Maximum score possible from an array with jumps of at most length K

Given an array arr[] and an integer K, the 0^th index, the task is to collect the maximum score possible by performing following operations:

• Start from the 0^th index of the array.
• Reach the last index of the array by jumping at most K indices in each move.
• Add the value of every index reached after each jump.
• Examples :

• Input: arr[] = {100, -30, -50, -15, -20, -30}, K = 3
  Output: 55
  Explanation: From 0^th index, jump 3 indices ahead to arr[3] (= -15). From 3^rd, jump 2 steps ahead to arr[5] (= -30). Therefore, the maximum score possible = (100 + (-15) + (-30)) = 55

  Input: arr[] = {-44, -17, -54, 79},  K = 2
  Output: 18

• Naive Approach: The problem can be solved using recursion with memoization. Follow the steps below to solve the problem:
• • Initialize an array dp[] to store the previously computed results.
  • Now, starting from the 0^th index, perform the following operations for every i^th index:
    • If the current index is greater than or equal to index of last element, return the last element of the array.
    • If the value for the current index is pre-calculated, then return the pre-calculated value.
    • Otherwise, calculate maximum score that can be obtained by moving to all steps in the range i + 1 to i + K and store results for respective indices in dp[] array using the following recurrence relation:
      • Start from the 0^th index of the array.
      • Reach the last index of the array by jumping at most K indices in each move.
      • Add the value of every index reached after each jump.
        • Initialize an array dp[] to store the previously computed results.
        • Now, starting from the 0^th index, perform the following operations for every i^th index:
          • If the current index is greater than or equal to the index of the last element, return the last element of the array.
          • If the value for the current index is pre-calculated, then return the pre-calculated value.
          • Otherwise, calculate maximum score that can be obtained by moving to all steps in the range i + 1 to i + K and store results for respective indices in dp[] array using the following recurrence relation: 

      dp[i] = max(dp[i + 1], dp[i + 2], dp[i + 3], ….., dp[i + K]) + A[i].

      dp[i] = max(dp[i + 1], dp[i + 2], dp[i + 3], ….., dp[i + K]) + A[i].

    • Now, print dp[0] as the required answer.
• Below is the implementation of the above approach:
• C++

  // C++ program for the above approach
  #include <bits/stdc++.h>
  using namespace std;
  
  // Function to count the maximum
  // score of an index
  int maxScore(int i, int A[], int K, int N, int dp[])
  {
      // Base Case
      if (i >= N - 1)
          return A[N - 1];
  
      // If the value for the current
      // index is pre-calculated
      if (dp[i] != -1)
          return dp[i];
  
      int score = INT_MIN;
  
      // Calculate maximum score
      // for all the steps in the
      // range from i + 1 to i + k
      for (int j = 1; j <= K; j++) {
  
          // Score for index (i + j)
          score = max(score, maxScore(i + j, A, K, N, dp));
      }
  
      // Update dp[i] and return
      // the maximum value
      return dp[i] = score + A[i];
  }
  
  // Function to get maximum score
  // possible from the array A[]
  int getScore(int A[], int N, int K)
  {
      // Array to store memoization
      int dp[N];
  
      // Initialize dp[] with -1
      for (int i = 0; i < N; i++)
          dp[i] = -1;
  
      cout << maxScore(0, A, K, N, dp);
  }
  
  // Driver Code
  int main()
  {
      int A[] = { 100, -30, -50, -15, -20, -30 };
      int K = 3;
      int N = sizeof(A) / sizeof(A[0]);
  
      getScore(A, N, K);
  
      return 0;
  }

  Java

  // JAVA program for the above approach
  import java.io.*;
  import java.math.*;
  import java.util.*;
  public class GFG
  {
  
    // Function to count the maximum
    // score of an index
    static int maxScore(int i, int A[], int K, int N,
                        int dp[])
    {
  
      // Base Case
      if (i >= N - 1)
        return A[N - 1];
  
      // If the value for the current
      // index is pre-calculated
      if (dp[i] != -1)
        return dp[i];
      int score = Integer.MIN_VALUE;
  
      // Calculate maximum score
      // for all the steps in the
      // range from i + 1 to i + k
      for (int j = 1; j <= K; j++)
      {
  
        // Score for index (i + j)
        score = Math.max(score,
                         maxScore(i + j, A, K, N, dp));
      }
  
