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: -1Input: nums = [10, -2, -10, -5, 20], K = 2
Output: 23
Approach:
- The optimal solution of this problem can be achieved by using the Sliding Window Maximum and Dynamic Programming .
- At any index i after calculating the maximum sum for i, nums[i] will now store the maximum possible sum that can be obtained from a subsequence ending at i.
- To calculate the maximum possible sum for every index i , 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.
- To use the values of calculated sums optimally we use a deque to store the sums along with their indexes in an increasing order of the sums . We also pop the element from the back when its index goes out of the window k of current index i.
CPP
// 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 codeint main(){ vector<int> nums = { 10, -2, -10, -5, 20 }; int K = 2; // Function call cout << ConstrainedSubsetSum(nums, K) << endl; return 0;} |
Java
// Java program to find the maximum sum// subsequence under given constraintimport java.io.*;import java.util.*;class GFG { // pair class static class pair { int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function to return the maximum sum static int ConstrainedSubsetSum(int[] nums, int k) { Deque<pair> d = new LinkedList<>(); // Iterating the given array for (int i = 0; i < nums.length; i++) { // Check if deque is empty nums[i] += (d.size() != 0 ? Math.max(d.peekLast().first, 0) : 0); while (d.size() != 0 && d.peekFirst().first < nums[i]) { d.removeFirst(); } d.offerFirst(new pair(nums[i], i)); if (d.peekLast().second == i - k) { d.removeLast(); } } int ans = nums[0]; for (int i = 0; i < nums.length; i++) { ans = Math.max(ans, nums[i]); } return ans; } // Driver code public static void main(String[] args) { int[] nums = { 10, -2, -10, -5, 20 }; int K = 2; // Function call System.out.print(ConstrainedSubsetSum(nums, K)); }}// This code is contributed by lokeshmvs21. |
Python3
# Python codefrom collections import dequedef ConstrainedSubsetSum(nums, k): d = deque() # Iterating the given array for i in range(len(nums)): nums[i] += d[-1][0] if d and d[-1][0] > 0 else 0 while d and d[0][0] < nums[i]: d.popleft() d.appendleft((nums[i], i)) if d and d[-1][1] == i - k: d.pop() ans=nums[0] for i in range(1,len(nums)): ans=max(ans,nums[i]) return ans# Driver codenums = [10, -2, -10, -5, 20]K = 2print(ConstrainedSubsetSum(nums, K))# This code is contributed by Utkarsh |
C#
using System;using System.Collections.Generic;class GFG { // pair class class Pair { public int First { get; set; } public int Second { get; set; } public Pair(int first, int second) { First = first; Second = second; } } // Function to return the maximum sum static int ConstrainedSubsetSum(int[] nums, int k) { LinkedList<Pair> d = new LinkedList<Pair>(); // Iterating the given array for (int i = 0; i < nums.Length; i++) { // Check if deque is empty nums[i] += (d.Count != 0 ? Math.Max(d.Last.Value.First, 0) : 0); while (d.Count != 0 && d.First.Value.First < nums[i]) { d.RemoveFirst(); } d.AddFirst(new Pair(nums[i], i)); if (d.Last.Value.Second == i - k) { d.RemoveLast(); } } int ans = nums[0]; foreach (Pair pair in d) { ans = Math.Max(ans, pair.First); } return ans; } // Driver code public static void Main(string[] args) { int[] nums = { 10, -2, -10, -5, 20 }; int k = 2; // Function call Console.Write(ConstrainedSubsetSum(nums, k)); }} |
Javascript
// JS program to find the maximum sum // subsequence under given constralet// Function return the maximum sum function ConstrainedSubsetSum(nums, k){ let d = []; // Iterating the given array for (let i = 0; i < nums.length; i++) { // Check if deque is empty nums[i] += d.length ? Math.max(d[d.length - 1][0], 0) : 0; while (d.length && d[0][0] < nums[i]) d.shift(); d.unshift([nums[i], i]); if (d[d.length - 1][1] === i - k) d.pop(); } let ans = nums[0]; for (let x of nums) ans = Math.max(ans, x); return ans;}// Driver codelet nums = [ 10, -2, -10, -5, 20 ];let K = 2;// Function callconsole.log(ConstrainedSubsetSum(nums, K)); |
Output
23
Time Complexity: O(N)
Auxiliary Space: O(K)
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