Leetcode_1269_Number of Ways to Stay in the Same Place After Some Steps

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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.

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 <= 500
• 1 <= arrLen <= 10^6

算法思路：利用动态规划的思想，设dp[steps][len]表示经过steps步到达len位置的方法的数量，dp[i][j] = dp[i - 1][j - 1] + dp[i -1][j] + dp[i + 1][j - 1]，且len <= steps + 1，所以len取steps + 1和arrLen中的较小值。

class Solution {
    private static int MOD = 1_000_000_007;

    public int numWays(int steps, int arrLen) {
        int len = Math.min(steps + 1, arrLen);

        int[][] dp = new int[steps+1][len];
        dp[0][0] = 1;
        for (int i = 1; i <= steps; i++) {
            for (int j = 0; j < len; j++) {
                dp[i][j] = dp[i-1][j];
                if (j - 1 >= 0){
                    dp[i][j] += dp[i-1][j-1];
                    dp[i][j] %= MOD;
                }
                if (j + 1 < len){ 
                    dp[i][j] += dp[i-1][j+1]; 
                    dp[i][j] %= MOD; 
                }
            }
        }
        return dp[steps][0];
    }
}