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Maximum subsequence sum such that no three are consecutive

Given a sequence of positive numbers, find the maximum sum that can be formed which has no three consecutive elements present.

Examples :

Input: arr[] = {1, 2, 3}
Output: 5
We can't take three of them, so answer is
2 + 3 = 5

Input: arr[] = {3000, 2000, 1000, 3, 10}
Output: 5013 
3000 + 2000 + 3 + 10 = 5013

Input: arr[] = {100, 1000, 100, 1000, 1}
Output: 2101
100 + 1000 + 1000 + 1 = 2101

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

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 27


This problem is mainly an extension of below problem.

Maximum sum such that no two elements are adjacent

We maintain an auxiliary array sum[] (of same size as input array) to find the result.

sum[i] : Stores result for subarray arr[0..i], i.e.,
         maximum possible sum in subarray arr[0..i]
         such that no three elements are consecutive.

sum[0] = arr[0]

// Note : All elements are positive
sum[1] = arr[0] + arr[1]

// We have three cases
// 1) Exclude arr[2], i.e., sum[2] = sum[1]
// 2) Exclude arr[1], i.e., sum[2] = sum[0] + arr[2]
// 3) Exclude arr[0], i.e., sum[2] = arr[1] + arr[2] 
sum[2] = max(sum[1], arr[0] + arr[2], arr[1] + arr[2])

In general,
// We have three cases
// 1) Exclude arr[i],  i.e.,  sum[i] = sum[i-1]
// 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
// 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1] 
sum[i] = max(sum[i-1], sum[i-2] + arr[i],
             sum[i-3] + arr[i] + arr[i-1])

Below is implementation of above idea.

C/C++

// C++ program to find the maximum sum such that
// no three are consecutive
#include <bits/stdc++.h>
using namespace std;
  
// Returns maximum subsequence sum such that no three
// elements are consecutive
int maxSumWO3Consec(int arr[], int n)
{
    // Stores result for subarray arr[0..i], i.e.,
    // maximum possible sum in subarray arr[0..i]
    // such that no three elements are consecutive.
    int sum[n];
  
    // Base cases (process first three elements)
    if (n >= 1)
        sum[0] = arr[0];
  
    if (n >= 2)
        sum[1] = arr[0] + arr[1];
  
    if (n > 2)
        sum[2] = max(sum[1], max(arr[1] + arr[2], arr[0] + arr[2]));
  
    // Process rest of the elements
    // We have three cases
    // 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
    // 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
    // 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
    for (int i = 3; i < n; i++)
        sum[i] = max(max(sum[i - 1], sum[i - 2] + arr[i]),
                     arr[i] + arr[i - 1] + sum[i - 3]);
  
    return sum[n - 1];
}
  
// Driver code
int main()
{
    int arr[] = { 100, 1000 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxSumWO3Consec(arr, n);
    return 0;
}

Java

// java program to find the maximum sum
// such that no three are consecutive
import java.io.*;
  
class GFG {
  
    // Returns maximum subsequence sum such that no three
    // elements are consecutive
    static int maxSumWO3Consec(int arr[], int n)
    {
        // Stores result for subarray arr[0..i], i.e.,
        // maximum possible sum in subarray arr[0..i]
        // such that no three elements are consecutive.
        int sum[] = new int[n];
  
        // Base cases (process first three elements)
        if (n >= 1)
            sum[0] = arr[0];
  
        if (n >= 2)
            sum[1] = arr[0] + arr[1];
  
        if (n > 2)
            sum[2] = Math.max(sum[1], Math.max(arr[1] + arr[2], arr[0] + arr[2]));
  
        // Process rest of the elements
        // We have three cases
        // 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
        // 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
        // 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
        for (int i = 3; i < n; i++)
            sum[i] = Math.max(Math.max(sum[i - 1], sum[i - 2] + arr[i]),
                              arr[i] + arr[i - 1] + sum[i - 3]);
  
        return sum[n - 1];
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 100, 1000, 100, 1000, 1 };
        int n = arr.length;
        System.out.println(maxSumWO3Consec(arr, n));
    }
}
  
// This code is contributed by vt_m

Python

# Python program to find the maximum sum such that
# no three are consecutive
  
# Returns maximum subsequence sum such that no three
# elements are consecutive
def maxSumWO3Consec(arr, n):
    # Stores result for subarray arr[0..i], i.e.,
    # maximum possible sum in subarray arr[0..i]
    # such that no three elements are consecutive.
    sum = [0 for k in range(n)]
  
