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Longest subsequence whose average is less than K

  • Difficulty Level : Hard
  • Last Updated : 06 May, 2021
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Given an array of N positive integers and Q queries consisting of an integer K, the task is to print the length of the longest subsequence whose average is less than K. 
Examples: 
 

Input: a[] = {1, 3, 2, 5, 4} 
Query1: K = 3 
Query2: K = 5
Output: 


Query1: The subsequence is: {1, 3, 2, 4} or {1, 3, 2, 5} 
Query2: The subsequence is: {1, 3, 2, 5, 4} 
 

 

A Naive Approach is to generate all subsequences using power-set and check for the longest subsequence whose average is less than K. 
Time Complexity: O(2N * N ) 
An efficient approach is to sort the array elements and find the average of elements starting from the left. Insert the average of elements computed from the left into the container(vector or arrays). Sort the container’s element and then use binary search to search for the number K in the container. The length of the longest subsequence will thus be the index number which upper_bound() returns for every query.
Below is the implementation of the above approach. 
 

C++




// C++ program to perform Q queries
// to find longest subsequence whose
// average is less than K
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the length for evey query
int longestSubsequence(int a[], int n, int q[], int m)
{
 
    // sort array of N elements
    sort(a, a + n);
    int sum = 0;
 
    // Array to store average from left
    int b[n];
 
    for (int i = 0; i < n; i++) {
        sum += a[i];
        double av = (double)(sum) / (double)(i + 1);
        b[i] = ((int)(av + 1));
    }
 
    // Sort array of average
    sort(b, b + n);
 
    // number of queries
 
    for (int i = 0; i < m; i++) {
        int k = q[i];
 
        // print answer to every query
        // using binary search
        int longest = upper_bound(b, b + n, k) - b;
 
        cout << "Answer to Query" << i + 1 << ": "
             << longest << endl;
    }
}
 
// Driver Code
int main()
{
    int a[] = { 1, 3, 2, 5, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // 4 queries
    int q[] = { 4, 2, 1, 5 };
    int m = sizeof(q) / sizeof(q[0]);
 
    longestSubsequence(a, n, q, m);
    return 0;
}

Java




// Java program to perform Q queries
// to find longest subsequence whose
// average is less than K
import java.util.Arrays;
 
class GFG
{
 
    // Function to print the length for evey query
    static void longestSubsequence(int a[], int n,
                                    int q[], int m)
    {
 
        // sort array of N elements
        Arrays.sort(a);
        int sum = 0;
 
        // Array to store average from left
        int []b = new int[n];
 
        for (int i = 0; i < n; i++)
        {
            sum += a[i];
            double av = (double)(sum) / (double)(i + 1);
            b[i] = ((int)(av + 1));
        }
 
        // Sort array of average
        Arrays.sort(b);
 
        // number of queries
 
        for (int i = 0; i < m; i++)
        {
            int k = q[i];
 
            // print answer to every query
            // using binary search
            int longest = upperBound(b,0, n, k);
 
            System.out.println("Answer to Query" + (i + 1) +": "
                + longest);
        }
    }
    private static int upperBound(int[] a, int low, int high, int element)
    {
        while(low < high)
        {
            int middle = low + (high - low)/2;
            if(a[middle] > element)
                high = middle;
            else
                low = middle + 1;
        }
        return low;
    }
     
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = { 1, 3, 2, 5, 4 };
        int n = a.length;
 
        // 4 queries
        int q[] = { 4, 2, 1, 5 };
        int m = q.length;
 
        longestSubsequence(a, n, q, m);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 program to perform Q queries to find
# longest subsequence whose average is less than K
import bisect
 
# Function to print the length for evey query
def longestSubsequence(a, n, q, m):
  
    # sort array of N elements
    a.sort()
    Sum = 0
 
    # Array to store average from left
    b = [None] * n
 
    for i in range(0, n): 
        Sum += a[i]
        av = Sum // (i + 1)
        b[i] = av + 1
 
    # Sort array of average
    b.sort()
 
    # number of queries
 
    for i in range(0, m): 
        k = q[i]
 
        # print answer to every query
        # using binary search
        longest = bisect.bisect_right(b, k)
 
        print("Answer to Query", i + 1, ":", longest)
 
# Driver Code
if __name__ == "__main__":
  
    a = [1, 3, 2, 5, 4
    n = len(a)
 
    # 4 queries
    q = [4, 2, 1, 5
    m = len(q)
 
    longestSubsequence(a, n, q, m)
     
# This code is contributed by Rituraj Jain

C#




// C# program to perform Q queries
// to find longest subsequence whose
// average is less than K
using System;
 
class GFG
{
     
    // Function to print the length for evey query
    static void longestSubsequence(int []a, int n,
                                    int []q, int m)
    {
 
        // sort array of N elements
        Array.Sort(a);
        int sum = 0;
 
        // Array to store average from left
        int []b = new int[n];
 
        for (int i = 0; i < n; i++)
        {
            sum += a[i];
            double av = (double)(sum) / (double)(i + 1);
            b[i] = ((int)(av + 1));
        }
 
        // Sort array of average
        Array.Sort(b);
 
        // number of queries
 
        for (int i = 0; i < m; i++)
        {
            int k = q[i];
 
            // print answer to every query
            // using binary search
            int longest = upperBound(b,0, n, k);
 
            Console.WriteLine("Answer to Query" + (i + 1) +": "
                + longest);
        }
    }
     
    private static int upperBound(int[] a, int low,
                                    int high, int element)
    {
        while(low < high)
        {
            int middle = low + (high - low)/2;
            if(a[middle] > element)
                high = middle;
            else
                low = middle + 1;
        }
        return low;
    }
     
    // Driver Code
    static public void Main ()
    {
        int []a = { 1, 3, 2, 5, 4 };
        int n = a.Length;
 
        // 4 queries
        int []q = { 4, 2, 1, 5 };
        int m = q.Length;
 
        longestSubsequence(a, n, q, m);
    }
}
     
/* This code contributed by ajit */

Javascript




<script>
 
    // Javascript program to perform Q queries
    // to find longest subsequence whose
    // average is less than K
     
    // Function to print the length for evey query
    function longestSubsequence(a, n, q, m)
    {
   
        // sort array of N elements
        a.sort(function(a, b){return a - b});
        let sum = 0;
   
        // Array to store average from left
        let b = new Array(n);
   
        for (let i = 0; i < n; i++)
        {
            sum += a[i];
            let av = parseInt((sum) / (i + 1), 10);
            b[i] = (av + 1);
        }
   
        // Sort array of average
        b.sort(function(a, b){return a - b});
   
        // number of queries
   
        for (let i = 0; i < m; i++)
        {
            let k = q[i];
   
            // print answer to every query
            // using binary search
            let longest = upperBound(b,0, n, k);
   
            document.write("Answer to Query" +
            (i + 1) +": "
                + longest + "</br>");
        }
    }
       
    function upperBound(a, low, high, element)
    {
        while(low < high)
        {
            let middle = low +
            parseInt((high - low)/2, 10);
            if(a[middle] > element)
                high = middle;
            else
                low = middle + 1;
        }
        return low;
    }
     
    let a = [ 1, 3, 2, 5, 4 ];
    let n = a.length;
 
    // 4 queries
    let q = [ 4, 2, 1, 5 ];
    let m = q.length;
 
    longestSubsequence(a, n, q, m);
     
</script>

Output:  



Answer to Query1: 5
Answer to Query2: 2
Answer to Query3: 0
Answer to Query4: 5

Time Complexity: O(N*log N + M*log N) 
Auxiliary Space: O(N)
 

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