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This 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.
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
1public class NumWays {
2 private static final int MOD = 1000000007;
3 public int numWays(int steps, int arrLen) {
4 int maxPos = Math.min(steps, arrLen - 1);
5 int[][] dp = new int[steps + 1][maxPos + 1];
6 dp[0][0] = 1;
7 for (int i = 1; i <= steps; ++i) {
8 for (int j = 0; j <= maxPos; ++j) {
9 dp[i][j] = dp[i - 1][j];
10 if (j > 0) dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
11 if (j < maxPos) dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
12 }
13 }
14 return dp[steps][0];
15 }
16 public static void main(String[] args) {
17 NumWays nw = new NumWays();
18 System.out.println(nw.numWays(3, 2)); // Output: 4
19 }
20}The Java solution follows the dynamic programming method to calculate the number of ways to stay at the initial position after the given steps. The transition between states considers all potential moves: staying, moving left, or right.
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.
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
The Java implementation leverages a HashMap to cache results of recursive calls, effectively managing state transitions by evaluating all possibilities and memoizing results.