# Find maximum possible stolen value from houses

• Difficulty Level : Medium
• Last Updated : 17 Jun, 2022

There are n houses build in a line, each of which contains some value in it. A thief is going to steal the maximal 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 side. 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: 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.

Implementation of recursion approach:

## C++

 `// CPP program to find the maximum stolen value` `#include ` `using` `namespace` `std;`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``// base case` `    ``if` `(n < 0) {` `        ``return` `0;` `    ``}`   `    ``if` `(n == 0) {` `        ``return` `hval;` `    ``}` `    ``// if current element is pick then previous cannot be` `    ``// picked` `    ``int` `pick = hval[n] + maxLoot(hval, n - 2);` `    ``// if current element is not picked then previous` `    ``// element is picked` `    ``int` `notPick = maxLoot(hval, n - 1);`   `    ``// return max of picked and not picked` `    ``return` `max(pick, notPick);` `}`   `// Driver to test above code` `int` `main()` `{` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``cout << ``"Maximum loot possible : "` `         ``<< maxLoot(hval, n - 1);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to find the maximum stolen value` `#include `   `// Find maximum between two numbers.` `int` `max(``int` `num1, ``int` `num2)` `{` `    ``return` `(num1 > num2) ? num1 : num2;` `}`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``// base case` `    ``if` `(n < 0)` `        ``return` `0;`   `    ``if` `(n == 0)` `        ``return` `hval;` `    ``// if current element is pick then previous cannot be` `    ``// picked` `    ``int` `pick = hval[n] + maxLoot(hval, n - 2);` `    ``// if current element is not picked then previous` `    ``// element is picked` `    ``int` `notPick = maxLoot(hval, n - 1);`   `    ``// return max of picked and not picked` `    ``return` `max(pick, notPick);` `}`   `// Driver to test above code` `int` `main()` `{` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``printf``(``"Maximum loot possible : %d "``,` `           ``maxLoot(hval, n - 1));` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `/*package whatever //do not write package name here */` `// Java program to find the maximum stolen value` `import` `java.io.*;`   `class` `GFG ` `{` `  `  `  ``// Function to calculate the maximum stolen value` `  ``static` `int` `maxLoot(``int` `hval[], ``int` `n)` `  ``{` `    `  `    ``// base case` `    ``if` `(n < ``0``) {` `      ``return` `0``;` `    ``}`   `    ``if` `(n == ``0``) {` `      ``return` `hval[``0``];` `    ``}` `    `  `    ``// if current element is pick then previous cannot` `    ``// be picked` `    ``int` `pick = hval[n] + maxLoot(hval, n - ``2``);` `    `  `    ``// if current element is not picked then previous` `    ``// element is picked` `    ``int` `notPick = maxLoot(hval, n - ``1``);`   `    ``// return max of picked and not picked` `    ``return` `Math.max(pick, notPick);` `  ``}`   `  ``// driver program` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int` `hval[] = { ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `};` `    ``int` `n = hval.length;` `    ``System.out.println(``"Maximum loot value : "` `                       ``+ maxLoot(hval, n-``1``));` `  ``}` `}`   `// This code is contributed by sanskar84.`

## Python3

 `# Python3 program to find the maximum stolen value`   `# calculate the maximum stolen value` `def` `maxLoot(hval,n):`   `    ``# base case` `    ``if` `(n < ``0``):` `        ``return` `0`   `    ``if` `(n ``=``=` `0``):` `        ``return` `hval[``0``]` `    `  `    ``# if current element is pick then previous cannot be` `    ``# picked` `    ``pick ``=` `hval[n] ``+` `maxLoot(hval, n ``-` `2``)` `    ``# if current element is not picked then previous` `    ``# element is picked` `    ``notPick ``=` `maxLoot(hval, n ``-` `1``)`   `    ``# return max of picked and not picked` `    ``return` `max``(pick, notPick)`   `# Driver to test above code` `hval ``=` `[ ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `]` `n ``=` `len``(hval)` `print``(``"Maximum loot possible : "``,maxLoot(hval, n ``-` `1``));`   `# This code is contributed by shinjanpatra`

## Javascript

 ``

Output

`Maximum loot possible : 19`

Complexity Analysis

Time Complexity:  O(2N). Every element has 2 choices to pick and not pick

Space Complexity: O(2N). A recursion stack space is required of size 2n, so space complexity is O(2N).

