Solution House thief
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?
Problem Statement
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.
Example 1:
Input: {2, 5, 1, 3, 6, 2, 4}
Output: 15
Explanation: The thief should steal from houses 5 + 6 + 4
Example 2:
Input: {2, 10, 14, 8, 1}
Output: 18
Explanation: The thief should steal from houses 10 + 8
Constraints:
1 <= wealth.length <= 1000 <= wealth[i] <= 400
Let's first start with a recursive brute-force solution.
Basic 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:
class Solution {
public int findMaxSteal(int[] wealth) {
return findMaxStealRecursive(wealth, 0);
}
private int findMaxStealRecursive(int[] wealth, int currentIndex) {
if (currentIndex >= wealth.length) return 0;
// steal from current house and skip one to steal from the next house
int stealCurrent =
wealth[currentIndex] + findMaxStealRecursive(wealth, currentIndex + 2);
// skip current house to steel from the adjacent house
int skipCurrent = findMaxStealRecursive(wealth, currentIndex + 1);
return Math.max(stealCurrent, skipCurrent);
}
public static void main(String[] args) {
Solution ht = new Solution();
int[] wealth = { 2, 5, 1, 3, 6, 2, 4 };
System.out.println(ht.findMaxSteal(wealth));
wealth = new int[] { 2, 10, 14, 8, 1 };
System.out.println(ht.findMaxSteal(wealth));
}
}
The time complexity of the above algorithm is exponential
Top-down Dynamic Programming with Memoization
To resolve overlapping subproblems, we can use an array to store the already solved subproblems.
Here is the code:
class Solution {
public int findMaxSteal(int[] wealth) {
int dp[] = new int[wealth.length];
return findMaxStealRecursive(dp, wealth, 0);
}
private int findMaxStealRecursive(int[] dp, int[] wealth, int currentIndex) {
if (currentIndex >= wealth.length) return 0;
if (dp[currentIndex] == 0) {
// steal from current house and skip one to steal next
int stealCurrent =
wealth[currentIndex] +
findMaxStealRecursive(dp, wealth, currentIndex + 2);
// skip current house to steel from the adjacent house
int skipCurrent = findMaxStealRecursive(dp, wealth, currentIndex + 1);
dp[currentIndex] = Math.max(stealCurrent, skipCurrent);
}
return dp[currentIndex];
}
public static void main(String[] args) {
Solution ht = new Solution();
int[] wealth = { 2, 5, 1, 3, 6, 2, 4 };
System.out.println(ht.findMaxSteal(wealth));
wealth = new int[] { 2, 10, 14, 8, 1 };
System.out.println(ht.findMaxSteal(wealth));
}
}
Bottom-up Dynamic Programming
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.
Code
Here is the code for our bottom-up dynamic programming approach:
class Solution {
public int findMaxSteal(int[] wealth) {
if (wealth.length == 0) return 0;
int dp[] = new int[wealth.length + 1]; // '+1' to handle the zero house
dp[0] = 0; // if there are no houses, the thief can't steal anything
dp[1] = wealth[0]; // only one house, so the thief have to steal from it
// please note that dp[] has one extra element to handle zero house
for (int i = 1; i < wealth.length; i++) dp[i + 1] = Math.max(
wealth[i] + dp[i - 1],
dp[i]
);
return dp[wealth.length];
}
public static void main(String[] args) {
Solution ht = new Solution();
int[] wealth = { 2, 5, 1, 3, 6, 2, 4 };
System.out.println(ht.findMaxSteal(wealth));
wealth = new int[] { 2, 10, 14, 8, 1 };
System.out.println(ht.findMaxSteal(wealth));
}
}
The above solution has time and space complexity of
Memory optimization
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:
class Solution {
public int findMaxSteal(int[] wealth) {
if (wealth.length == 0) return 0;
int n1 = 0, n2 = wealth[0], temp;
for (int i = 1; i < wealth.length; i++) {
temp = Math.max(n1 + wealth[i], n2);
n1 = n2;
n2 = temp;
}
return n2;
}
public static void main(String[] args) {
Solution ht = new Solution();
int[] wealth = { 2, 5, 1, 3, 6, 2, 4 };
System.out.println(ht.findMaxSteal(wealth));
wealth = new int[] { 2, 10, 14, 8, 1 };
System.out.println(ht.findMaxSteal(wealth));
}
}
The above solution has a time complexity of
Fibonacci number pattern
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.
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