# Super Egg Drop This problem is equivalent to Drop Eggs II at Lintcode but harder to pass.

## # Solution

See this video for perfect explanation on this problem and the previous easier problem.

### # Iterative DP (TLE)

Let i be the one of m eggs and j be one of the n floors. The recurrence relation is:

$dp(i, j) = \min_{1 \le k \le j}\left(\max(dp(i-1, k-1), dp(i, j-k)) \right)$

#### # Explanation on Recurrence Relation

If ith egg breaks at kth floor, there are i-1 eggs left and critical floor exists below k, so problem is reduced to res[i-1][k-1].

Else, ith egg doesn't break and critical floor exists btw k and j, so problem is reduced to res[i][k-j].

Note:

• plus 1 for initial drop
• max takes into account the worst case
• min takes the number of drops for the best way to drop ith egg at optimal kth floor

Complexity:

• time: $O(mn^2)$
• space: $O(mn)$
class Solution:
"""
@param m: the number of eggs
@param n: the number of floors
@return: the number of drops in the worst case
"""
def dropEggs2(self, m, n):
res = [[sys.maxsize]*(n+1) for _ in range(m+1)]

for i in range(1, m+1):
res[i] = 0
res[i] = 1
for j in range(1, n+1):
res[j] = j
for i in range(2, m+1):
for j in range(2, n+1):
for k in range(1, j+1):
res[i][j] = min(res[i][j], 1 + max(res[i-1][k-1], res[i][j-k]))
return res[m][n]


### # Recursive DP (TLE)

The recursive DP w/ memoization also TLE.

    def superEggDrop(self, K, N):
if N == 0 or N == 1:
return N
if K == 1:
return N
min = sys.maxsize

# Consider all droppings from 1st
# floor to kth floor and return
# the minimum of these values plus 1.
for x in range(1, N + 1):

res = 1 + max(self.superEggDrop(K - 1, x - 1),
self.superEggDrop(K, N - x))
if (res < min):
min = res

return min


### # DP w/ Optimality Criterion (REDO)

class Solution:
"""
@param m: the number of eggs
@param n: the number of floors
@return: the number of drops in the worst case
"""
def dropEggs2(self, m, n):
# Right now, dp[i] represents dp(1, i)
dp = range(N+1)

for k in range(2, K+1):
# Now, we will develop dp2[i] = dp(k, i)
dp2 = 
x = 1
for n in range(1, N+1):
# Let's find dp2[n] = dp(k, n)
# Increase our optimal x while we can make our answer better.
# Notice max(dp[x-1], dp2[n-x]) > max(dp[x], dp2[n-x-1])
# is simply max(T1(x-1), T2(x-1)) > max(T1(x), T2(x)).
while x < n and max(dp[x-1], dp2[n-x]) > max(dp[x], dp2[n-x-1]:
x += 1

# The final answer happens at this x.
dp2.append(1 + max(dp[x-1], dp2[n-x]))

dp = dp2

return dp[-1]


### # Math (REDO)

def superEggDrop(self, K, N):
dp = [[0 for col in range(K + 1)] for row in range(N + 1)]
m = 0
while dp[m][K] < N
m = m + 1
for i in range(1,K + 1)
dp[m][i] = dp[m - 1][i - 1] + dp[m - 1][i] + 1
return m

Last Updated: 9/21/2020, 4:44:16 PM