      // Update dp[i] and return
      // the maximum value
      return dp[i] = score + A[i];
    }
  
    // Function to get maximum score
    // possible from the array A[]
    static void getScore(int A[], int N, int K)
    {
  
      // Array to store memoization
      int dp[] = new int[N];
  
      // Initialize dp[] with -1
      for (int i = 0; i < N; i++)
        dp[i] = -1;
      System.out.println(maxScore(0, A, K, N, dp));
    }
  
    // Driver Code
    public static void main(String args[])
    {
      int A[] = { 100, -30, -50, -15, -20, -30 };
      int K = 3;
      int N = A.length;
  
      getScore(A, N, K);
    }
  }
  
  // This code is contributed by jyoti369

  Python3

  # Python program for the above approach
  import sys
  
  # Function to count the maximum
  # score of an index
  def maxScore(i, A, K, N, dp):
  
      # Base Case
      if (i >= N - 1):
          return A[N - 1];
  
      # If the value for the current
      # index is pre-calculated
      if (dp[i] != -1):
          return dp[i];
      score = 1-sys.maxsize;
  
      # Calculate maximum score
      # for all the steps in the
      # range from i + 1 to i + k
      for j in range(1, K + 1):
  
          # Score for index (i + j)
          score = max(score, maxScore(i + j, A, K, N, dp));
  
      # Update dp[i] and return
      # the maximum value
      dp[i] = score + A[i];
      return dp[i];
  
  
  # Function to get maximum score
  # possible from the array A
  def getScore(A, N, K):
      # Array to store memoization
      dp = [0]*N;
  
      # Initialize dp with -1
      for i in range(N):
          dp[i] = -1;
      print(maxScore(0, A, K, N, dp));
  
  # Driver Code
  if __name__ == '__main__':
      A = [100, -30, -50, -15, -20, -30];
      K = 3;
      N = len(A);
  
      getScore(A, N, K);
  
  # This code contributed by shikhasingrajput

  C#

  // C# program for the above approach
  using System;
  class GFG
  {
  
    // Function to count the maximum
    // score of an index
    static int maxScore(int i, int []A, int K, int N,
                        int []dp)
    {
  
      // Base Case
      if (i >= N - 1)
        return A[N - 1];
  
      // If the value for the current
      // index is pre-calculated
      if (dp[i] != -1)
        return dp[i];
      int score = int.MinValue;
  
      // Calculate maximum score
      // for all the steps in the
      // range from i + 1 to i + k
      for (int j = 1; j <= K; j++)
      {
  
        // Score for index (i + j)
        score = Math.Max(score,
                         maxScore(i + j, A, K, N, dp));
      }
  
      // Update dp[i] and return
      // the maximum value
      return dp[i] = score + A[i];
    }
  
    // Function to get maximum score
    // possible from the array A[]
    static void getScore(int []A, int N, int K)
    {
  
      // Array to store memoization
      int []dp = new int[N];
  
      // Initialize dp[] with -1
      for (int i = 0; i < N; i++)
        dp[i] = -1;
      Console.WriteLine(maxScore(0, A, K, N, dp));
    }
  
  // Driver Code
  static public void Main()
  {
      int []A = { 100, -30, -50, -15, -20, -30 };
      int K = 3;
      int N = A.Length;
  
      getScore(A, N, K);
  }
  }
  
  // This code is contributed by jana_sayantan.

  Javascript

  <script>
  // javascript program of the above approach
  
    // Function to count the maximum
    // score of an index
    function maxScore(i, A, K, N,
                        dp)
    {
  
      // Base Case
      if (i >= N - 1)
        return A[N - 1];
  
      // If the value for the current
      // index is pre-calculated
      if (dp[i] != -1)
        return dp[i];
      let score = -1000;
  
      // Calculate maximum score
      // for all the steps in the
      // range from i + 1 to i + k
      for (let j = 1; j <= K; j++)
      {
  
        // Score for index (i + j)
        score = Math.max(score,
                         maxScore(i + j, A, K, N, dp));
      }
  
      // Update dp[i] and return
      // the maximum value
      return dp[i] = score + A[i];
    }
  
    // Function to get maximum score
    // possible from the array A[]
    function getScore(A, N, K)
    {
  
      // Array to store memoization
      let dp = new Array(N).fill(-1);
  
      document.write(maxScore(0, A, K, N, dp));
    }
  
      // Driver Code
  
      let A = [ 100, -30, -50, -15, -20, -30 ];
      let K = 3;
      let N = A.length;
  
      getScore(A, N, K);
  
