Find maximum possible stolen value from houses

There are N houses built in a line, each of which contains some value in it. A thief is going to steal the maximum value of these houses, but he can’t steal in two adjacent houses because the owner of the stolen houses will tell his two neighbors left and right sides. The task is to find what is the maximum stolen value.

Examples: 

Input: hval[] = {6, 7, 1, 3, 8, 2, 4}
Output: 19
Explanation: The thief will steal 6, 1, 8 and 4 from the house.

Input: hval[] = {5, 3, 4, 11, 2}
Output: 16
Explanation: Thief will steal 5 and 11

Naive Approach: Below is the idea to solve the problem:

Given an array, the solution is to find the maximum sum subsequence where no two selected elements are adjacent. So the approach to the problem is a recursive solution. 

So there are two cases:

1. If an element is selected then the next element cannot be selected.
2. if an element is not selected then the next element can be selected.

Use recursion to find the result for both cases.

Below is the Implementation of the above approach:

Maximum loot possible : 19

Time Complexity:  O(2^N). Every element has 2 choices to pick and not pick
Auxiliary Space: O(2^N). A recursion stack space is required of size 2ⁿ, so space complexity is O(2^N).

Find the maximum possible stolen value from houses using O(n) extra space

Here, we are finding which will be the best pattern to steal a house without getting any police alert 

the approach has one edge case there will be only one/two houses then the output will be 

maximum money that is there in any house

else will be traverse for all houses and try to find the maximum amount that can get by coming from the previous house to current house

and at the end, we will just output maximum amount along the vector.

Maximum loot possible : 19

Time Complexity: O(n^2). for the worst case for the last element it will traverse over all elements of the vector. 
Space Complexity: O(n). the only used space is dp vector of o(n).

Find maximum possible stolen value from houses Dynamic Programming(Top-Down Approach):

The sub-problems can be stored thus reducing the complexity and converting the recursive solution to a Dynamic programming problem.

Follow the below steps to Implement the idea:

• Create a DP vector of size n+1 and value -1 and variables pick and not pick. 
• Create a recursive function 
  • If n < 0 possible stolen value is 0.
  • If n = 0 possible stolen value is hval[0].
  • If DP[n] != -1 possible stolen value is DP[n].
  • Set pick as pick = hval[n] +  maxLoot(hval, n-2, DP) 
  • Set not pick as notPick = maxLoot(hval, n-1, DP).
  • Set Dp[n] = max(pick, notPick) and return DP[n].

Below is the Implementation of the above approach:

Maximum loot possible : 19

Time Complexity: O(n). Only one traversal of the original array is needed. So the time complexity is O(n)
Space Complexity: O(n). Recursive stack space is required of size n, so space complexity is O(n).

Find maximum possible stolen value from houses Dynamic Programming(Bottom Up Approach):

The sub-problems can be stored thus reducing the complexity and converting the recursive solution to a Dynamic programming problem.

Follow the below steps to Implement the idea:

• Create an extra space DP array to store the sub-problems.
• Tackle the following basic cases, 
  • If the length of the array is 0, print 0. 
  • If the length of the array is 1, print the first element. 
  • If the length of the array is 2, print a maximum of two elements.
• Update dp[0] as array[0] and dp[1] as maximum of array[0] and array[1]
• Traverse the array from the second element (2nd index) to the end of the array.
• For every index, update dp[i] as a maximum of dp[i-2] + array[i] and dp[i-1], this step defines the two cases if an element is selected then the previous element cannot be selected and if an element is not selected then the previous element can be selected.
• Print the value dp[n-1]

Below is the Implementation of the above approach:

Maximum loot possible : 19

Time Complexity: O(N). Only one traversal of the original array is needed. So the time complexity is O(n)
Auxiliary Space: O(N). An array is required of size n, so space complexity is O(n).

Find maximum possible stolen value from houses using constant Space:

By carefully observing the DP array, it can be seen that the values of the previous two indices matter while calculating the value for an index. To replace the total DP array with two variables.

Follow the below steps to Implement the idea:

1. Tackle some basic cases, if the length of the array is 0, print 0, if the length of the array is 1, print the first element, if the length of the array is 2, print a maximum of two elements.
2. Create two variables value1 and value2, value1 as array[0] and value2 as maximum of array[0] and array[1] and a variable max_val to store the answer
3. Traverse the array from the second element (2nd index) to the end of the array.
4. For every index, update max_val as the maximum of value1 + array[i] and value2, this step defines the two cases, if an element is selected then the previous element cannot be selected and if an element is not selected then the previous element can be selected.
5. For every index, Update value1 = value2 and value2 = max_val
6. Print the value of max_val

Below is the Implementation of the above approach:

Maximum loot possible : 19

Time Complexity: O(N), Only one traversal of the original array is needed. So the time complexity is O(N).
Auxiliary Space: O(1), No extra space is required so the space complexity is constant.

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