Maximum sum such that exactly half of the elements are selected and no two adjacent

Given an array A containing N integers. Find the maximum sum possible such that exact floor(N/2) elements are selected and no two selected elements are adjacent to each other. (if N = 5, then exactly 2 elements should be selected as floor(5/2) = 2)
For a simpler version of this problem check out this.

Examples:

Input: A = [1, 2, 3, 4, 5, 6]
Output: 12
Explanation:
Select 2, 4 and 6 making the sum 12.

Input : A = [-1000, -100, -10, 0, 10]
Output : 0
Explanation:
Select -10 and 10, making the sum 0.

Approach:

We will use the concept of dynamic programming. The following is how the dp states are defined:

dp[i][j] = maximum sum till index i such that j elements are selected

Since no two adjacent elements can be selected :

dp[i][j] = max(a[i] + dp[i-2][j-1], dp[i-1][j])

Below is the implementation of the above approach.

C++

// C++ program to find maximum sum possible
// such that exactly floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other
#include <bits/stdc++.h>

using namespace std;
 
// Function return the maximum sum 
// possible under given condition

int MaximumSum(int a[], int n)
{

    int dp[n + 1][n + 1];
 
    // Intitialising the dp table

    for (int i = 0; i < n + 1; i++)

    {

        for (int j = 0; j < n + 1; j++)

            dp[i][j] = INT_MIN;

    }
 
    // Base case

    for (int i = 0; i < n + 1; i++)

        dp[i][0] = 0;
 
    for (int i = 1; i <= n; i++) 

    {

        for (int j = 1; j <= i; j++) 

        {

            int val = INT_MIN;
 
            // Condition to select the current 

            // element

            if ((i - 2 >= 0

                 && dp[i - 2][j - 1] != INT_MIN)

                                   || i - 2 < 0)

            {

                val = a[i - 1] + 

                    (i - 2 >= 0 ? 

                     dp[i - 2][j - 1] : 0);

            }

            // Condition to not select the 

            // current element if possible

            if (i - 1 >= j)

            {

                val = max(val, dp[i - 1][j]);

            }

            dp[i][j] = val;

        }

    }
 
    return dp[n][n / 2];
}
 
//Driver code

int main()
{

    int A[] = {1, 2, 3, 4, 5, 6};

    int N = sizeof(A) / sizeof(A[0]);
 
    cout << MaximumSum(A, N);
 
    return 0;
}

Java

// Java program to find maximum sum possible
// such that exactly Math.floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other

class GFG{
 
// Function return the maximum sum 
// possible under given condition

static int MaximumSum(int a[], int n)
{

    int [][]dp = new int[n + 1][n + 1];
 
    // Intitialising the dp table

    for(int i = 0; i < n + 1; i++)

    {

       for(int j = 0; j < n + 1; j++)

          dp[i][j] = Integer.MIN_VALUE;

    }
 
    // Base case

    for(int i = 0; i < n + 1; i++)

       dp[i][0] = 0;
 
    for(int i = 1; i <= n; i++) 

    {

       for(int j = 1; j <= i; j++) 

       {

          int val = Integer.MIN_VALUE;

          // Condition to select the current 

          // element

          if ((i - 2 >= 0 && 

            dp[i - 2][j - 1] != Integer.MIN_VALUE) || 

               i - 2 < 0)

          {

              val = a[i - 1] + (i - 2 >= 0 ? 

                             dp[i - 2][j - 1] : 0);

          }

          // Condition to not select the 

          // current element if possible

          if (i - 1 >= j)

          {

              val = Math.max(val, dp[i - 1][j]);

          }

          dp[i][j] = val;

        }

    }

    return dp[n][n / 2];
}
 
// Driver code

public static void main(String[] args)
{

    int A[] = { 1, 2, 3, 4, 5, 6 };

    int N = A.length;
 
    System.out.print(MaximumSum(A, N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find maximum sum possible
# such that exactly floor(N/2) elements 
# are selected and no two selected 
# elements are adjacent to each other
 
import sys
# Function return the maximum sum 
# possible under given condition

def MaximumSum(a,n):

    dp = [[0 for i in range(n+1)] for j in range(n+1)]
 
    # Intitialising the dp table

    for i in range(n + 1):

        for j in range(n + 1):

            dp[i][j] = -sys.maxsize-1
 
    # Base case

    for i in range(n+1):

        dp[i][0] = 0
 
    for i in range(1,n+1,1):

        for j in range(1,i+1,1):

            val = -sys.maxsize-1
 
            # Condition to select the current 

            # element

            if ((i - 2 >= 0 and dp[i - 2][j - 1] != -sys.maxsize-1) or i - 2 < 0):

                if (i - 2 >= 0):

                    val = a[i - 1] + dp[i - 2][j - 1] 

                else:

                    val = a[i - 1] 

