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 = 5Output:
4
5
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 every queryint 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 Codeint 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 Kimport java.util.Arrays;class GFG { // Function to print the length for every 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 every 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 Kusing System;class GFG{ // Function to print the length for every 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 every 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
Complexity Analysis:
- Time Complexity: O(N*log N + M*log N)
- Auxiliary Space: O(N)
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