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 Codeint 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 Kimport sysdef 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 codeif __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 Kfunction 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 Codevar 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:
5
Time Complexity: O (2N)
Auxiliary Space: O (1)
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