Method 2: Dynamic Programming : Top Down Approach

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

## C++

 `// CPP program to find the maximum stolen value` `#include ` `using` `namespace` `std;`   `// calculate the maximum stolen value` `int` `maxLoot(``int` `*hval, ``int` `n, vector<``int``> &dp){` `  `  `   ``// base case` `   ``if``(n < 0){` `           ``return` `0 ;` `   ``}` `   `  `   ``if``(n == 0){` `           ``return` `hval ;` `   ``}` `   ``// If the subproblem is already solved` `   ``// then return its value` `   ``if``(dp[n] != -1 ){` `          ``return` `dp[n] ;` `   ``}` `  `  `   ``//if current element is pick then previous cannot be picked` `   ``int` `pick = hval[n] +  maxLoot(hval, n-2, dp) ;` `   ``//if current element is not picked then previous element is picked` `   ``int` `notPick = maxLoot(hval, n-1, dp)  ;` `  `  `   ``// return max of picked and not picked` `   ``return` `dp[n] = max(pick, notPick) ;` `  `  `}`   `// Driver to test above code` `int` `main()` `{` `    ``int` `hval[] = {6, 7, 1, 3, 8, 2, 4};` `    ``int` `n = ``sizeof``(hval)/``sizeof``(hval);` `    ``// Initialize a dp vector` `    ``vector<``int``> dp(n+1, -1) ;` `    ``cout << ``"Maximum loot possible : "` `        ``<< maxLoot(hval, n-1, dp);` `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */`   `// Java program to find the maximum stolen value` `import` `java.io.*;` `import` `java.util.Arrays;`   `class` `GFG {` `  ``// Function to calculate the maximum stolen value` `  ``static` `int` `maxLoot(``int` `hval[], ``int` `n, ``int` `dp[])` `  ``{` `    ``// base case` `    ``if` `(n < ``0``) {` `      ``return` `0``;` `    ``}`   `    ``if` `(n == ``0``) {` `      ``return` `hval[``0``];` `    ``}` `    ``// If the subproblem is already solved` `    ``// then return its value` `    ``if` `(dp[n] != -``1``) {` `      ``return` `dp[n];` `    ``}`   `    ``// if current element is pick then previous cannot` `    ``// be picked` `    ``int` `pick = hval[n] + maxLoot(hval, n - ``2``, dp);` `    ``// if current element is not picked then previous` `    ``// element is picked` `    ``int` `notPick = maxLoot(hval, n - ``1``, dp);`   `    ``// return max of picked and not picked` `    ``return` `dp[n] = Math.max(pick, notPick);` `  ``}`   `  ``// Driver program` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int` `hval[] = { ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `};` `    ``int` `n = hval.length;` `    ``int` `dp[] = ``new` `int``[n + ``1``];` `    ``Arrays.fill(dp, -``1``);` `    ``System.out.println(``"Maximum loot value : "` `                       ``+ maxLoot(hval, n - ``1``, dp));` `  ``}` `}`   `// This code is contributed by sanskar84.`

## Python3

 `# Python3 program to find the maximum stolen value`   `# calculate the maximum stolen value` `def` `maxLoot(hval,n,dp):` `  `  `    ``# base case` `    ``if``(n < ``0``):` `        ``return` `0` `   `  `    ``if``(n ``=``=` `0``):` `        ``return` `hval[``0``]` `   `  `   ``# If the subproblem is already solved` `   ``# then return its value` `    ``if``(dp[n] !``=` `-``1``):` `        ``return` `dp[n]`   `  `  `    ``# if current element is pick then previous cannot be picked` `    ``pick ``=` `hval[n] ``+`  `maxLoot(hval, n``-``2``, dp)` `    `  `    ``# if current element is not picked then previous element is picked` `    ``notPick ``=` `maxLoot(hval, n``-``1``, dp)` `  `  `    ``# return max of picked and not picked` `    ``dp[n] ``=` `max``(pick, notPick) ` `    ``return` `dp[n]` `  `  `# Driver to test above code` `hval ``=` `[``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4``]` `n ``=` `len``(hval)`   `# Initialize a dp vector` `dp ``=` `[``-``1` `for` `i ``in` `range``(n``+``1``)]` `print``(``"Maximum loot possible : "``+``str``(maxLoot(hval, n``-``1``, dp)))`   `# This code is contributed by shinjanpatra`

## Javascript

 ``

Output

`Maximum loot possible : 19`

Complexity Analysis:

Time Complexity: O(n) . Only one traversal of 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).

