Drop a Ball Down the Stairs - Google Top Interview Questions
Problem Statement :
You are given an integer height and a list of integers blacklist. You are currently standing at height height and you are playing a game to move a small ball down to height 0. In even rounds (0, 2, 4, 6 etc.) you can move the ball down 1, 2, or 4 stairs down. In odd rounds, you can move the ball down 1, 3, or 4 stairs. There are some levels of the stairs where if the ball lands there, it will die as labelled in blacklist. Return number of ways you can move the ball down to height 0. Mod the result by 10 ** 9 + 7. Constraints 1 ≤ height ≤ 100,000 0 ≤ b ≤ 100,000 where b is the length of blacklist Example 1 Input height = 4 blacklist = [2] Output 2 Explanation We can move 4 stairs down on round 0. We can move 1 stair down on round 0 and then 3 stairs down on round 1. Example 2 Input height = 5 blacklist = [0] Output 0 Explanation There's no way to reach height 0 since it's on the blacklist.
Solution :
Solution in C++ :
const int MOD = 1e9 + 7;
const int even[3] = {1, 2, 4};
const int odd[3] = {1, 3, 4};
unordered_set<int> s;
vector<vector<int>> dp;
int f(int h, int round) {
// base case
if (h < 0 or s.count(h)) return 0;
// found a way
if (h == 0) return 1;
// memoization
int& ans = dp[round][h];
if (ans != -1) return ans;
ans = 0;
// even round
if (!round) {
for (auto x : even) {
ans = (ans + f(h - x, round ^ 1)) % MOD;
}
}
// odd round
else {
for (auto x : odd) {
ans = (ans + f(h - x, round ^ 1)) % MOD;
}
}
return ans;
}
int solve(int height, vector<int>& blacklist) {
s.clear();
dp.clear();
dp.resize(2, vector<int>(height + 1, -1));
for (auto x : blacklist) s.insert(x);
return f(height, 0);
}
Solution in Java :
import java.util.*;
class Solution {
long MOD = 1000000007;
public int solve(int height, int[] bList) {
DP = new Long[height + 1][2];
Set<Integer> set = new HashSet<>();
for (int num : bList) set.add(num);
return (int) dp(set, height, 0);
}
Long[][] DP;
private long dp(Set<Integer> bList, int curr, int odd) {
if (bList.contains(curr) || curr < 0)
return 0;
if (curr == 0)
return 1;
if (DP[curr][odd] != null)
return DP[curr][odd];
if (odd == 1)
return DP[curr][odd] =
(dp(bList, curr - 1, 0) + dp(bList, curr - 3, 0) + dp(bList, curr - 4, 0))
% MOD;
return DP[curr][odd] =
(dp(bList, curr - 1, 1) + dp(bList, curr - 2, 1) + dp(bList, curr - 4, 1)) % MOD;
}
}
Solution in Python :
class Solution:
def solve(self, height, blacklist):
MOD = 10 ** 9 + 7
found = set(blacklist)
if 0 in found:
return 0
@lru_cache(None)
def dp(h, rnd):
if h < 0 or h in found:
return 0
if h == 0:
return 1
ans = 0
if rnd % 2 == 0:
ans += dp(h - 2, rnd ^ 1)
else:
ans += dp(h - 3, rnd ^ 1)
ans += dp(h - 1, rnd ^ 1)
ans += dp(h - 4, rnd ^ 1)
return ans
return dp(height, 0) % MOD
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