    # Base cases (process first three elements)
      
    if n >= 1 :
        sum[0] = arr[0]
      
    if n >= 2 :
        sum[1] = arr[0] + arr[1]
      
    if n > 2 :
        sum[2] = max(sum[1], max(arr[1] + arr[2], arr[0] + arr[2]))
  
    # Process rest of the elements
    # We have three cases
    # 1) Exclude arr[i], i.e., sum[i] = sum[i-1]
    # 2) Exclude arr[i-1], i.e., sum[i] = sum[i-2] + arr[i]
    # 3) Exclude arr[i-2], i.e., sum[i-3] + arr[i] + arr[i-1]
    for i in range(3, n):
        sum[i] = max(max(sum[i-1], sum[i-2] + arr[i]), arr[i] + arr[i-1] + sum[i-3])
  
    return sum[n-1]
  
# Driver code
arr = [100, 1000, 100, 1000, 1]
n = len(arr)
print maxSumWO3Consec(arr, n)
  
# This code is contributed by Afzal Ansari

C#

// C# program to find the maximum sum
// such that no three are consecutive
using System;
class GFG {
  
    // Returns maximum subsequence
    // sum such that no three
    // elements are consecutive
    static int maxSumWO3Consec(int[] arr,
                               int n)
    {
        // Stores result for subarray
        // arr[0..i], i.e., maximum
        // possible sum in subarray
        // arr[0..i] such that no
        // three elements are consecutive.
        int[] sum = new int[n];
  
        // Base cases (process
        // first three elements)
        if (n >= 1)
            sum[0] = arr[0];
  
        if (n >= 2)
            sum[1] = arr[0] + arr[1];
  
        if (n > 2)
            sum[2] = Math.Max(sum[1], Math.Max(arr[1] + arr[2], arr[0] + arr[2]));
  
        // Process rest of the elements
        // We have three cases
        // 1) Exclude arr[i], i.e.,
        // sum[i] = sum[i-1]
        // 2) Exclude arr[i-1], i.e.,
        // sum[i] = sum[i-2] + arr[i]
        // 3) Exclude arr[i-2], i.e.,
        // sum[i-3] + arr[i] + arr[i-1]
        for (int i = 3; i < n; i++)
            sum[i] = Math.Max(Math.Max(sum[i - 1],
                                       sum[i - 2] + arr[i]),
                              arr[i] + arr[i - 1] + sum[i - 3]);
  
        return sum[n - 1];
    }
  
    // Driver code
    public static void Main()
    {
        int[] arr = { 100, 1000, 100, 1000, 1 };
        int n = arr.Length;
        Console.Write(maxSumWO3Consec(arr, n));
    }
}
  
// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to find the maximum
// sum such that no three are consecutive
  
// Returns maximum subsequence
// sum such that no three
// elements are consecutive
function maxSumWO3Consec($arr, $n)
{
    // Stores result for subarray
    // arr[0..i], i.e., maximum 
    // possible sum in subarray
    // arr[0..i] such that no three 
    // elements are consecutive$.
    $sum = array();
  
    // Base cases (process 
    // first three elements)
    if ( $n >= 1)
    $sum[0] = $arr[0];
      
    if ($n >= 2)
    $sum[1] = $arr[0] + $arr[1];
      
    if ( $n > 2)
    $sum[2] = max($sum[1], max($arr[1] + $arr[2], 
                            $arr[0] + $arr[2]));
  
    // Process rest of the elements
    // We have three cases
    // 1) Exclude arr[i], i.e., 
    // sum[i] = sum[i-1]
    // 2) Exclude arr[i-1], i.e., 
    // sum[i] = sum[i-2] + arr[i]
    // 3) Exclude arr[i-2], i.e., 
    // sum[i-3] + arr[i] + arr[i-1]
    for ($i = 3; $i < $n; $i++)
        $sum[$i] = max(max($sum[$i - 1], 
                        $sum[$i - 2] + $arr[$i]), 
                        $arr[$i] + $arr[$i - 1] + 
                                    $sum[$i - 3]);
  
    return $sum[$n-1];
}
  
// Driver code
$arr = array(100, 1000, 100, 1000, 1);
$n =count($arr);
echo maxSumWO3Consec($arr, $n);
  
// This code is contributed by anuj_67.
?>

Output :

2101

Time Complexity : O(n)
Auxiliary Space : O(n)

Another approach: (Using recursion)

C++

// C++ program to find the maximum sum such that 
// no three are consecutive using recursion.
#include<bits/stdc++.h>
using namespace std;
  
int arr[] = {100, 1000, 100, 1000, 1};
int sum[10000];
  
// Returns maximum subsequence sum such that no three 
// elements are consecutive 
int maxSumWO3Consec(int n)
{
    if(sum[n]!=-1)
    return sum[n];
      