Method 3: Dynamic Programming : Bottom Up Approach

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

Algorithm:

1. Create an extra space dp, DP array to store the sub-problems.
2. 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 maximum of two elements.
3. Update dp as array and dp as maximum of array and array
4. Traverse the array from the second element (2nd index) to the end of array.
5. For every index, update dp[i] as 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.
6. Print the value dp[n-1]

Implementation:

## C++

 `// CPP program to find the maximum stolen value` `#include ` `using` `namespace` `std;`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``if` `(n == 0)` `        ``return` `0;` `    ``if` `(n == 1)` `        ``return` `hval;` `    ``if` `(n == 2)` `        ``return` `max(hval, hval);`   `    ``// dp[i] represent the maximum value stolen` `    ``// so far after reaching house i.` `    ``int` `dp[n];`   `    ``// Initialize the dp and dp` `    ``dp = hval;` `    ``dp = max(hval, hval);`   `    ``// Fill remaining positions` `    ``for` `(``int` `i = 2; i < n; i++)` `        ``dp[i] = max(hval[i] + dp[i - 2], dp[i - 1]);`   `    ``return` `dp[n - 1];` `}`   `// Driver to test above code` `int` `main()` `{` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``cout << ``"Maximum loot possible : "` `<< maxLoot(hval, n);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to find the maximum stolen value` `#include `   `// Find maximum between two numbers.` `int` `max(``int` `num1, ``int` `num2)` `{` `    ``return` `(num1 > num2) ? num1 : num2;` `}`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``if` `(n == 0)` `        ``return` `0;` `    ``if` `(n == 1)` `        ``return` `hval;` `    ``if` `(n == 2)` `        ``return` `max(hval, hval);`   `    ``// dp[i] represent the maximum value stolen` `    ``// so far after reaching house i.` `    ``int` `dp[n];`   `    ``// Initialize the dp and dp` `    ``dp = hval;` `    ``dp = max(hval, hval);`   `    ``// Fill remaining positions` `    ``for` `(``int` `i = 2; i < n; i++)` `        ``dp[i] = max(hval[i] + dp[i - 2], dp[i - 1]);`   `    ``return` `dp[n - 1];` `}`   `// Driver to test above code` `int` `main()` `{` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``printf``(``"Maximum loot possible : %d"``, maxLoot(hval, n));` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to find the maximum stolen value` `import` `java.io.*;`   `class` `GFG {` `    ``// Function to calculate the maximum stolen value` `    ``static` `int` `maxLoot(``int` `hval[], ``int` `n)` `    ``{` `        ``if` `(n == ``0``)` `            ``return` `0``;` `        ``if` `(n == ``1``)` `            ``return` `hval[``0``];` `        ``if` `(n == ``2``)` `            ``return` `Math.max(hval[``0``], hval[``1``]);`   `        ``// dp[i] represent the maximum value stolen` `        ``// so far after reaching house i.` `        ``int``[] dp = ``new` `int``[n];`   `        ``// Initialize the dp and dp` `        ``dp[``0``] = hval[``0``];` `        ``dp[``1``] = Math.max(hval[``0``], hval[``1``]);`   `        ``// Fill remaining positions` `        ``for` `(``int` `i = ``2``; i < n; i++)` `            ``dp[i]` `                ``= Math.max(hval[i] + dp[i - ``2``], dp[i - ``1``]);`   `        ``return` `dp[n - ``1``];` `    ``}`   `    ``// Driver program` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `hval[] = { ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `};` `        ``int` `n = hval.length;` `        ``System.out.println(``"Maximum loot value : "` `+ maxLoot(hval, n));` `    ``}` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Python3 program to find the maximum stolen value`   `# calculate the maximum stolen value` `def` `maximize_loot(hval, n):` `    ``if` `n ``=``=` `0``:` `        ``return` `0` `    ``if` `n ``=``=` `1``:` `        ``return` `hval[``0``]` `    ``if` `n ``=``=` `2``:` `        ``return` `max``(hval[``0``], hval[``1``])`   `    ``# dp[i] represent the maximum value stolen so` `    ``# for after reaching house i.` `    ``dp ``=` `[``0``]``*``n`   `    ``# Initialize the dp and dp` `    ``dp[``0``] ``=` `hval[``0``]` `    ``dp[``1``] ``=` `max``(hval[``0``], hval[``1``])` `    `  `    ``# Fill remaining positions` `    ``for` `i ``in` `range``(``2``, n):` `        ``dp[i] ``=` `max``(hval[i]``+``dp[i``-``2``], dp[i``-``1``])`   `    ``return` `dp[``-``1``]`   `# Driver to test above code ` `def` `main():`   `    ``# Value of houses` `    ``hval ``=` `[``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4``]`   `    ``# number of houses` `    ``n ``=` `len``(hval)` `    ``print``(``"Maximum loot value : {}"``.` `        ``format``(maximize_loot(hval, n)))`   `if` `__name__ ``=``=` `'__main__'``:` `    ``main()`