  </script>
• Output

  55
  
  
  
  
  

• Time Complexity: O(N * K)
  Auxiliary Space: O(N * K)

• Maximum score possible from an array with jumps of at most length K using Dynamic Programming (Bottom up/Tabulation):

• The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
• Step-by-step approach:
• • Initialize a DP array of size N, where DP[i] represents the maximum score that can be obtained starting from position i.
  • Initialize DP[N-1] with the value of the last element of the array A.
  • Traverse the DP array from the second last index to the first.
  • For each index i, consider all possible steps that can be taken from that position, i.e., 1 to K steps forward.
  • For each step, calculate the maximum score that can be obtained by adding the score at the current position i to the maximum score that can be obtained from the destination position j.
  • Update DP[i] with the maximum score obtained among all the possible steps.
  • Return DP[0], which represents the maximum score that can be obtained starting from the first position.
• Below is the implementation of the above approach:
• C++

  // C++ program for above approach
  
  #include <bits/stdc++.h>
  using namespace std;
  
  int getScore(int A[], int N, int K) {
        // Initialize dp array with values for the last element
      int dp[N];
      dp[N - 1] = A[N - 1];
  
        // Traverse dp array from the second last index to the first
      for (int i = N - 2; i >= 0; i--) {
            // initial score
          int score = INT_MIN;
          for (int j = 1; j <= K && i + j < N; j++) {
                // Update the score with the maximum value
                // found among the possible steps
              score = max(score, dp[i + j]);
          }
            // Calculate the maximum score possible
          dp[i] = score + A[i];
      }
  
        // return answer
      return dp[0];
  }
  
  // Driver code
  int main() {
      int A[] = { 100, -30, -50, -15, -20, -30 };
      int K = 3;
      int N = sizeof(A) / sizeof(A[0]);
  
        //function call
      cout << getScore(A, N, K);
  
      return 0;
  }
  
  // this code is contributed by bhardwajji

  Java

  // Java program for above approach
  import java.util.Arrays;
  
  public class Main {
      public static int getScore(int[] A, int N, int K) {
          // Initialize dp array with values for the last element
          int[] dp = new int[N];
          dp[N - 1] = A[N - 1];
  
          // Traverse dp array from the second last index to the first
          for (int i = N - 2; i >= 0; i--) {
              // initial score
              int score = Integer.MIN_VALUE;
              for (int j = 1; j <= K && i + j < N; j++) {
                  // Update the score with the maximum value
                  // found among the possible steps
                  score = Math.max(score, dp[i + j]);
              }
              // Calculate the maximum score possible
              dp[i] = score + A[i];
          }
  
          // return answer
          return dp[0];
      }
  
        // Driver code
      public static void main(String[] args) {
          int[] A = { 100, -30, -50, -15, -20, -30 };
          int K = 3;
          int N = A.length;
  
          // function call
          System.out.println(getScore(A, N, K));
      }
  }

  Python3

  class Main:
      @staticmethod
      def getScore(A, N, K):
          # Initialize dp array with values for the last element
          dp = [0] * N
          dp[N - 1] = A[N - 1]
  
          # Traverse dp array from the second last index to the first
          for i in range(N - 2, -1, -1):
              # initial score
              score = float('-inf')
              for j in range(1, K + 1):
                  # Update the score with the maximum value
                  # found among the possible steps
                  if i + j < N:
                      score = max(score, dp[i + j])
              # Calculate the maximum score possible
              dp[i] = score + A[i]
  
          # return answer
          return dp[0]
  
  if __name__ == '__main__':
      A = [100, -30, -50, -15, -20, -30]
      K = 3
      N = len(A)
  
      # function call
      print(Main.getScore(A, N, K))

  C#

  using System;
  
  public class Program {
      public static int GetScore(int[] A, int N, int K) {
          // Initialize dp array with values for the last element
          int[] dp = new int[N];
          dp[N - 1] = A[N - 1];
  
          // Traverse dp array from the second last index to the first
          for (int i = N - 2; i >= 0; i--) {
              // initial score
              int score = int.MinValue;
              for (int j = 1; j <= K && i + j < N; j++) {
                  // Update the score with the maximum value
                  // found among the possible steps
                  score = Math.Max(score, dp[i + j]);
              }
              // Calculate the maximum score possible
              dp[i] = score + A[i];
          }
  
