# Maximum subsequence sum such that no K elements are consecutive

Given an array arr[] of N positive integers, the task is to find the maximum sum of a subsequence consisting of no K consecutive array elements.

Examples:

Input: arr[] = {10, 5, 8, 16, 21}, K = 4
Output: 55
Explanation:
Maximum sum is obtained by picking 10, 8, 16, 21.

Input: arr[] = {4, 12, 22, 18, 34, 12, 25}, K = 5
Output: 111
Explanation:
Maximum sum is obtained by picking 12, 22, 18, 34, 25

Naive Approach: The simplest approach is to generate all the subsets of the given array and for each subset, check if it contains K consecutive array elements or not. For subsets found to be not containing K consecutive array elements, calculate their sum. Find the maximum of the sums of all such subsequences.

Time Complexity: O(N*2N)
Auxiliary Space: O(1)

Efficient Approach: There are many overlapping subproblems in the above solution which are calculated again and again. To avoid recomputation of the same subproblems, the idea is to use Memoization or Tabulation. Follow the steps below to solve the problem:

1. Initialize an array dp[] to memorize the maximum value of the sum up to each index.
2. Now, dp[i] gives the maximum value of the sum that can be picked such that no K elements are consecutive from the 0th Index till ith index.
3. The base case is when i < K :
• Since the array elements are all positive, pick all the elements before (K )th Index.
• So dp[1] = arr [0] and dp[i] = dp[i -1] + arr[i-1], (1 ≤ i < k ).
4. Now for i ≥ K :
• Since K consecutive elements cannot be picked, so skip at least one element from i to (i – K + 1) inclusive so to make sure that no K elements are consecutive.
• Since any element can contribute to the result, so skip every element from i to (i – K + 1) inclusive and will keep track of the maximum sum.
• To skip the jth Element, add maximum sum till (j – 1)th Index which is given by dp[j – 1] with the sum of all the elements from (j + 1)th index to ith index which can be calculated in O(1) time using prefix array sum.
• Therefore update the current dp state as: dp[i] = max (dp[i], dp[j -1] + prefix[i] – prefix [j]), (i ≤ j ≤ (i – K + 1)), where prefix array stores the prefix sum.
5. Print the maximum sum after the above steps.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;   // Function to find the maximum sum // of a subsequence consisting of // no K consecutive array elements int Max_Sum(int arr[], int K, int N) {         // Stores states of dp     int dp[N + 1];       // Initialise dp state     memset(dp, 0, sizeof(dp));       // Stores the prefix sum     int prefix[N + 1];       prefix[0] = 0;       // Update the prefix sum     for(int i = 1; i <= N; i++)     {         prefix[i] = prefix[i - 1] + arr[i-1];     }       // Base case for i < K     dp[0] = 0;       // For indices less than k     // take all the elements     for(int i = 1; i < K ; i++)     {         dp[i] = prefix[i];     }       // For i >= K  case     for(int i = K ; i <= N; ++i)     {                 // Skip each element from i to         // (i - K + 1) to ensure that         // no K elements are consecutive         for(int j = i; j >= (i - K + 1); j--)         {                         // j-th element is skipped               // Update the current dp state             dp[i] = max(dp[i], dp[j - 1] +                     prefix[i] - prefix[j]);         }     }       // dp[N] stores the maximum sum     return dp[N]; }   // Driver Code int main() {         // Given array arr[]     int arr[] = { 4, 12, 22, 18, 34, 12, 25 };       int N = sizeof(arr) / sizeof(int);     int K = 5;       // Function Call     cout << Max_Sum(arr, K, N);       return 0; }