## C#

 `// C# program to find the ` `// maximum stolen value` `using` `System;` `        `  `class` `GFG ` `{` `   ``// Function to calculate the ` `   ``// maximum stolen value` `    ``static` `int` `maxLoot(``int` `[]hval, ``int` `n)` `    ``{` `        ``if` `(n == 0)` `        ``return` `0;` `        ``if` `(n == 1)` `            ``return` `hval;` `        ``if` `(n == 2)` `            ``return` `Math.Max(hval, hval);`   `        ``// dp[i] represent the maximum value stolen` `        ``// so far after reaching house i.` `        ``int``[] dp = ``new` `int``[n];`   `        ``// Initialize the dp and dp` `        ``dp = hval;` `        ``dp = Math.Max(hval, hval);`   `        ``// Fill remaining positions` `        ``for` `(``int` `i = 2; i

## PHP

 ``

## Javascript

 ``

Output

`Maximum loot possible : 19`

Complexity Analysis:

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

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

Algorithm:

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 maximum of two elements.
2. Create two variables value1 and value2 value1 as array and value2 as maximum of array and array and a variable max_val to store the answer
3. Traverse the array from the second element (2nd index)  to the end of array.
4. For every index, update max_val as 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

Implementation:

## C++

 `// C++ program to find the maximum stolen value` `#include ` `using` `namespace` `std;`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``if` `(n == 0)` `        ``return` `0;`   `    ``int` `value1 = hval;` `    ``if` `(n == 1)` `        ``return` `value1;`   `    ``int` `value2 = max(hval, hval);` `    ``if` `(n == 2)` `        ``return` `value2;`   `    ``// contains maximum stolen value at the end` `    ``int` `max_val;`   `    ``// Fill remaining positions` `    ``for` `(``int` `i = 2; i < n; i++) {` `        ``max_val = max(hval[i] + value1, value2);` `        ``value1 = value2;` `        ``value2 = max_val;` `    ``}`   `    ``return` `max_val;` `}`   `// Driver to test above code` `int` `main()` `{` `    ``// Value of houses` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``cout << ``"Maximum loot possible : "` `<< maxLoot(hval, n);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to find the maximum stolen value` `#include `   `//Find maximum between two numbers.` `int` `max(``int` `num1, ``int` `num2)` `{` `  ``return` `(num1 > num2) ? num1 : num2;` `}`   `// calculate the maximum stolen value` `int` `maxLoot(``int``* hval, ``int` `n)` `{` `    ``if` `(n == 0)` `        ``return` `0;`   `    ``int` `value1 = hval;` `    ``if` `(n == 1)` `        ``return` `value1;`   `    ``int` `value2 = max(hval, hval);` `    ``if` `(n == 2)` `        ``return` `value2;`   `    ``// contains maximum stolen value at the end` `    ``int` `max_val;`   `    ``// Fill remaining positions` `    ``for` `(``int` `i = 2; i < n; i++) {` `        ``max_val = max(hval[i] + value1, value2);` `        ``value1 = value2;` `        ``value2 = max_val;` `    ``}`   `    ``return` `max_val;` `}`   `// Driver to test above code` `int` `main()` `{` `    ``// Value of houses` `    ``int` `hval[] = { 6, 7, 1, 3, 8, 2, 4 };` `    ``int` `n = ``sizeof``(hval) / ``sizeof``(hval);` `    ``printf``(``"Maximum loot possible : %d"``, maxLoot(hval, n));` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to find the maximum stolen value` `import` `java.io.