          // return answer
          return dp[0];
      }
  
      // Driver code
      public static void Main() {
          int[] A = { 100, -30, -50, -15, -20, -30 };
          int K = 3;
          int N = A.Length;
  
          // function call
          Console.WriteLine(GetScore(A, N, K));
      }
  }

  Javascript

  // JavaScript program for above approach
  
  function getScore(A, N, K) {
  // Initialize dp array with values for the last element
  let dp = new Array(N);
  dp[N - 1] = A[N - 1];
  // Traverse dp array from the second last index to the first
  for (let i = N - 2; i >= 0; i--) {
      // initial score
      let score = -Infinity;
      for (let j = 1; j <= K && i + j < N; j++) {
          // Update the score with the maximum value
          // found among the possible steps
          score = Math.max(score, dp[i + j]);
      }
      // Calculate the maximum score possible
      dp[i] = score + A[i];
  }
  
  // return answer
  return dp[0];
  }
  
  // Driver code
  let A = [100, -30, -50, -15, -20, -30];
  let K = 3;
  let N = A.length;
  
  // function call
  console.log(getScore(A, N, K));
• Output

  55
  
  
  
  
  

• Time Complexity: O(N * K), where N is the size of the array and K is the input variable
  Auxiliary Space: O(N)

• Maximum score possible from an array with jumps of at most length K using Heap:

• Step-by-step approach:
• • Initialize a Max Heap to store the result of previous K indices.
  • Now, traverse the array A[] to calculate the maximum score for all indices.
    • For 0^th index, the score will be the value at the 0^th index.
    • Now, for every i^th index in the range [1, N – 1].
      • Firstly, remove maximum scores from the Max Heap for indices less than i – K.
      • Now calculate the maximum score for i^th index. 
        Maximum score = A[i] + score at the top of the Max Heap.
      • Now insert the maximum score into the max heap with its index.
  • Return the maximum score obtained.
• Below is the implementation of the above approach:
• C++

  // C++ program for the above approach
  #include <bits/stdc++.h>
  using namespace std;
  
  // Structure to sort a priority queue on
  // the basis of first element of the pair
  struct mycomp {
      bool operator()(pair<int, int> p1,
                      pair<int, int> p2)
      {
          return p1.first < p2.first;
      }
  };
  
  // Function to calculate maximum
  // score possible from the array A[]
  int maxScore(int A[], int K, int N)
  {
      // Stores the score of previous k indices
      priority_queue<pair<int, int>,
                  vector<pair<int, int> >, mycomp>
          maxheap;
  
      // Stores the maximum
      // score for current index
      int maxScore = 0;
  
      // Maximum score at first index
      maxheap.push({ A[0], 0 });
  
      // Traverse the array to calculate
      // maximum score for all indices
      for (int i = 1; i < N; i++) {
  
          // Remove maximum scores for
          // indices less than i - K
          while (maxheap.top().second < (i - K)) {
              maxheap.pop();
          }
  
          // Calculate maximum score for current index
          maxScore = A[i] + maxheap.top().first;
  
          // Push maximum score of
          // current index along
          // with index in maxheap
          maxheap.push({ maxScore, i });
      }
  
      // Return the maximum score
      return maxScore;
  }
  
  // Driver Code
  int main()
  {
      int A[] = { -44, -17, -54, 79 };
      int K = 2;
      int N = sizeof(A) / sizeof(A[0]);
  
      // Function call to calculate
      // maximum score from the array A[]
      cout << maxScore(A, K, N);
  
      return 0;
  }

  Java

  // Java code for the above approach:
  
  import java.util.*;
  
  public class Main {
      static class IntPair {
          int first;
          int second;
          IntPair(int x, int y)
          {
              this.first = x;
              this.second = y;
          }
      }
      static class HeapComparator
          implements Comparator<IntPair> {
          public int compare(IntPair p1, IntPair p2)
          {
              if (p1.first < p2.first)
                  return 1;
              else if (p1.first > p2.first)
                  return -1;
              return 0;
          }
      }
      // Function to calculate maximum
      // score possible from the array A[]
      static int maxScore(int A[], int K, int N)
      {
          // Stores the score of previous k indices
          PriorityQueue<IntPair> maxheap
              = new PriorityQueue<IntPair>(
                  new HeapComparator());
  