            # Condition to not select the 

            # current element if possible

            if (i - 1 >= j):

                val = max(val, dp[i - 1][j])

            dp[i][j] = val
 
    return dp[n][n // 2]
 
#Driver code

if __name__ == '__main__':

    A = [1, 2, 3, 4, 5, 6]

    N = len(A)
 
    print(MaximumSum(A,N))
 
# This code is contributed by Bhupendra_Singh

// C# program to find maximum sum possible
// such that exactly Math.floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other

using System;
 
class GFG{
 
// Function return the maximum sum 
// possible under given condition

static int MaximumSum(int []a, int n)
{

    int [,]dp = new int[n + 1, n + 1];
 
    // Intitialising the dp table

    for(int i = 0; i < n + 1; i++)

    {

       for(int j = 0; j < n + 1; j++)

          dp[i, j] = Int32.MinValue;

    }
 
    // Base case

    for(int i = 0; i < n + 1; i++)

       dp[i, 0] = 0;
 
    for(int i = 1; i <= n; i++) 

    {

       for(int j = 1; j <= i; j++) 

       {

          int val = Int32.MinValue;

          // Condition to select the current 

          // element

          if ((i - 2 >= 0 && 

            dp[i - 2, j - 1] != Int32.MinValue) ||

               i - 2 < 0)

          {

              val = a[i - 1] + (i - 2 >= 0 ? 

                             dp[i - 2, j - 1] : 0);

          }

          // Condition to not select the 

          // current element if possible

          if (i - 1 >= j)

          {

              val = Math.Max(val, dp[i - 1, j]);

          }

          dp[i, j] = val;

       }

    }

    return dp[n, n / 2];
}
 
// Driver code

public static void Main()
{

    int []A = { 1, 2, 3, 4, 5, 6 };

    int N = A.Length;
 
    Console.Write(MaximumSum(A, N));
}
}
 
// This code is contributed by Nidhi_biet

Javascript

<script>
 
// JavaScript program to find maximum sum possible
// such that exactly Math.floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other
 
// Function return the maximum sum 
// possible under given condition

function MaximumSum(a,n)
{

    let dp = [];
 
    for(let i=0;i<n+1;i++) dp.push(Array(n+1));
 
    // Intitialising the dp table

    for(let i = 0; i < n + 1; i++)

    {

       for(let j = 0; j < n + 1; j++)

          dp[i][j] = -10000;

    }
 
    // Base case

    for(let i = 0; i < n + 1; i++)

       dp[i][0] = 0;
 
    for(let i = 1; i <= n; i++) 

    {

       for(let j = 1; j <= i; j++) 

       {

          let val = -10000;

          // Condition to select the current 

          // element

          if ((i - 2 >= 0 && 

            dp[i - 2, j - 1] != -10000) ||

               i - 2 < 0)

          {

              val = a[i - 1] + (i - 2 >= 0 ? 

                             dp[i - 2][j - 1] : 0);

          }

          // Condition to not select the 

          // current element if possible

          if (i - 1 >= j)

          {

              val = Math.max(val, dp[i - 1][j]);

          }

          dp[i][j] = val;

       }

    }

    return dp[n][n / 2]
}
 
// Driver code
 
let A = [1, 2, 3, 4, 5, 6];
let N = A.length;
 
document.write(MaximumSum(A, N));
 
// This code is contributed by Nidhi_biet
 
</script>

Output:

Time Complexity : O(N²)

Efficient Approach

We will use dynamic programming but with slightly modified states. Storing both the index and is number of elements taken till now are futile since we always need to take exact floor(i/2) elements, so at i’th position for the dp storage we will assume floor(i/2) elements in the subset till now.
The following is the dp table states:

dp[i][1] = maximum sum till i’th index, picking element a[i], with floor(i/2) elements, none adjacent to each other.
dp[i][0] = maximum sum till i’th index, not picking element a[i], with floor(i/2) elements, none adjacent to each other.

