You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).
Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: steps = 3, arrLen = 2 Output: 4 Explanation: There are 4 differents ways to stay at index 0 after 3 steps. Right, Left, Stay Stay, Right, Left Right, Stay, Left Stay, Stay, Stay
Example 2:
Input: steps = 2, arrLen = 4 Output: 2 Explanation: There are 2 differents ways to stay at index 0 after 2 steps Right, Left Stay, Stay
Example 3:
Input: steps = 4, arrLen = 2 Output: 8
Constraints:
1 <= steps <= 5001 <= arrLen <= 106This approach leverages dynamic programming to find the number of ways to stay at index 0 after a given number of steps. We define a 2D table dp[i][j] where i represents the number of steps remaining, and j represents the current position of the pointer.
To optimize computation, we can limit the table size to the minimum of steps and arrLen since going beyond these positions is unnecessary.
This code implements a dynamic programming solution in C. It initializes a DP table where dp[i][j] represents the number of ways to be at position j after i steps. The transition involves considering staying at the same position, moving left, or moving right.
C++
Java
Python
C#
JavaScript
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
This approach utilizes recursion combined with memoization to optimize the recursive calls. Here, recursion is used to explore all possible paths dynamically adjusting by staying at, moving left, or moving right from each position in every step.
The results of the recursive calls are stored in a memoization table to avoid redundant calculations.
This C implementation uses recursion with memoization. The recursive function explores all directions (stay, left, right) and stores results in a cache to prevent redundant computation, ensuring efficiency.
C++
Java
Python
C#
JavaScript
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
| Approach | Complexity |
|---|---|
| Dynamic Programming Approach | Time Complexity: O(steps * min(steps, arrLen)) Space Complexity: O(steps * min(steps, arrLen)) |
| Recursive with Memoization Approach | Time Complexity: O(steps * min(steps, arrLen)) Space Complexity: O(steps * min(steps, arrLen)) |
Number of Ways to Stay in the Same Place After Some Steps - Leetcode 1269 - Python • NeetCodeIO • 6,304 views views
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