# 1269. Number of Ways to Stay in the Same Place After Some Steps
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.
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
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
Input: steps = 4, arrLen = 2 Output: 8
Approach 1: DP, with optimization by limiting the range of motion.
# Code (Python)
def numWays(self, steps: int, arrLen: int) -> int: # DP. But can limit the range of motion, because you can't get back after you move away for stes / 2 steps. size = min(steps // 2 + 1, arrLen) # max number of positions can move to get back ways = [0, 1] +  * size # leave room on both sides, for positions -1 and size for step in range(steps): new_ways =  * len(ways) for i in range(1, size + 1): # new_ways and new_ways[-1] are always 0 because they don't get updated; they're just here to cut new_ways[i] = (ways[i] + ways[i-1] + ways[i+1]) % (10 ** 9 + 7) ways = new_ways return ways
# Code (C++)