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There are n
houses built in a line. A thief wants to steal the maximum possible money from these houses. The only restriction the thief has is that he can’t steal from two consecutive houses, as that would alert the security system. How should the thief maximize his stealing?
🙋 Question
Given a number array representing the wealth of n
houses, determine the maximum amount of money the thief can steal without alerting the security system.
Input: {2, 5, 1, 3, 6, 2, 4}
Output: 15
Explanation: The thief should steal from houses 5 + 6 + 4
Input: {2, 10, 14, 8, 1}
Output: 18
Explanation: The thief should steal from houses 10 + 8
Let’s first start with a recursive brute-force solution.
For every house i
, we have two options:
Steal from the current house (i
), skip one and steal from (i+2
).
Skip the current house (i
), and steal from the adjacent house (i+1
).
The thief should choose the one with the maximum amount from the above two options. So our algorithm will look like this:
def find_max_steal(wealth):
return find_max_steal_recursive(wealth, 0)
def find_max_steal_recursive(wealth, currentIndex):
if currentIndex >= len(wealth):
return 0
stealCurrent = wealth[currentIndex] + \
find_max_steal_recursive(wealth, currentIndex + 2)
skipCurrent = find_max_steal_recursive(wealth, currentIndex + 1)
return max(stealCurrent, skipCurrent)
def main():
print(find_max_steal([2, 5, 1, 3, 6, 2, 4]))
print(find_max_steal([2, 10, 14, 8, 1]))
main()
👉 Complexity
The time complexity of the above algorithm is exponential
O(2n). The space complexity is
O(n) which is used to store the recursion stack.
To resolve overlapping subproblems, we can use an array to store the already solved subproblems.
Here is the code:
def find_max_steal(wealth):
dp = [0 for x in range(len(wealth))]
return find_max_steal_recursive(dp, wealth, 0)
def find_max_steal_recursive(dp, wealth, currentIndex):
if currentIndex >= len(wealth):
return 0
if dp[currentIndex] == 0:
stealCurrent = wealth[currentIndex] + \
find_max_steal_recursive(dp, wealth, currentIndex + 2)
skipCurrent = find_max_steal_recursive(dp, wealth, currentIndex + 1)
dp[currentIndex] = max(stealCurrent, skipCurrent)
return dp[currentIndex]
def main():
print(find_max_steal([2, 5, 1, 3, 6, 2, 4]))
print(find_max_steal([2, 10, 14, 8, 1]))
main()
Let’s try to populate our dp[]
array from the above solution, working in a bottom-up fashion. As we saw in the above code, every findMaxStealRecursive()
is the maximum of the two recursive calls; we can use this fact to populate our array.
Here is the code for our bottom-up dynamic programming approach:
def find_max_steal(wealth):
n = len(wealth)
if n == 0:
return 0
dp = [0 for x in range(n+1)]
dp[0] = 0
dp[1] = wealth[0]
for i in range(1, n):
dp[i + 1] = max(wealth[i] + dp[i - 1], dp[i])
return dp[n]
def main():
print(find_max_steal([2, 5, 1, 3, 6, 2, 4]))
print(find_max_steal([2, 10, 14, 8, 1]))
main()
👉 Complexity
The above solution has time and space complexity of O(n)
.
We can optimize the space used in our previous solution. We don’t need to store all the previous numbers up to n
, as we only need two previous numbers to calculate the next number in the sequence. Let’s use this fact to further improve our solution:
def find_max_steal(wealth):
n = len(wealth)
if n == 0:
return 0
n1, n2 = 0, wealth[0]
for i in range(1, n):
n1, n2 = n2, max(n1 + wealth[i], n2)
return n2
def main():
print(find_max_steal([2, 5, 1, 3, 6, 2, 4]))
print(find_max_steal([2, 10, 14, 8, 1]))
main()def find_max_steal(wealth):
n = len(wealth)
if n == 0:
return 0
n1, n2 = 0, wealth[0]
for i in range(1, n):
n1, n2 = n2, max(n1 + wealth[i], n2)
return n2
def main():
print(find_max_steal([2, 5, 1, 3, 6, 2, 4]))
print(find_max_steal([2, 10, 14, 8, 1]))
main()
👉 Complexity
The above solution has a time complexity of O(n)
and a constant space complexity O(1)
.
We can clearly see that this problem follows the Fibonacci number pattern. The only difference is that every Fibonacci number is a sum of the two preceding numbers, whereas in this problem every number (total wealth) is the maximum of previous two numbers.