Longest Increasing Subsequence having sum value atmost K

Last Updated : 28 Sep, 2021

Given an integer array arr[] of size N and an integer K. The task is to find the length of the longest subsequence whose sum is less than or equal to K.

Example:

Input: arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15} K = 40
Output: 5
Explanation:
If we select subsequence {0, 1, 3, 7, 15} then total sum will be 26, which is less than 40. Hence, the longest increasing possible subsequence length is 5.

Input: arr[] = {5, 8, 3, 7, 9, 1} K = 4
Output: 1

Approach:

The above problem can be solved using recursion.
- Choose the element at that position if the total sum is less than K and explore the rest items.
- Leave the element at that position and explore the rest.

Recurrence relation will be given as:

Recurrence relation:
T(N) = max(solve(arr, N, arr[i], i+1, K-arr[i])+1, solve(arr, N, prevele, i+1, K));
Base conditions:
if(i >= N || K <= 0)
return 0

Here is the implementation of the above approach:

C++

// C++ program to find the Longest
// Increasing Subsequence having sum
// value atmost K
#include <bits/stdc++.h>
using namespace std;

int solve(int arr[], int N,
          int prevele, int i, int K)
{
    // check for base cases
    if (i >= N || K <= 0)
        return 0;

    // check if it is possible to take
    // current elements
    if (arr[i] <= prevele
        || (K - arr[i] < 0)) {

        return solve(arr, N, prevele,
                     i + 1, K);
    }

    // if current element is ignored
    else {
        int ans = max(
            solve(arr, N, arr[i],
                  i + 1, K - arr[i])
                + 1,
            solve(arr, N, prevele,
                  i + 1, K));
        return ans;
    }
}

// Driver Code
int main()
{
    int N = 16;
    int arr[N]
        = { 0, 8, 4, 12,
            2, 10, 6, 14,
            1, 9, 5, 13,
            3, 11, 7, 15 };
    int K = 40;

    cout << solve(arr, N,
                  INT_MIN, 0, K)
         << endl;
}

Java

// Java program to find the Longest 
// Increasing Subsequence having sum 
// value atmost K 
import java.io.*; 

class GFG{ 
    
static int solve(int arr[], int N,
                 int prevele, int i, int K)
{
    
    // Check for base cases
    if (i >= N || K <= 0)
        return 0;

    // Check if it is possible to take
    // current elements
    if (arr[i] <= prevele ||
       (K - arr[i] < 0))
    {
        return solve(arr, N, prevele,
                     i + 1, K);
    }

    // If current element is ignored
    else
    {
        int ans = Math.max(solve(arr, N, arr[i],
                              i + 1, K - arr[i]) + 1,
                           solve(arr, N, prevele,
                                 i + 1, K));
                                 
        return ans;
    }
} 

// Driver code 
public static void main (String[] args) 
{ 
    int N = 16;
    int arr[] = new int[]{ 0, 8, 4, 12,
                           2, 10, 6, 14,
                           1, 9, 5, 13,
                           3, 11, 7, 15 };
    int K = 40;

    System.out.print(solve(arr, N,
          Integer.MIN_VALUE, 0, K)); 
} 
} 

// This code is contributed by Pratima Pandey

Python3

# Python3 program to find the Longest
# Increasing Subsequence having sum
# value atmost K
import sys

def solve(arr, N, prevele, i, K):
    
    # Check for base cases
    if (i >= N or K <= 0):
        return 0;

    # Check if it is possible to take
    # current elements
    if (arr[i] <= prevele or 
       (K - arr[i] < 0)):
        return solve(arr, N, prevele,
                     i + 1, K);

    # If current element is ignored
    else:
        ans = max(solve(arr, N, arr[i],
                     i + 1, K - arr[i]) + 1,
                  solve(arr, N, prevele, 
                        i + 1, K));

        return ans;

# Driver code
if __name__ == '__main__':
    
    N = 16;
    arr = [ 0, 8, 4, 12,
            2, 10, 6, 14,
            1, 9, 5, 13,
            3, 11, 7, 15 ];
    K = 40;

    print(solve(arr, N, -sys.maxsize, 0, K));

# This code is contributed by 29AjayKumar

// C# program to find the Longest 
// Increasing Subsequence having sum 
// value atmost K 
using System; 

class GFG{ 
    
static int solve(int[] arr, int N,
                 int prevele, int i, int K)
{
    
    // Check for base cases
    if (i >= N || K <= 0)
        return 0;

    // Check if it is possible to take
    // current elements
    if (arr[i] <= prevele ||
       (K - arr[i] < 0))
    {
        return solve(arr, N, prevele,
                     i + 1, K);
    }

    // If current element is ignored
    else
    {
        int ans = Math.Max(solve(arr, N, arr[i],
                                 i + 1, K - arr[i]) + 1,
                           solve(arr, N, prevele,
                                 i + 1, K));
                                
        return ans;
    }
} 

// Driver code 
public static void Main () 
{ 
    int N = 16;
    int[] arr = new int[]{ 0, 8, 4, 12,
                           2, 10, 6, 14,
                           1, 9, 5, 13,
                           3, 11, 7, 15 };
    int K = 40;

    Console.Write(solve(arr, N,
        Int32.MinValue, 0, K)); 
} 
}

// This code is contributed by sanjoy_62

JavaScript

<script>

// Javascript program to find the Longest
// Increasing Subsequence having sum
// value atmost K

function solve(arr, N, prevele, i, K)
{
    // check for base cases
    if (i >= N || K <= 0)
        return 0;

    // check if it is possible to take
    // current elements
    if (arr[i] <= prevele
        || (K - arr[i] < 0)) {

        return solve(arr, N, prevele,
                     i + 1, K);
    }

    // if current element is ignored
    else {
        var ans = Math.max(
            solve(arr, N, arr[i],
                  i + 1, K - arr[i])
                + 1,
            solve(arr, N, prevele,
                  i + 1, K));
        return ans;
    }
}

// Driver Code
var N = 16;
var arr
    = [0, 8, 4, 12,
        2, 10, 6, 14,
        1, 9, 5, 13,
        3, 11, 7, 15];
var K = 40;
document.write( solve(arr, N,
              -1000000000, 0, K));


</script>

Output:

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

Longest Increasing Subsequence having sum value atmost K

sunilkannur98

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