Maximum sum such that no two elements are adjacent

Given an array of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array. So 3 2 7 10 should return 13 (sum of 3 and 10) or 3 2 5 10 7 should return 15 (sum of 3, 5 and 7).Answer the question in most efficient way.

Examples :

Input : arr[] = {5, 5, 10, 100, 10, 5}
Output : 110

Input : arr[] = {1, 2, 3}
Output : 4

Input : arr[] = {1, 20, 3}
Output : 20

Algorithm:
Loop for all elements in arr[] and maintain two sums incl and excl where incl = Max sum including the previous element and excl = Max sum excluding the previous element.

Max sum excluding the current element will be max(incl, excl) and max sum including the current element will be excl + current element (Note that only excl is considered because elements cannot be adjacent).

At the end of the loop return max of incl and excl.

Example:

  arr[] = {5,  5, 10, 40, 50, 35}

  inc = 5 
  exc = 0

  For i = 1 (current element is 5)
  incl =  (excl + arr[i])  = 5
  excl =  max(5, 0) = 5

  For i = 2 (current element is 10)
  incl =  (excl + arr[i]) = 15
  excl =  max(5, 5) = 5

  For i = 3 (current element is 40)
  incl = (excl + arr[i]) = 45
  excl = max(5, 15) = 15

  For i = 4 (current element is 50)
  incl = (excl + arr[i]) = 65
  excl =  max(45, 15) = 45

  For i = 5 (current element is 35)
  incl =  (excl + arr[i]) = 80
  excl =  max(65, 45) = 65

And 35 is the last element. So, answer is max(incl, excl) =  80

Thanks to Debanjan for providing code.

Implementation:

C/C++

#include<stdio.h>

/*Function to return max sum such that no two elements
 are adjacent */
int FindMaxSum(int arr[], int n)
{
  int incl = arr[0];
  int excl = 0;
  int excl_new;
  int i;

  for (i = 1; i < n; i++)
  {
     /* current max excluding i */
     excl_new = (incl > excl)? incl: excl;

     /* current max including i */
     incl = excl + arr[i];
     excl = excl_new;
  }

   /* return max of incl and excl */
   return ((incl > excl)? incl : excl);
}

/* Driver program to test above function */
int main()
{
  int arr[] = {5, 5, 10, 100, 10, 5};
  int n = sizeof(arr) / sizeof(arr[0]);
  printf("%d n", FindMaxSum(arr, n));
  return 0;
}

Java

class MaximumSum
{
    /*Function to return max sum such that no two elements
      are adjacent */
    int FindMaxSum(int arr[], int n)
    {
        int incl = arr[0];
        int excl = 0;
        int excl_new;
        int i;

        for (i = 1; i < n; i++)
        {
            /* current max excluding i */
            excl_new = (incl > excl) ? incl : excl;

            /* current max including i */
            incl = excl + arr[i];
            excl = excl_new;
        }

        /* return max of incl and excl */
        return ((incl > excl) ? incl : excl);
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
        MaximumSum sum = new MaximumSum();
        int arr[] = new int[]{5, 5, 10, 100, 10, 5};
        System.out.println(sum.FindMaxSum(arr, arr.length));
    }
}

// This code has been contributed by Mayank Jaiswal

Python

# Function to return max sum such that 
# no two elements are adjacent
def find_max_sum(arr):
    incl = 0
    excl = 0
   
    for i in arr:
        
        # Current max excluding i (No ternary in 
        # Python)
        new_excl = excl if excl>incl else incl
       
        # Current max including i
        incl = excl + i
        excl = new_excl
    
    # return max of incl and excl
    return (excl if excl>incl else incl)

# Driver program to test above function
arr = [5, 5, 10, 100, 10, 5]
print find_max_sum(arr)

# This code is contributed by Kalai Selvan


Output:

110

Time Complexity: O(n)

Refer Find maximum possible stolen value from houses for more explanation.

Now try the same problem for array with negative numbers also.

Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.

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