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| 1 | +/* Top-Down Approach |
| 2 | +------------------------------------- |
| 3 | + Time Complexity: O(n*min(n,m)) |
| 4 | + Space Complexity: O(n*min(n,m)) |
| 5 | +------------------------------------*/ |
| 6 | +class Solution { |
| 7 | + Map<String, Integer> memo; |
| 8 | + int MOD = (int)1e9+7; |
| 9 | + public int numWays(int steps, int arrLen){ |
| 10 | + memo = new HashMap<>(); |
| 11 | + return dfs(0, steps, arrLen); |
| 12 | + } |
| 13 | + private int dfs(int i, int steps, int n){ |
| 14 | + String pos = i + "," + steps; |
| 15 | + if(memo.containsKey(pos)) |
| 16 | + return memo.get(pos); |
| 17 | + if(steps == 0 && i == 0) |
| 18 | + return 1; |
| 19 | + if(i < 0 || i > n-1 || steps == 0) |
| 20 | + return 0; |
| 21 | + |
| 22 | + long left = dfs(i-1, steps-1, n); |
| 23 | + long stay = dfs(i, steps-1, n); |
| 24 | + long right = dfs(i+1, steps-1, n); |
| 25 | + int res = (int)((left + stay + right)%MOD); |
| 26 | + memo.put(pos, res); |
| 27 | + return res; |
| 28 | + } |
| 29 | +} |
| 30 | + |
| 31 | +/* Bottom-Up Approach |
| 32 | +------------------------------------ |
| 33 | + Time Complexity: O(n*min(n,m)) |
| 34 | + Space Complexity: O(n*min(n,m)) |
| 35 | +-----------------------------------*/ |
| 36 | +class Solution { |
| 37 | + public int numWays(int steps, int arrLen) { |
| 38 | + int MOD = (int)1e9+7; |
| 39 | + arrLen = Math.min(arrLen, steps); |
| 40 | + int[][] dp = new int[steps + 1][arrLen]; |
| 41 | + |
| 42 | + dp[0][0] = 1; |
| 43 | + |
| 44 | + for (int step = 1; step <= steps; step++) { |
| 45 | + for (int pos = 0; pos < arrLen; pos++) { |
| 46 | + dp[step][pos] = dp[step - 1][pos]; |
| 47 | + |
| 48 | + if (pos - 1 >= 0) { |
| 49 | + dp[step][pos] = (dp[step][pos] + dp[step - 1][pos - 1]) % MOD; |
| 50 | + } |
| 51 | + if (pos + 1 < arrLen) { |
| 52 | + dp[step][pos] = (dp[step][pos] + dp[step - 1][pos + 1]) % MOD; |
| 53 | + } |
| 54 | + } |
| 55 | + } |
| 56 | + |
| 57 | + return dp[steps][0]; |
| 58 | + } |
| 59 | +} |
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