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)

This article is contributed by Roshni Agarwal. 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.



My Personal Notes arrow_drop_up

Improved By : nitin mittal, vt_m




Article Tags :
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



2.8 Average Difficulty : 2.8/5.0
Based on 42 vote(s)






User Actions