We have two cases :
- When i is odd : If we have to pick a[i], then we can’t pick a[i-1], so the only options left are (i – 2)th and (i – 3) rd state (because floor((i – 2)/2) = floor((i – 3)/2) = floor(i/2) – 1, and since we pick a[i], total picked elements will be floor(i/2) ). If we don’t pick a[i], then a sum will be formed by taking a[i-1] and using states i – 1, i – 2 and i – 3 or a[i – 2] using state i – 3 as these will only give the total of floor(i/2).

dp[i][1] = arr[i] + max({dp[i – 3][1], dp[i – 3][0], dp[i – 2][1], dp[i – 2][0]})
dp[i][0] = max({arr[i – 1] + dp[i – 2][0], arr[i – 1] + dp[i – 3][1], arr[i – 1] + dp[i – 3][0],
arr[i – 2] + dp[i – 3][0]})

When i is even : If we pick a[i], then using states i – 1 and i – 2, else picking a[i – 1] and using state i – 2.

dp[i][1] = arr[i] + max({dp[i – 2][1], dp[i – 2][0], dp[i – 1][0]})
dp[i][0] = arr[i – 1] + dp[i – 2][0]

Below is the implementation of the above approach.

C++

// C++ program to find maximum sum possible
// such that exactly floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other
#include <bits/stdc++.h>

using namespace std;
 
// Function return the maximum sum 
// possible under given condition

int MaximumSum(int a[], int n)
{
 
    int dp[n + 1][2];

    // Intitialising the dp table

    memset(dp, 0, sizeof(dp));
 
    // Base case

    dp[2][1] = a[1];

    dp[2][0] = a[0];
 
    for (int i = 3; i < n + 1; i++) {

        // When i is odd

        if (i & 1) {

            int temp = max({ dp[i - 3][1],

                             dp[i - 3][0], 

                             dp[i - 2][1],

                             dp[i - 2][0] });

            dp[i][1] = a[i - 1] + temp;

            dp[i][0] = max({ a[i - 2] + dp[i - 2][0],

                             a[i - 2] + dp[i - 3][1],

                             a[i - 2] + dp[i - 3][0],

                             a[i - 3] + dp[i - 3][0] });

        }
 
        // When i is even

        else {

            dp[i][1] = a[i - 1] + max({ dp[i - 2][1],

                                        dp[i - 2][0],

                                        dp[i - 1][0] });

            dp[i][0] = a[i - 2] + dp[i - 2][0];

        }

    }

    // Maximum of if we pick last element or not

    return max(dp[n][1], dp[n][0]);
}
 
// Driver code

int main()
{

    int A[] = {1, 2, 3, 4, 5, 6};

    int N = sizeof(A) / sizeof(A[0]);
 
    cout << MaximumSum(A, N);
 
    return 0;
}

Java

// Java program to find maximum sum possible
// such that exactly Math.floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other

import java.util.*;
 
class GFG{
 
// Function return the maximum sum 
// possible under given condition

static int MaximumSum(int a[], int n)
{
 
    int [][]dp = new int[n + 1][2];
 
    // Base case

    dp[2][1] = a[1];

    dp[2][0] = a[0];
 
    for(int i = 3; i < n + 1; i++)

    {

       // When i is odd

       if (i % 2 == 1)

       {

           int temp = Math.max((Math.max(dp[i - 3][1], 

                                         dp[i - 3][0])), 

                                Math.max(dp[i - 2][1], 

                                         dp[i - 2][0]));

           dp[i][1] = a[i - 1] + temp;

           dp[i][0] = Math.max((Math.max(a[i - 2] + 

                                        dp[i - 2][0], 

                                         a[i - 2] + 

                                        dp[i - 3][1])),

                                Math.max(a[i - 2] + 

                                        dp[i - 3][0], 

                                         a[i - 3] + 

                                        dp[i - 3][0]));

       }

       // When i is even

       else

       {

           dp[i][1] = a[i - 1] + (Math.max(( 

                                  Math.max(dp[i - 2][1], 

                                           dp[i - 2][0])), 

                                           dp[i - 1][0]));

           dp[i][0] = a[i - 2] + dp[i - 2][0];

       }

    }

    // Maximum of if we pick last element or not

    return Math.max(dp[n][1], dp[n][0]);
}
 
static int max(int []arr)
{

    return 1;
}
 
// Driver code

public static void main(String[] args)
{

    int A[] = {1, 2, 3, 4, 5, 6};

    int N = A.length;
 
    System.out.print(MaximumSum(A, N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find maximum sum possible
# such that exactly floor(N/2) elements 
# are selected and no two selected 
# elements are adjacent to each other
 
# Function return the maximum sum 
# possible under given condition

def MaximumSum(a, n):
 
    dp = [[0 for x in range (2)]

             for y in range(n + 1)]

    # Base case

    dp[2][1] = a[1]

    dp[2][0] = a[0]
 
    for i in range (3, n + 1):