    //Base cases (process first three elements) 
      
    if(n==0)
    return sum[n] = 0;
      
    if(n==1)
    return sum[n] = arr[0];
      
    if(n==2)
    return sum[n] = arr[1]+arr[0];
      
    // Process rest of the elements 
    // We have three cases 
    return sum[n] = max(max(maxSumWO3Consec(n-1),
                    maxSumWO3Consec(n-2) + arr[n-1]), 
                    arr[n-2] + arr[n-1] + maxSumWO3Consec(n-3));
      
      
}
  
// Driver code
int main()
{
      
    int n = sizeof(arr) / sizeof(arr[0]);
    memset(sum,-1,sizeof(sum));
    cout << maxSumWO3Consec(n); 
  
// this code is contributed by Kushdeep Mittal
    return 0; 
}

Java

// Java program to find the maximum 
// sum such that no three are
// consecutive using recursion. 
import java.util.Arrays;
  
class GFG 
{
      
static int arr[] = {100, 1000, 100, 1000, 1}; 
static int sum[] = new int[10000]; 
  
// Returns maximum subsequence 
// sum such that no three 
// elements are consecutive 
static int maxSumWO3Consec(int n) 
    if(sum[n] != -1
        return sum[n]; 
      
    //Base cases (process first three elements) 
      
    if(n == 0
        return sum[n] = 0
      
    if(n == 1
        return sum[n] = arr[0]; 
      
    if(n == 2
        return sum[n] = arr[1] + arr[0]; 
      
    // Process rest of the elements 
    // We have three cases 
    return sum[n] = Math.max(Math.max(maxSumWO3Consec(n - 1), 
                    maxSumWO3Consec(n - 2) + arr[n - 1]), 
                    arr[n - 2] + arr[n - 1] + maxSumWO3Consec(n - 3)); 
      
      
  
// Driver code 
public static void main(String[] args) 
{
    int n = arr.length; 
        Arrays.fill(sum, -1);
    System.out.println(maxSumWO3Consec(n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the maximum 
# sum such that no three are consecutive
# using recursion. 
arr = [100, 1000, 100, 1000, 1]
sum = [-1] * 10000
  
# Returns maximum subsequence sum such 
# that no three elements are consecutive 
def maxSumWO3Consec(n) :
  
    if(sum[n] != -1):
        return sum[n]
      
    # 3 Base cases (process first 
    # three elements) 
    if(n == 0) : 
        sum[n] = 0
        return sum[n]
      
    if(n == 1) : 
        sum[n] = arr[0]
        return sum[n]
      
    if(n == 2) : 
        sum[n] = arr[1] + arr[0
        return sum[n]
      
    # Process rest of the elements 
    # We have three cases 
    sum[n] = max(max(maxSumWO3Consec(n - 1), 
                     maxSumWO3Consec(n - 2) + arr[n - 1]), 
                     arr[n - 2] + arr[n - 1] + 
                     maxSumWO3Consec(n - 3))
      
    return sum[n]
  
# Driver code 
if __name__ == "__main__" :
  
    n = len(arr)
      
    print(maxSumWO3Consec(n))
  
# This code is contributed by Ryuga 

C#

// C# program to find the maximum 
// sum such that no three are 
// consecutive using recursion.
using System;
  
class GFG 
  
    static int []arr = {100, 1000,
                        100, 1000, 1}; 
    static int []sum = new int[10000]; 
  
    // Returns maximum subsequence 
    // sum such that no three 
    // elements are consecutive 
    static int maxSumWO3Consec(int n) 
    
        if(sum[n] != -1) 
            return sum[n]; 
  
        //Base cases (process first
        // three elements) 
        if(n == 0) 
            return sum[n] = 0; 
  
        if(n == 1) 
            return sum[n] = arr[0]; 
  
        if(n == 2) 
            return sum[n] = arr[1] + arr[0]; 
  
        // Process rest of the elements 
        // We have three cases 
        return sum[n] = Math.Max(Math.Max(maxSumWO3Consec(n - 1), 
                        maxSumWO3Consec(n - 2) + arr[n - 1]), 
                        arr[n - 2] + arr[n - 1] + maxSumWO3Consec(n - 3)); 
  
  
    
  
    // Driver code 
    public static void Main(String[] args) 
    
        int n = arr.Length; 
        for(int i = 0; i < sum.Length; i++)
            sum[i] = -1;
        Console.WriteLine(maxSumWO3Consec(n)); 
    
}
  
// This code is contributed by 29AjayKumar

PHP

Output :

2101

Time Complexity : O(n)
Auxiliary Space : O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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