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Find maximum possible stolen value from houses
  • Difficulty Level : Medium
  • Last Updated : 03 May, 2021

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.

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[0] as array[0] and dp[1] as maximum of array[0] and array[1]
  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 <iostream>
using namespace std;
 
// calculate the maximum stolen value
int maxLoot(int *hval, int 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 far after reaching house i.
    int dp[n];
 
    // Initialize the dp[0] and dp[1]
    dp[0] = hval[0];
    dp[1] = max(hval[0], hval[1]);
 
    // 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[0]);
    cout << "Maximum loot possible : "
        << maxLoot(hval, n);
    return 0;
}

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[0] and dp[1]
        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));
    }
}
 
// Contributed by Pramod Kumar

Python




# 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[0] and dp[1]
    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[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[0] and dp[1]
        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 ()
    {
        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




<?php
// PHP program to find
// the maximum stolen value
 
// calculate the maximum
// stolen value
function maxLoot($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 far after
    // reaching house i.
    $dp = array();
 
    // Initialize the
    // dp[0] and dp[1]
    $dp[0] = $hval[0];
    $dp[1] = max($hval[0],
                 $hval[1]);
 
    // Fill remaining positions
    for ($i = 2; $i < $n; $i++)
        $dp[$i] = max($hval[$i] +
                      $dp[$i - 2],
                      $dp[$i - 1]);
 
    return $dp[$n - 1];
}
 
// Driver Code
$hval = array(6, 7, 1,
              3, 8, 2, 4);
$n = sizeof($hval);
echo "Maximum loot possible : ",
             maxLoot($hval, $n);
     
// This code is contributed by aj_36
?>

Javascript




<script>
 
    // Javascript program to find
    // the maximum stolen value
     
    // Function to calculate the
       // maximum stolen value
    function maxLoot(hval, 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.
        let dp = new Array(n);
   
        // Initialize the dp[0] and dp[1]
        dp[0] = hval[0];
        dp[1] = Math.max(hval[0], hval[1]);
   
        // Fill remaining positions
        for (let i = 2; i<n; i++)
            dp[i] = Math.max(hval[i]+dp[i-2], dp[i-1]);
   
        return dp[n-1];
    }
     
    let hval = [6, 7, 1, 3, 8, 2, 4];
    let n = hval.length;
    document.write(
    "Maximum loot value : " + maxLoot(hval, n)
    );
     
</script>

Output: 

Maximum loot value : 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)
    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[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 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 <iostream>
using namespace std;
 
// calculate the maximum stolen value
int maxLoot(int *hval, int n)
{
    if (n == 0)
        return 0;
 
    int value1 = hval[0];
    if (n == 1)
        return value1;
 
    int value2 = max(hval[0], hval[1]);
    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[0]);
    cout << "Maximum loot possible : "
        << maxLoot(hval, n);
    return 0;
}

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));
    }
}
 
// Contributed by Pramod kumar

Python




# 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[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 ()
    {
        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




<?php
// PHP program to find
// the maximum stolen value
 
// calculate the
// maximum stolen value
function maxLoot($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;
 
    // Fill remaining positions
    for ($i = 2; $i < $n; $i++)
    {
        $max_val = max($hval[$i] +
                       $value1, $value2);
        $value1 = $value2;
        $value2 = $max_val;
    }
 
    return $max_val;
}
 
// Driver code
$hval = array(6, 7, 1, 3, 8, 2, 4);
$n = sizeof($hval);
echo "Maximum loot value : ",
          maxLoot($hval, $n);
     
// This code is contributed by ajit
?>

Javascript




<script>
 
    // Javascript program to find the
    // maximum stolen value
     
    // Function to calculate the
    // maximum stolen value
    function maxLoot(hval, n)
    {
        if (n == 0)
            return 0;
  
        let value1 = hval[0];
        if (n == 1)
            return value1;
  
        let value2 = Math.max(hval[0], hval[1]);
        if (n == 2)
            return value2;
      
        // contains maximum stolen value at the end
        let max_val = 0;
  
        // Fill remaining positions
        for (let i = 2; i < n; i++)
        {
            max_val = Math.max(hval[i] +
                               value1, value2);
            value1 = value2;
            value2 = max_val;
        }
  
        return max_val;
    }
     
    let hval = [6, 7, 1, 3, 8, 2, 4];
    let n = hval.length;
    document.write("Maximum loot value : "
    + maxLoot(hval, n));
     
</script>

Output: 

Maximum loot value : 19

Complexity Analysis: 

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

This article is contributed by Atul Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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