          // Stores the maximum
          // score for current index
          int maxScore = 0;
  
          // Maximum score at first index
          maxheap.add(new IntPair(A[0], 0));
  
          // Traverse the array to calculate
          // maximum score for all indices
          for (int i = 1; i < N; i++) {
  
              // Remove maximum scores for
              // indices less than i - K
              while (maxheap.peek().second < (i - K)) {
                  maxheap.remove();
              }
  
              // Calculate maximum score for current index
              maxScore = A[i] + maxheap.peek().first;
  
              // Push maximum score of
              // current index along
              // with index in maxheap
              maxheap.add(new IntPair(maxScore, i));
          }
  
          // Return the maximum score
          return maxScore;
      }
  
      // Driver Code
      public static void main(String args[])
      {
          int A[] = { -44, -17, -54, 79 };
          int K = 2;
          int N = A.length;
  
          // Function call to calculate
          // maximum score from the array A[]
          System.out.println(maxScore(A, K, N));
      }
  }
  
  // This code has been contributed by Sachin Sahara
  // (sachin801)

  Python3

  # Python program for the above approach
  
  # Function to calculate maximum
  # score possible from the array A[]
  def maxScore(A, K, N):
  
      # Stores the score of previous k indices
      maxheap = []
  
      # Stores the maximum
      # score for current index
      maxScore = 0
  
      # Maximum score at first index
      maxheap.append([A[0], 0])
  
      # Traverse the array to calculate
      # maximum score for all indices
      for i in range(1, N):
  
          # Remove maximum scores for
          # indices less than i - K
          while(maxheap[len(maxheap) - 1][1] < (i - K)):
              maxheap.pop()
  
          # Calculate maximum score for current index
          maxScore = A[i] + maxheap[len(maxheap) - 1][0]
  
          # Push maximum score of
          # current index along
          # with index in maxheap
          maxheap.append([maxScore, i])
          maxheap.sort()
  
      # Return the maximum score
      return maxScore
  
  # Driver Code
  A = [-44, -17, -54, 79]
  K = 2
  N = len(A)
  
  # Function call to calculate
  # maximum score from the array A[]
  print(maxScore(A, K, N))
  
  # This code is contributed by Pushpesh Raj.

  C#

  // C# program for the above approach
  
  using System;
  using System.Collections.Generic;
  
  class GFG
  
  {
    // Function to calculate maximum
    // score possible from the array A[]
    static int maxScore(int[] A, int K, int N)
    {
      // Stores the score of previous k indices
      List<Tuple<int, int>> maxheap = new List<Tuple<int, int>> ();
  
      // Stores the maximum
      // score for current index
      int maxScore = 0;
  
      // Maximum score at first index
      maxheap.Add(Tuple.Create(A[0], 0));
  
      // Traverse the array to calculate
      // maximum score for all indices
      for (int i = 1; i < N; i++) {
  
        // Remove maximum scores for
        // indices less than i - K
        while (maxheap[0].Item2 < (i - K)) {
          maxheap.RemoveAt(0);
        }
  
        // Calculate maximum score for current index
        maxScore = A[i] + maxheap[0].Item1;
  
        // Add maximum score of
        // current index along
        // with index in maxheap
        maxheap.Add(Tuple.Create(maxScore, i ));
        maxheap.Sort();
        maxheap.Reverse();
      }
  
      // Return the maximum score
      return maxScore;
    }
  
    // Driver Code
    public static void Main(string[] args)
    {
      int[] A = { -44, -17, -54, 79 };
      int K = 2;
      int N = A.Length;
  
      // Function call to calculate
      // maximum score from the array A[]
      Console.WriteLine(maxScore(A, K, N));
  
    }
  }
  
  // This code is contributed by phasing17.