## Java

 // Java program for the above approach import java.io.*; import java.util.*;   class GFG{   // Function to find the maximum sum // of a subsequence consisting of // no K consecutive array elements public static int Max_Sum(int[] arr, int K,                           int N) {           // Stores states of dp     int[] dp = new int[N + 1];       // Initialise dp state     Arrays.fill(dp, 0);       // Stores the prefix sum     int[] prefix = new int[N + 1];       prefix[0] = 0;       // Update the prefix sum     for(int i = 1; i <= N; i++)     {         prefix[i] = prefix[i - 1] + arr[i-1];     }       // Base case for i < K     dp[0] = 0;       // For indices less than k     // take all the elements     for(int i = 1; i <= K - 1; i++)     {         dp[i] = prefix[i];     }       // For i >= K  case     for(int i = K ; i <= N; ++i)     {                   // Skip each element from i to         // (i - K + 1) to ensure that         // no K elements are consecutive         for(int j = i; j >= (i - K + 1); j--)         {                           // j-th element is skipped               // Update the current dp state             dp[i] = Math.max(dp[i], dp[j - 1] +                          prefix[i] - prefix[j]);         }     }       // dp[N] stores the maximum sum     return dp[N]; }   // Driver Code public static void main(String[] args) {           // Given array arr[]     int[] arr = { 4, 12, 22, 18, 34, 12, 25 };       int N = arr.length;     int K = 5;       // Function Call     System.out.println(Max_Sum(arr, K, N)); } }   // This code is contributed by akhilsaini

## Python3

 # Python3 program for the above approach   # Function to find the maximum sum # of a subsequence consisting of # no K consecutive array elements def Max_Sum(arr, K, N):           # Stores states of dp     dp = [0] * (N + 1)           # Stores the prefix sum     prefix = [None] * (N + 1)           prefix[0] = 0           # Update the prefix sum     for i in range(1, N + 1):         prefix[i] = prefix[i - 1] + arr[i - 1]             # Base case for i < K     dp[0] = 0           # For indices less than k     # take all the elements     for i in range(1, K):         dp[i] = prefix[i]           # For i >= K case     for i in range(K, N + 1):                   # Skip each element from i to         # (i - K + 1) to ensure that         # no K elements are consecutive         for j in range(i, i - K, -1):                           # j-th element is skipped                           # Update the current dp state             dp[i] = max(dp[i], dp[j - 1] +                     prefix[i] - prefix[j])           # dp[N] stores the maximum sum     return dp[N]   # Driver Code if __name__ == "__main__":           # Given array arr[]     arr = [ 4, 12, 22, 18, 34, 12, 25 ]           N = len(arr)     K = 5           # Function call     print(Max_Sum(arr, K, N))       # This code is contributed by akhilsaini

## C#

 // C# program for the above approach using System;   class GFG{   // Function to find the maximum sum // of a subsequence consisting of // no K consecutive array elements static int Max_Sum(int[] arr, int K, int N) {           // Stores states of dp     int[] dp = new int[N + 1];       // Initialise dp state     Array.Fill(dp, 0);       // Stores the prefix sum     int[] prefix = new int[N + 1];       prefix[0] = 0;       // Update the prefix sum     for(int i = 1; i <= N; i++)     {         prefix[i] = prefix[i - 1] + arr[i - 1];     }       // Base case for i < K     dp[0] = 0;       // For indices less than k     // take all the elements     for(int i = 1; i <= K - 1; i++)     {         dp[i] = prefix[i];     }       // For i >= K case     for(int i = K; i <= N; ++i)     {                   // Skip each element from i to         // (i - K + 1) to ensure that         // no K elements are consecutive         for(int j = i; j >= (i - K + 1); j--)         {                           // j-th element is skipped               // Update the current dp state             dp[i] = Math.Max(dp[i], dp[j - 1] +                          prefix[i] - prefix[j]);         }     }       // dp[N] stores the maximum sum     return dp[N]; }   // Driver Code static public void Main() {           // Given array arr[]     int[] arr = { 4, 12, 22, 18, 34, 12, 25 };       int N = arr.Length;     int K = 5;       // Function Call     Console.WriteLine(Max_Sum(arr, K, N)); } }   // This code is contributed by akhilsaini

Output:

111

Time Complexity: O(N*K) where N is the number of elements in the array and K is the input such that no K elements are consecutive.
Auxiliary Space: O(N)

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Improved By : akhilsaini