        # When i is odd

        if (i & 1):

            temp = max([dp[i - 3][1],

                        dp[i - 3][0], 

                        dp[i - 2][1],

                        dp[i - 2][0]])

            dp[i][1] = a[i - 1] + temp

            dp[i][0] = max([a[i - 2] + dp[i - 2][0],

                            a[i - 2] + dp[i - 3][1],

                            a[i - 2] + dp[i - 3][0],

                            a[i - 3] + dp[i - 3][0]])
 
        # When i is even

        else:

            dp[i][1] = (a[i - 1] + max([dp[i - 2][1],

                                        dp[i - 2][0],

                                        dp[i - 1][0]]))

            dp[i][0] = a[i - 2] + dp[i - 2][0]

    # Maximum of if we pick last 

    # element or not

    return max(dp[n][1], dp[n][0])
 
# Driver code

if __name__ == "__main__":  

    A = [1, 2, 3, 4, 5, 6]    

    N = len(A)

    print(MaximumSum(A, N))
 
# This code is contributed by Chitranayal

// C# program to find maximum sum possible
// such that exactly Math.Floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other

using System;
 
class GFG{
 
// Function return the maximum sum 
// possible under given condition

static int MaximumSum(int []a, int n)
{

    int [,]dp = new int[n + 1, 2];
 
    // Base case

    dp[2, 1] = a[1];

    dp[2, 0] = a[0];
 
    for(int i = 3; i < n + 1; i++)

    {
 
       // When i is odd

       if (i % 2 == 1)

       {

           int temp = Math.Max((Math.Max(dp[i - 3, 1], 

                                         dp[i - 3, 0])), 

                                Math.Max(dp[i - 2, 1], 

                                         dp[i - 2, 0]));

           dp[i, 1] = a[i - 1] + temp;

           dp[i, 0] = Math.Max((Math.Max(a[i - 2] + 

                                        dp[i - 2, 0], 

                                         a[i - 2] + 

                                        dp[i - 3, 1])),

                                Math.Max(a[i - 2] + 

                                        dp[i - 3, 0], 

                                         a[i - 3] + 

                                        dp[i - 3, 0]));

       }

       // When i is even

       else

       {

           dp[i, 1] = a[i - 1] + (Math.Max((

                                  Math.Max(dp[i - 2, 1], 

                                           dp[i - 2, 0])), 

                                           dp[i - 1, 0]));

           dp[i, 0] = a[i - 2] + dp[i - 2, 0];

       }

    }

    // Maximum of if we pick last element or not

    return Math.Max(dp[n, 1], dp[n, 0]);
}
 
static int max(int []arr)
{

    return 1;
}
 
// Driver code

public static void Main(String[] args)
{

    int []A = { 1, 2, 3, 4, 5, 6 };

    int N = A.Length;
 
    Console.Write(MaximumSum(A, N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program to find maximum sum possible
// such that exactly floor(N/2) elements 
// are selected and no two selected 
// elements are adjacent to each other
 
// Function return the maximum sum 
// possible under given condition

function MaximumSum(a, n)
{
 
    var dp = Array.from(Array(n+1), ()=>Array(2).fill(0));
 
    // Base case

    dp[2][1] = a[1];

    dp[2][0] = a[0];
 
    for (var i = 3; i < n + 1; i++) {

        // When i is odd

        if (i & 1) {

            var temp = ([dp[i - 3][1],

                             dp[i - 3][0], 

                             dp[i - 2][1],

                             dp[i - 2][0] ].reduce((a,b)=>

                               Math.max(a,b)));

            dp[i][1] = a[i - 1] + temp;

            dp[i][0] = ([ a[i - 2] + dp[i - 2][0],

                             a[i - 2] + dp[i - 3][1],

                             a[i - 2] + dp[i - 3][0],

                             a[i - 3] + dp[i - 3][0] ].reduce((a,b)=>

                               Math.max(a,b)));

        }
 
        // When i is even

        else {

            dp[i][1] = a[i - 1] + ([ dp[i - 2][1],

                                        dp[i - 2][0],

                                        dp[i - 1][0] ].reduce((a,b)=>

                                          Math.max(a,b)));

            dp[i][0] = a[i - 2] + dp[i - 2][0];

        }

    }

    // Maximum of if we pick last element or not

    return Math.max(dp[n][1], dp[n][0]);
}
 
// Driver code

var A = [1, 2, 3, 4, 5, 6];
 
var N = A.length;
document.write( MaximumSum(A, N));
 
</script>

Output:

Time Complexity: O(N)

Article Tags :

Algorithms

Arrays

Competitive Programming

DSA

Dynamic Programming