Given an array arr[] of size N and a number K, the task is to find the length of the smallest subsequence such that the sum of the subsequence is greater than or equal to number K.
Example:
Input: arr[] = {2, 3, 1, 5, 6, 3, 7, 9, 14, 10, 2, 5}, K = 35
Output: 4
Smallest subsequence with the sum greater than or equal to the given sum K is {7, 9, 14, 10}
Input: arr[] = {1, 2, 2, 2, 3, 4, 5, 4, 7, 6, 5, 12}, K = 70
Output:-1
Subsequence with sum greater than equal to the given sum is not possible.
Approach:
- This problem can be solved with the help of priority queue
- Traverse input array and insert every array element into priority queue.
- Initialize variables that holds the sum of picked element from priority queue and the variable to get the count of picked element from priority queue to 0
- Pop the elements out from the priority queue until the priority queue is not empty
- Add the element into the sum
- Increase the count because the element is picked to contribute to the total sum
- Check if the sum is greater than the given number K, If yes then stop checking and output the count.
Below is the implementation of above approach.
C++
// C++ implementation to find length of smallest // subsequence such that sum of elements // is greater than equal to given number K #include <bits/stdc++.h> using namespace std;
// Function to find the smallest // subsequence such that sum of elements // is greater than equal to given number K int lengthOfSmallestSubsequence( int K, vector< int > v)
{ // Initialize priority queue
priority_queue< int > pq;
// Loop to insert all elements into
// the priority queue
for ( int i = 0; i < v.size(); i++) {
pq.push(v[i]);
}
int sum = 0, count = 0;
// Loop to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
while (!pq.empty() && sum < K) {
sum += pq.top();
pq.pop();
count++;
}
// If sum is less than K
// then return -1 else return count.
if (sum < K) {
return -1;
}
return count;
} // Driver code int main()
{ vector< int > v{ 2, 3, 1, 5,
6, 3, 7, 9,
14, 10, 2, 5 };
int K = 35;
cout << lengthOfSmallestSubsequence(K, v);
return 0;
} |
Java
// Java implementation to find length of smallest // subsequence such that sum of elements // is greater than equal to given number K import java.util.*;
class GFG
{ // Function to find the smallest // subsequence such that sum of elements // is greater than equal to given number K static int lengthOfSmallestSubsequence( int K, int []v)
{ // Initialize priority queue
Queue<Integer> pq =
new PriorityQueue<Integer>(Collections.reverseOrder());
// Loop to insert all elements into
// the priority queue
for ( int i = 0 ; i < v.length; i++)
{
pq.add(v[i]);
}
int sum = 0 , count = 0 ;
// Loop to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
while (!pq.isEmpty() && sum < K)
{
sum += pq.peek();
pq.remove();
count++;
}
// If sum is less than K
// then return -1 else return count.
if (sum < K)
{
return - 1 ;
}
return count;
} // Driver code public static void main(String[] args)
{ int []v = { 2 , 3 , 1 , 5 ,
6 , 3 , 7 , 9 ,
14 , 10 , 2 , 5 };
int K = 35 ;
System.out.print(lengthOfSmallestSubsequence(K, v));
} } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation to find length of smallest # subsequence such that sum of elements # is greater than equal to given number K # Function to find the smallest # subsequence such that sum of elements # is greater than equal to given number K def lengthOfSmallestSubsequence(K, v):
# Initialize priority queue
pq = []
# Loop to insert all elements into
# the priority queue
for i in v:
pq.append(i)
pq.sort()
sum = 0
count = 0
# Loop to find the smallest
# subsequence such that sum of elements
# is greater than equal to given number K
while ( len (pq) > 0 and sum < K):
sum + = pq[ - 1 ]
del pq[ - 1 ]
count + = 1
# If sum is less than K
# then return -1 else return count.
if ( sum < K):
return - 1
return count
# Driver code v = [ 2 , 3 , 1 , 5 ,
6 , 3 , 7 , 9 ,
14 , 10 , 2 , 5 ]
K = 35
print (lengthOfSmallestSubsequence(K, v))
# This code is contributed by mohit kumar 29 |
C#
// C# implementation to find length of smallest // subsequence such that sum of elements // is greater than equal to given number K using System; using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{ // Function to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
static int lengthOfSmallestSubsequence( int K, int []v)
{
// Initialize List
List< int > pq = new List< int >();
// Loop to insert all elements into
// the List
for ( int i = 0; i < v.Length; i++)
{
pq.Add(v[i]);
}
// Sort list in reverse order
pq.Sort();
pq.Reverse();
int sum = 0;
int count = 0;
// Loop to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
while (pq.Count > 0 && sum < K)
{
sum += pq[0];
pq.RemoveAt(0);
count++;
}
// If sum is less than K
// then return -1 else return count.
if (sum < K)
{
return -1;
}
return count;
}
// Driver code
static public void Main ()
{
int []v = { 2, 3, 1, 5,6, 3, 7, 9, 14, 10, 2, 5 };
int K = 35;
Console.WriteLine(lengthOfSmallestSubsequence(K, v));
}
} // This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // JavaScript implementation to find length of smallest // subsequence such that sum of elements // is greater than equal to given number K using System; // Function to find the smallest // subsequence such that sum of elements // is greater than equal to given number K function lengthOfSmallestSubsequence(K, v) {
// Initialize List
let pq = new Array();
// Loop to insert all elements into
// the List
for (let i = 0; i < v.length; i++) {
pq.push(v[i]);
}
// Sort list in reverse order
pq.sort((a, b) => b - a);
let sum = 0;
let count = 0;
// Loop to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
while (pq.length > 0 && sum < K) {
sum += pq[0];
pq.splice(0, 1);
count++;
}
// If sum is less than K
// then return -1 else return count.
if (sum < K) {
return -1;
}
return count;
} // Driver code let v = [2, 3, 1, 5, 6, 3, 7, 9, 14, 10, 2, 5]; let K = 35; document.write(lengthOfSmallestSubsequence(K, v)); // This code is contributed by gfgking </script> |
Output:
4