Longest Increasing Subsequence having sum value atmost K

0 0 4 minutes read

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#

// 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)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!