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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))
1#include <stdio.h>
2#define MOD 1000000007
3
4int numWays(int steps, int arrLen) {
5 int maxPos = (arrLen < steps) ? arrLen - 1 : steps;
6 int dp[steps+1][maxPos+1];
7 for(int i = 0; i <= steps; i++)
8 for(int j = 0; j <= maxPos; j++)
9 dp[i][j] = 0;
10
11 dp[0][0] = 1;
12 for (int i = 1; i <= steps; i++) {
13 for (int j = 0; j <= maxPos; j++) {
14 dp[i][j] = dp[i-1][j];
15 if (j > 0)
16 dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD;
17 if (j < maxPos)
18 dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD;
19 }
20 }
21 return dp[steps][0];
22}
23
24int main() {
25 printf("%d\n", numWays(3, 2)); // Output: 4
26 return 0;
27}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.
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.