# Maximum subsequence sum such that no three are consecutive in O(1) space

Given an array A[] of N positive numbers, the task is to find the maximum sum that can be formed which has no three consecutive elements present.

**Examples:**

Input:A[] = {1, 2, 3}, N=3Output:5Explanation:Three of them can’t be taken together so answer is 2 + 3 = 5

Input:A[] = {3000, 2000, 1000, 3, 10}, N=5Output:5013

A **O(N) **approach that takes **O(N) **auxiliary space has been discussed here. This can be further optimized with the following approach that takes **O(1)** extra space.

**O(1) Space Approach: **From the above approach, we can conclude that for calculating **sum[i], **only the values of **sum[i-1], sum[i-2] **and **sum[i-3] **are relevant.This observation can help to discard the sum array completely and instead just maintain some variables to solve the problem using O(1) auxiliary space.

Follow the steps below to solve the problem:

- Initialize the following variables to be used:
**sum:**This stores the final sum such that no three elements are consecutive.**first:**This stores the subsequence sum up to index**i-1**.**second:**This stores the subsequence sum up to index**i-2**.**third:**This stores the subsequence sum up to index**i-3**.

- If
**N<3**, the answer would be the sum of all the elements as there will be no consecutives. - Otherwise, do the following:
- Initialize
**third**with**A[0]** - Initialize
**second**with**A[0]+A[1]** - Initialize
**first**with**max(second, A[1]+A[2])** - Initialize
**sum**with the maximum among**first**,**second**, and**third**. - Iterate from
**3**to**N-1**, and do the following for each current index**i**:- There can be the following three cases:
- Exclude
**A[i],**i.e.**sum = first** - Exclude
**A[i-1],**i.e.,**sum = second + A[i]** - Exclude
**A[i-2]**, i.e.,**sum = third + A[i] + A[i-1]**

- Exclude
- Thus,
**sum**is updated as the maximum between**first**,**(second+A[i])**and**(third+A[i]+A[i-1])** - Update
**third**with**second**,**second**with**first**and**first**with**sum**.

- There can be the following three cases:

- Initialize
- Finally, return
**sum**.

Below is the implementation of the above approach:

## C++

`// C++ implementation for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to calculate the maximum subsequence sum such` `// that no three elements are consecutive` `int` `maxSumWO3Consec(` `int` `A[], ` `int` `N)` `{` ` ` `// when N is 1, answer would be the only element present` ` ` `if` `(N == 1)` ` ` `return` `A[0];` ` ` `// when N is 2, answer would be sum of elements` ` ` `if` `(N == 2)` ` ` `return` `A[0] + A[1];` ` ` `// variable to store sum up to i - 3` ` ` `int` `third = A[0];` ` ` `// variable to store sum up to i - 2` ` ` `int` `second = third + A[1];` ` ` `// variable to store sum up to i - 1` ` ` `int` `first = max(second, A[1] + A[2]);` ` ` `// variable to store the final sum of the subsequence` ` ` `int` `sum = max(max(third, second), first);` ` ` `for` `(` `int` `i = 3; i < N; i++) {` ` ` `// find the maximum subsequence sum up to index i` ` ` `sum = max(max(first, second + A[i]),` ` ` `third + A[i] + A[i - 1]);` ` ` `// update first, second and third` ` ` `third = second;` ` ` `second = first;` ` ` `first = sum;` ` ` `}` ` ` `// return ans;` ` ` `return` `sum;` `}` `// Driver code` `int` `main()` `{` ` ` `// Input` ` ` `int` `A[] = { 3000, 2000, 1000, 3, 10 };` ` ` `int` `N = ` `sizeof` `(A) / ` `sizeof` `(A[0]);` ` ` `// Function call` ` ` `cout << maxSumWO3Consec(A, N);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG` `{` ` ` ` ` `// Function to calculate the maximum subsequence sum` ` ` `// such` ` ` `// that no three elements are consecutive` ` ` `public` `static` `int` `maxSumWO3Consec(` `int` `A[], ` `int` `N)` ` ` `{` ` ` ` ` `// when N is 1, answer would be the only element` ` ` `// present` ` ` `if` `(N == ` `1` `)` ` ` `return` `A[` `0` `];` ` ` `// when N is 2, answer would be sum of elements` ` ` `if` `(N == ` `2` `)` ` ` `return` `A[` `0` `] + A[` `1` `];` ` ` `// variable to store sum up to i - 3` ` ` `int` `third = A[` `0` `];` ` ` `// variable to store sum up to i - 2` ` ` `int` `second = third + A[` `1` `];` ` ` `// variable to store sum up to i - 1` ` ` `int` `first = Math.max(second, A[` `1` `] + A[` `2` `]);` ` ` `// variable to store the final sum of the` ` ` `// subsequence` ` ` `int` `sum = Math.max(Math.max(third, second), first);` ` ` `for` `(` `int` `i = ` `3` `; i < N; i++)` ` ` `{` ` ` ` ` `// find the maximum subsequence sum up to index` ` ` `// i` ` ` `sum = Math.max(Math.max(first, second + A[i]),` ` ` `third + A[i] + A[i - ` `1` `]);` ` ` `// update first, second and third` ` ` `third = second;` ` ` `second = first;` ` ` `first = sum;` ` ` `}` ` ` `// return ans;` ` ` `return` `sum;` ` ` `}` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Input` ` ` `int` `A[] = { ` `3000` `, ` `2000` `, ` `1000` `, ` `3` `, ` `10` `};` ` ` `int` `N = A.length;` ` ` `// Function call` ` ` `int` `res = maxSumWO3Consec(A, N);` ` ` `System.out.println(res);` ` ` ` ` `}` `}` `//This code is contributed by Potta Lokesh` |

