**A Maniacal Walk- Google Top Interview Questions**

### Problem Statement :

A person is placed on a list of length length, at index 0, and on each step, they can move right one index or left one index (without going out of bounds), or stay on that index. Given that the person can take exactly n steps, how many unique walks can the person take and reach back to index 0? Mod the result by 10 ** 9 + 7. Constraints length ≤ 1,000 n ≤ 500 Example 1 Input length = 5 n = 3 Output 4 Explanation The four actions are: stay at index 0 3 times in a row. right, stay, left. right, left, stay. stay, right, left.

### Solution :

` ````
Solution in C++ :
int solve(int length, int n) {
if (!length) return 0;
vector<int> dp(length);
// 1 way to be at the index 0 with 0 steps
dp[0] = 1;
const int MOD = 1e9 + 7;
// calculate answer for each steps
while (n--) {
auto dpc = dp;
for (int i = 0; i < length; ++i) {
if (i > 0) dpc[i - 1] = (dpc[i - 1] + dp[i]) % MOD;
if (i < length - 1) dpc[i + 1] = (dpc[i + 1] + dp[i]) % MOD;
}
dp = dpc;
}
return dp[0] % MOD;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int length, int n) {
if (n == 0)
return 1;
int[] dp1 = new int[length];
int[] dp2 = new int[length];
final int MOD = (int) 1e9 + 7;
dp1[0] = 1;
for (int left = 1; left <= n; left++) {
for (int i = 0; i < length; i++) {
dp2[i] = ((dp1[i] + (i - 1 >= 0 ? dp1[i - 1] : 0)) % MOD
+ (i + 1 < length ? dp1[i + 1] : 0))
% MOD;
}
for (int i = 0; i < length; i++) {
dp1[i] = dp2[i];
}
}
return dp1[0];
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, length, n):
len = min(n + 1, length)
start = [0] * len
start[0] = 1
for i in range(n):
new = [0] * len
for j in range(len):
if j != 0 and j != len - 1:
new[j] = start[j - 1] + start[j] + start[j + 1]
elif j == 0:
new[j] = start[j] + start[j + 1]
else:
new[j] = start[j - 1] + start[j]
start = new
return start[0] % (10 ** 9 + 7)
```

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