1269. Number of Ways to Stay in the Same Place After Some Steps
https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/
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Intuition
Number of ways => DFS with backtracking
Recurrence
Bounds check - if current pos is out of array bounds
Recurrence termination
If steps = 0
If position = 0
return 1
return 0
dfs(stay at current pos, steps - 1) + dfs(move left, steps - 1) + dfs(move right, steps - 1)
some of these calls are repeated => use dp (current pos, steps remaining)
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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 still at index 0 after exactly steps steps.
Since the answer may be too large, return it modulo
10^9 + 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 <= 10^6
Intuition
Number of ways => DFS with backtracking
Recurrence
Bounds check - if current pos is out of array bounds
Recurrence termination
If steps = 0
If position = 0
return 1
return 0
dfs(stay at current pos, steps - 1) + dfs(move left, steps - 1) + dfs(move right, steps - 1)
some of these calls are repeated => use dp (current pos, steps remaining)
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