  Javascript

  <script>
  
  // Javascript program for the above approach
  
  // Function to calculate maximum
  // score possible from the array A[]
  function maxScore(A, K, N)
  {
  
      // Stores the score of previous k indices
      var maxheap = [];
  
      // Stores the maximum
      // score for current index
      var maxScore = 0;
  
      // Maximum score at first index
      maxheap.push([A[0], 0]);
  
      // Traverse the array to calculate
      // maximum score for all indices
      for (var i = 1; i < N; i++) {
  
          // Remove maximum scores for
          // indices less than i - K
          while (maxheap[maxheap.length-1][1] < (i - K)) {
              maxheap.pop();
          }
  
          // Calculate maximum score for current index
          maxScore = A[i] + maxheap[maxheap.length-1][0];
  
          // Push maximum score of
          // current index along
          // with index in maxheap
          maxheap.push([maxScore, i]);
          maxheap.sort();
      }
  
      // Return the maximum score
      return maxScore;
  }
  
  // Driver Code
  var A = [-44, -17, -54, 79];
  var K = 2;
  var N = A.length;
  
  // Function call to calculate
  // maximum score from the array A[]
  document.write(maxScore(A, K, N));
  
  // This code is contributed by noob2000.
  </script>
• Output

  18
  
  
  
  
  

• Time Complexity: O(N * log K)
  Auxiliary Space: O(N)

• Finding the maximum score using Deque:

• In the previous approaches, we used a dp[] array such that dp[i] represents the maximum score that can be obtained starting from position i. And then we iterated over the last k positions to find the maximum among them. In order to find the maximum efficiently, we can simply maintain a Deque in sorted order to reduce the max previous score lookup from O(K) to O(1), reducing the overall time complexity to O(N).
• Approach:

• We maintain the deque such that we will pop (i-K-1)^th index from queue since they are at a distance greater than K. Along with that, we will also pop those indices which will never have any chance of being chosen in the future. So for eg., if the score for current index – dp[i] is greater than some indices stored in the queue, it will always be optimal to choose dp[i] instead of those other indices. So, we will just pop those indices from queue since they won’t ever be used.

• Step by step algorithm:
• • Create a deque maxQueue and add 0 to it.
  • Iterate from i = 1 to the size of array and for every element arr[i],
    • Check if the front element of maxQueue is less than i - k. If so, remove the front element.
    • Calculate the current result dp[i] as nums[i] + dp[maxQueue.front()].
    • Remove the back element from maxQueue, till dp[i] is greater than or equal to the value of dp[maxQueue.back()].
    • Add the current index i to the back of maxQueue.
  • Return the value of dp[n-1], which represents the maximum result.
• Below is the implementation of the above approach:
• C++

  #include <bits/stdc++.h>
  using namespace std;
  
  // Method to return the max score
  int getScore(vector<int>& arr, int N, int K)
  {
      // Max Queue to choose the maximum value of dp[] from
      // subarray dp[i-K...i-1]
      deque<int> maxQueue{ 0 };
      // dp array to store maximum score that can be obtained
      // starting from position
      vector<int> dp(N);
      dp[0] = arr[0];
      for (int i = 1; i < N; i++) {
          // Check if the element is at a distance greater
          // than K
          if (maxQueue.front() < i - K) {
              maxQueue.pop_front();
          }
          dp[i] = arr[i] + dp[maxQueue.front()];
          // Pop all those indices which will never have any
          // chance of being chosen in the future
          while (!maxQueue.empty()
                 && dp[i] >= dp[maxQueue.back()])
              maxQueue.pop_back();
          maxQueue.push_back(i);
      }
      // Return dp[n-1], which is the maximum result
      return dp[N - 1];
  }
  
  int main()
  {
      // Sample Input
      vector<int> A{ 100, -30, -50, -15, -20, -30 };
      int K = 3;
      int N = A.size();
  
      // function call
      cout << (getScore(A, N, K));
      return 0;
  }

  Java

  import java.util.*;
  
  public class MaxScore {
      // Method to calculate the maximum score based on the rules
      public static int getScore(List<Integer> arr, int N, int K) {
          Deque<Integer> maxQueue = new LinkedList<>(); // Create a deque to store indices
          maxQueue.offer(0); // Initialize the deque with the first index
  
          int[] dp = new int[N]; // Create an array to store maximum scores
          dp[0] = arr.get(0); // Initialize the first score with the first element
  
          for (int i = 1; i < N; i++) {
              if (maxQueue.peekFirst() < i - K) {
                  maxQueue.pollFirst(); // Remove indices that are outside the window of K
              }
              dp[i] = arr.get(i) + dp[maxQueue.peekFirst()]; // Calculate the maximum score at index i
  
              // Remove indices from the back of the deque that have scores less than the current score
              while (!maxQueue.isEmpty() && dp[i] >= dp[maxQueue.peekLast()]) {
                  maxQueue.pollLast();
              }
              maxQueue.offer(i); // Add the current index to the deque
          }
  
          return dp[N - 1]; // Return the maximum score from the last index
      }
  
      public static void main(String[] args) {
          List<Integer> A = Arrays.asList(100, -30, -50, -15, -20, -30); // Sample input list
          int K = 3; // Window size
          int N = A.size(); // Size of the input list
  
