for the same subproblems.

Special Values:

• If `startPos` and `endPos` are the same, we must ensure that we move in such a way that we still perform exactly `k` steps.

Constraints:

• Dynamic programming or memoization should be used to efficiently handle inputs close to the upper limit.

int numberOfWays(int start, int end, int k) {
    return dfs(k, abs(start - end));
}

int dfs(int k, int d) {
    if (d >= k) return d == k;
    if(dp[k][d] == 0)
        dp[k][d] = (1 + dfs(k-1, d + 1) + dfs(k -1, abs(d -1))) % mod;
    
    return dp[k][d] -1;
    
}

1 : Function Definition

int numberOfWays(int start, int end, int k) {

Defines the function `numberOfWays` which takes three parameters: `start`, `end`, and `k`. The function will return the number of ways to reach from `start` to `end` with `k` steps.

2 : Recursive Call

    return dfs(k, abs(start - end));

This line calls the helper function `dfs`, passing `k` and the absolute difference between `start` and `end` as arguments.

3 : Helper Function Definition

int dfs(int k, int d) {

Defines the helper function `dfs`, which takes two parameters: `k` (number of steps remaining) and `d` (the current distance to cover). It will return the number of ways to complete the journey recursively.

4 : Base Case Check

    if (d >= k) return d == k;

Checks if the distance `d` is greater than or equal to `k`. If it is, it checks if `d` is exactly equal to `k` and returns 1 if true (base case), otherwise returns 0.

5 : Memoization Check

    if(dp[k][d] == 0)

Checks if the value of `dp[k][d]` has been previously computed. If it hasn't been computed (i.e., it is 0), the code proceeds to calculate it.

6 : Recursion and Memoization

        dp[k][d] = (1 + dfs(k-1, d + 1) + dfs(k -1, abs(d -1))) % mod;

Calculates the number of ways recursively by making two recursive calls to `dfs`: one where the distance increases by 1 (`d + 1`) and another where the distance decreases by 1 (`abs(d - 1)`). The result is stored in `dp[k][d]` and taken modulo `mod`.

7 : Return Value

    return dp[k][d] -1;

Returns the computed result from the `dp` array, decrementing it by 1 (to account for the last unnecessary computation step).

Best Case: O(k), where k is the number of steps. This is when the number of recursive calls is minimized due to memoization.

Average Case: O(k * d), where d is the distance between `startPos` and `endPos`. This depends on the recursive branching.

Worst Case: O(k * d), where d is the maximum possible distance. Memoization ensures that redundant calculations are avoided.

Description: The time complexity is determined by the number of recursive calls and the distance between `startPos` and `endPos`.

Best Case: O(1), if no memoization is used and the problem is solved directly.

Worst Case: O(k * d), where d is the distance between `startPos` and `endPos`. This is the space used by the memoization table.

Description: The space complexity is primarily determined by the size of the memoization table.

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

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