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Longest non-decreasing subsequence having difference between adjacent elements less than D

  • Last Updated : 27 Oct, 2021

Given an array arr[] of N integers and an integer D, the task is to find the length of the longest non-decreasing subsequence such that the difference between every adjacent element is less than D.

Examples:

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Input: arr[] = {1, 3, 2, 4, 5}, D = 2
Output: 3
Explanation:
Consider the subsequence as {3, 4, 5}, which is of maximum length = 3 satisfying the given criteria.



Input: arr[] = {1, 5, 3, 2, 7}, D = 2
Output: 2

Approach: The given problem is a variation of Longest Increasing Subsequence with criteria for the difference between adjacent array elements as less than D, this idea can be implemented using Dynamic Programming. Follow the steps below to solve the given problem:

  • Initialize a dp array, where dp[i] will store the maximum length of non-decreasing subsequence after including the ith element such that the difference between every adjacent pair of elements is less than D.
  • Initialize all values of the array dp[] as 1.
  • Iterate a loop over the range [0, N] and in each iteration, i traverse the given array arr[] over the range [0, i – 1] using the variable j and if the value of arr[j] is at least arr[i] and the difference between them is less than D, then update the value of dp[i] to the maximum of dp[i] and (1 + dp[j]).
  • After completing the above steps, print the maximum value of the array dp[] as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the length of the
// longest non-decreasing subsequence
// having the difference as D for every
// adjacent elements
int longestSubsequence(vector<int> arr,
                       int d)
{
    // Store the size of array
    int n = arr.size();
 
    // Stores the maximum length of the
    // subsequence after including the
    // ith element
    vector<int> dp(n, 1);
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
 
            // If it satisfies the
            // given condition
            if (arr[i] - d < arr[j]
                and arr[i] >= arr[j]) {
 
                // Update dp[i]
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
    }
 
    // Maximum value in the dp
    // table is the answer
    return *max_element(
        dp.begin(), dp.end());
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 3, 2, 4, 5 };
    int D = 2;
    cout << longestSubsequence(arr, D);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG {
     
    // Function to return the length of the
    // longest non-decreasing subsequence
    // having the difference as D for every
    // adjacent elements
    static int longestSubsequence(int  []arr,
                           int d)
    {
       
        // Store the size of array
        int n = arr.length;
     
        // Stores the maximum length of the
        // subsequence after including the
        // ith element
        int []dp = new int[n];
         
        for(int i = 0; i < n ; i++)
            dp[i] = 1;
     
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
     
                // If it satisfies the
                // given condition
                if (arr[i] - d < arr[j] && arr[i] >= arr[j]) {
     
                    // Update dp[i]
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }
     
        // Maximum value in the dp
        // table is the answer
        Arrays.sort(dp);
        return dp[n - 1];
    }
     
    // Driver Code
    public static void main (String[] args) {
        int arr[] = { 1, 3, 2, 4, 5 };
        int D = 2;
        System.out.println(longestSubsequence(arr, D));
    }
}
 
// This code is contributed by AnkThon

Python3




# python program for the above approach
 
# Function to return the length of the
# longest non-decreasing subsequence
# having the difference as D for every
# adjacent elements
def longestSubsequence(arr, d):
 
    # Store the size of array
    n = len(arr)
 
    # Stores the maximum length of the
    # subsequence after including the
    # ith element
    dp = [1 for _ in range(n)]
 
    for i in range(0, n):
        for j in range(0, i):
 
            # If it satisfies the
            # given condition
            if (arr[i] - d < arr[j] and arr[i] >= arr[j]):
 
                # Update dp[i]
                dp[i] = max(dp[i], dp[j] + 1)
 
    # Maximum value in the dp
    # table is the answer
    return max(dp)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 3, 2, 4, 5]
    D = 2
    print(longestSubsequence(arr, D))
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
 
public class GFG {
     
    // Function to return the length of the
    // longest non-decreasing subsequence
    // having the difference as D for every
    // adjacent elements
    static int longestSubsequence(int  []arr,
                           int d)
    {
       
        // Store the size of array
        int n = arr.Length;
     
        // Stores the maximum length of the
        // subsequence after including the
        // ith element
        int []dp = new int[n];
         
        for(int i = 0; i < n ; i++)
            dp[i] = 1;
     
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
     
                // If it satisfies the
                // given condition
                if (arr[i] - d < arr[j] && arr[i] >= arr[j]) {
     
                    // Update dp[i]
                    dp[i] = Math.Max(dp[i], dp[j] + 1);
                }
            }
        }
     
        // Maximum value in the dp
        // table is the answer
        Array.Sort(dp);
        return dp[n - 1];
    }
     
    // Driver Code
    public static void Main (string[] args) {
        int []arr = { 1, 3, 2, 4, 5 };
        int D = 2;
        Console.WriteLine(longestSubsequence(arr, D));
    }
}
 
// This code is contributed by AnkThon

Javascript




<script>
// Javascript program for the above approach
 
// Function to return the length of the
// longest non-decreasing subsequence
// having the difference as D for every
// adjacent elements
function longestSubsequence(arr, d)
{
 
  // Store the size of array
  let n = arr.length;
 
  // Stores the maximum length of the
  // subsequence after including the
  // ith element
  let dp = new Array(n).fill(1);
 
  for (let i = 0; i < n; i++)
  {
    for (let j = 0; j < i; j++)
    {
     
      // If it satisfies the
      // given condition
      if (arr[i] - d < arr[j] && arr[i] >= arr[j])
      {
       
        // Update dp[i]
        dp[i] = Math.max(dp[i], dp[j] + 1);
      }
    }
  }
 
  // Maximum value in the dp
  // table is the answer
  return dp.sort((a, b) => b - a)[0];
}
 
// Driver Code
let arr = [1, 3, 2, 4, 5];
let D = 2;
document.write(longestSubsequence(arr, D));
 
// This code is contributed by gfgking.
</script>

 
 

Output: 
3

 

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

 




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