          // Call the getScore method to find the maximum score
          System.out.println(getScore(A, N, K));
      }
  }

  Python3

  from collections import deque
  
  def get_score(arr, N, K):
      max_queue = deque()  # Create a deque to store indices
      max_queue.append(0)  # Initialize the deque with the first index
  
      dp = [0] * N  # Create a list to store maximum scores
      dp[0] = arr[0]  # Initialize the first score with the first element
  
      for i in range(1, N):
          while max_queue and max_queue[0] < i - K:
              max_queue.popleft()  # Remove indices that are outside the window of K
  
          dp[i] = arr[i] + dp[max_queue[0]]  # Calculate the maximum score at index i
  
          while max_queue and dp[i] >= dp[max_queue[-1]]:
              max_queue.pop()  # Remove indices from the back of the deque with scores less than the current score
  
          max_queue.append(i)  # Add the current index to the deque
  
      return dp[N - 1]  # Return the maximum score from the last index
  
  def main():
      A = [100, -30, -50, -15, -20, -30]  # Sample input list
      K = 3  # Window size
      N = len(A)  # Size of the input list
  
      # Call the get_score function to find the maximum score
      print(get_score(A, N, K))
  
  if __name__ == "__main__":
      main()

  C#

  using System;
  using System.Collections.Generic;
  
  class MaxScore
  {
      // Method to calculate the maximum score based on the rules
      public static int GetScore(List<int> arr, int N, int K)
      {
          LinkedList<int> maxQueue = new LinkedList<int>(); // Create a deque to store indices
          maxQueue.AddLast(0); // Initialize the deque with the first index
  
          int[] dp = new int[N]; // Create an array to store maximum scores
          dp[0] = arr[0]; // Initialize the first score with the first element
  
          for (int i = 1; i < N; i++)
          {
              if (maxQueue.First.Value < i - K)
              {
                  maxQueue.RemoveFirst(); // Remove indices that are outside the window of K
              }
              dp[i] = arr[i] + dp[maxQueue.First.Value]; // Calculate the maximum score at index i
  
              // Remove indices from the back of the deque that have scores less than the current score
              while (maxQueue.Count > 0 && dp[i] >= dp[maxQueue.Last.Value])
              {
                  maxQueue.RemoveLast();
              }
              maxQueue.AddLast(i); // Add the current index to the deque
          }
  
          return dp[N - 1]; // Return the maximum score from the last index
      }
  
      public static void Main()
      {
          List<int> A = new List<int> { 100, -30, -50, -15, -20, -30 }; // Sample input list
          int K = 3; // Window size
          int N = A.Count; // Size of the input list
  
          // Call the GetScore method to find the maximum score
          Console.WriteLine(GetScore(A, N, K));
      }
  }

  Javascript

  function getScore(arr, N, K) {
      const maxQueue = []; // Array to store indices, acting as a deque
      maxQueue.push(0); // Initialize the deque with the first index
  
      const dp = new Array(N).fill(0); // Array to store maximum scores
      dp[0] = arr[0]; // Initialize the first score with the first element
  
      for (let i = 1; i < N; i++) {
          if (maxQueue[0] < i - K) {
              maxQueue.shift(); // Remove indices that are outside the window of K
          }
          dp[i] = arr[i] + dp[maxQueue[0]]; // Calculate the maximum score at index i
  
          // Remove indices from the back of the deque that have scores less than the current score
          while (maxQueue.length > 0 && dp[i] >= dp[maxQueue[maxQueue.length - 1]]) {
              maxQueue.pop();
          }
          maxQueue.push(i); // Add the current index to the deque
      }
  
      return dp[N - 1]; // Return the maximum score from the last index
  }
  
  // Sample input
  const A = [100, -30, -50, -15, -20, -30];
  const K = 3; // Window size
  const N = A.length; // Size of the input list
  
  // Call the getScore function to find the maximum score
  console.log(getScore(A, N, K));
• Output

  55
  
  
  
  
  

• Time Complexity: O(N), where N is the number of elements in the array arr[]
  Auxiliary Space: O(K), where K is the maximum length of jump.