Find maximum Subsequence Sum according to given conditions

Last Updated : 03 May, 2020

Given an integer array nums and an integer K, The task is to find the maximum sum of a non-empty subsequence of the array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j – i <= K is satisfied.

A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.

Examples:

Input: nums = [10, 2, -10, 5, 20], K = 2
Output: 37
Explanation: 
The subsequence is [10, 2, 5, 20].

Input: nums = [-1, -2, -3], K = 1
Output: -1

Input: nums = [10, -2, -10, -5, 20], K = 2
Output: 23

Aprroach:

The optimal solution of this problem can be achieved by using the Sliding Window Maximum .
For every index, check what is the maximum value that can be obtained from a window of size K before it. If the maximum value is negative, use zero instead.

// C++ program to find the maximum sum  
// subsequence under given constraint 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function return the maximum sum  
int ConstrainedSubsetSum(vector<int>& nums, 
                         int k) 
{ 
    deque<pair<int, int> > d; 
      
    // Iterating the given array 
    for (int i = 0; i < nums.size(); i++) 
    { 
        // Check if deque is empty 
        nums[i] += d.size() 
            ? max(d.back().first, 0) : 0; 
          
        while (d.size() && 
               d.front().first < nums[i]) 
            d.pop_front(); 
          
        d.push_front({ nums[i], i }); 
          
        if (d.back().second == i - k) 
            d.pop_back(); 
    } 
  
    int ans = nums[0]; 
      
    for (auto x : nums) 
        ans = max(ans, x); 
  
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    vector<int> nums = { 10, -2, -10,  
                        -5, 20 }; 
    int K = 2; 
      
    // Function call 
    cout << ConstrainedSubsetSum(nums, K)  
        << endl; 
    return 0; 
} 

Output:

Time Complexity: O(N)
Space Complexity: O(K)

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