*;`   `class` `GFG {` `    ``// Function to calculate the maximum stolen value` `    ``static` `int` `maxLoot(``int` `hval[], ``int` `n)` `    ``{` `        ``if` `(n == ``0``)` `            ``return` `0``;`   `        ``int` `value1 = hval[``0``];` `        ``if` `(n == ``1``)` `            ``return` `value1;`   `        ``int` `value2 = Math.max(hval[``0``], hval[``1``]);` `        ``if` `(n == ``2``)` `            ``return` `value2;`   `        ``// contains maximum stolen value at the end` `        ``int` `max_val = ``0``;`   `        ``// Fill remaining positions` `        ``for` `(``int` `i = ``2``; i < n; i++) {` `            ``max_val = Math.max(hval[i] + value1, value2);` `            ``value1 = value2;` `            ``value2 = max_val;` `        ``}`   `        ``return` `max_val;` `    ``}`   `    ``// driver program` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `hval[] = { ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `};` `        ``int` `n = hval.length;` `        ``System.out.println(``"Maximum loot value : "` `                           ``+ maxLoot(hval, n));` `    ``}` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Python3 program to find the maximum stolen value`   `# calculate the maximum stolen value` `def` `maximize_loot(hval, n):` `    ``if` `n ``=``=` `0``:` `        ``return` `0`   `    ``value1 ``=` `hval[``0``]` `    ``if` `n ``=``=` `1``:` `        ``return` `value1`   `    ``value2 ``=` `max``(hval[``0``], hval[``1``])` `    ``if` `n ``=``=` `2``:` `        ``return` `value2`   `    ``# contains maximum stolen value at the end` `    ``max_val ``=` `None`   `    ``# Fill remaining positions` `    ``for` `i ``in` `range``(``2``, n):` `        ``max_val ``=` `max``(hval[i]``+``value1, value2)` `        ``value1 ``=` `value2` `        ``value2 ``=` `max_val`   `    ``return` `max_val`   `# Driver to test above code ` `def` `main():`   `    ``# Value of houses` `    ``hval ``=` `[``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4``]`   `    ``# number of houses` `    ``n ``=` `len``(hval)` `    ``print``(``"Maximum loot value : {}"``.``format``(maximize_loot(hval, n)))`   `if` `__name__ ``=``=` `'__main__'``:` `    ``main()`

## C#

 `// C# program to find the ` `// maximum stolen value` `using` `System;` `        `  `public` `class` `GFG ` `{` `    ``// Function to calculate the ` `    ``// maximum stolen value` `    ``static` `int` `maxLoot(``int` `[]hval, ``int` `n)` `    ``{` `        ``if` `(n == 0)` `        ``return` `0;`   `        ``int` `value1 = hval;` `        ``if` `(n == 1)` `            ``return` `value1;`   `        ``int` `value2 = Math.Max(hval, hval);` `        ``if` `(n == 2)` `            ``return` `value2;` `    `  `        ``// contains maximum stolen value at the end` `        ``int` `max_val = 0;`   `        ``// Fill remaining positions` `        ``for` `(``int` `i = 2; i < n; i++)` `        ``{` `            ``max_val = Math.Max(hval[i] + value1, value2);` `            ``value1 = value2;` `            ``value2 = max_val;` `        ``}`   `        ``return` `max_val;` `    ``}` `    `  `    ``// Driver program` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `[]hval = {6, 7, 1, 3, 8, 2, 4};` `        ``int` `n = hval.Length;` `        ``Console.WriteLine(``"Maximum loot value : "` `+` `                                 ``maxLoot(hval, n));` `    ``}` `}`   `// This code is contributed by Sam007`

## PHP

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## Javascript

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Output

`Maximum loot possible : 19`

Complexity Analysis:

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

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