## Python3

`# Python 3 implementation for the above approach` `# Function to calculate the maximum subsequence sum such` `# that no three elements are consecutive` `def` `maxSumWO3Consec(A, N):` ` ` ` ` `# when N is 1, answer would be the only element present` ` ` `if` `(N ` `=` `=` `1` `):` ` ` `return` `A[` `0` `]` ` ` ` ` `# when N is 2, answer would be sum of elements` ` ` `if` `(N ` `=` `=` `2` `):` ` ` `return` `A[` `0` `] ` `+` `A[` `1` `]` ` ` ` ` `# variable to store sum up to i - 3` ` ` `third ` `=` `A[` `0` `]` ` ` `# variable to store sum up to i - 2` ` ` `second ` `=` `third ` `+` `A[` `1` `]` ` ` `# variable to store sum up to i - 1` ` ` `first ` `=` `max` `(second, A[` `1` `] ` `+` `A[` `2` `])` ` ` `# variable to store the final sum of the subsequence` ` ` `sum` `=` `max` `(` `max` `(third, second), first)` ` ` `for` `i ` `in` `range` `(` `3` `,N,` `1` `):` ` ` ` ` `# find the maximum subsequence sum up to index i` ` ` `sum` `=` `max` `(` `max` `(first, second ` `+` `A[i]), third ` `+` `A[i] ` `+` `A[i ` `-` `1` `])` ` ` `# update first, second and third` ` ` `third ` `=` `second` ` ` `second ` `=` `first` ` ` `first ` `=` `sum` ` ` `# return ans;` ` ` `return` `sum` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Input` ` ` `A ` `=` `[` `3000` `, ` `2000` `, ` `1000` `, ` `3` `, ` `10` `]` ` ` `N ` `=` `len` `(A)` ` ` `# Function call` ` ` `print` `(maxSumWO3Consec(A, N))` ` ` ` ` `# This code is contributed by ipg2016107.` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG {` ` ` `// Function to calculate the maximum subsequence sum` ` ` `// such` ` ` `// that no three elements are consecutive` ` ` `public` `static` `int` `maxSumWO3Consec(` `int` `[] A, ` `int` `N)` ` ` `{` ` ` `// when N is 1, answer would be the only element` ` ` `// present` ` ` `if` `(N == 1)` ` ` `return` `A[0];` ` ` `// when N is 2, answer would be sum of elements` ` ` `if` `(N == 2)` ` ` `return` `A[0] + A[1];` ` ` `// variable to store sum up to i - 3` ` ` `int` `third = A[0];` ` ` `// variable to store sum up to i - 2` ` ` `int` `second = third + A[1];` ` ` `// variable to store sum up to i - 1` ` ` `int` `first = Math.Max(second, A[1] + A[2]);` ` ` `// variable to store the final sum of the` ` ` `// subsequence` ` ` `int` `sum = Math.Max(Math.Max(third, second), first);` ` ` `for` `(` `int` `i = 3; i < N; i++) {` ` ` `// find the maximum subsequence sum up to index` ` ` `// i` ` ` `sum = Math.Max(Math.Max(first, second + A[i]),` ` ` `third + A[i] + A[i - 1]);` ` ` `// update first, second and third` ` ` `third = second;` ` ` `second = first;` ` ` `first = sum;` ` ` `}` ` ` `// return ans;` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `// Input` ` ` `int` `[] A = { 3000, 2000, 1000, 3, 10 };` ` ` `int` `N = A.Length;` ` ` `// Function call` ` ` `int` `res = maxSumWO3Consec(A, N);` ` ` `Console.Write(res);` ` ` `}` `}` |

## Javascript

`<script>` ` ` `// JavaScript implementation for the above approach` ` ` `// Function to calculate the maximum subsequence sum such` ` ` `// that no three elements are consecutive` ` ` `function` `maxSumWO3Consec(A, N)` ` ` `{` ` ` ` ` `// when N is 1, answer would be the only element present` ` ` `if` `(N == 1)` ` ` `return` `A[0];` ` ` `// when N is 2, answer would be sum of elements` ` ` `if` `(N == 2)` ` ` `return` `A[0] + A[1];` ` ` `// variable to store sum up to i - 3` ` ` `let third = A[0];` ` ` `// variable to store sum up to i - 2` ` ` `let second = third + A[1];` ` ` `// variable to store sum up to i - 1` ` ` `let first = Math.max(second, A[1] + A[2]);` ` ` `// variable to store the final sum of the subsequence` ` ` `let sum = Math.max(Math.max(third, second), first);` ` ` `for` `(let i = 3; i < N; i++) {` ` ` `// find the maximum subsequence sum up to index i` ` ` `sum = Math.max(Math.max(first, second + A[i]),` ` ` `third + A[i] + A[i - 1]);` ` ` `// update first, second and third` ` ` `third = second;` ` ` `second = first;` ` ` `first = sum;` ` ` `}` ` ` `// return ans;` ` ` `return` `sum;` ` ` `}` ` ` `// Driver code` ` ` `// Input` ` ` `let A = [3000, 2000, 1000, 3, 10];` ` ` `let N = A.length;` ` ` `// Function call` ` ` `document.write(maxSumWO3Consec(A, N));` `//This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output:**

5013

**Time complexity: **O(N)**Auxiliary Space: **O(1)

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