Maximize count of persons receiving a chocolate
Last Updated :
02 Mar, 2023
Given two arrays A[], consisting of N integers, and B[], consisting of taste values of M chocolates and an integer X, the task is to find the maximum number of people who can receive a chocolate based on the condition that one person can have only one chocolate and with taste value in the range [A[i] – X, A[i] + X].
Note: Once a chocolate is given to a person, it cannot be given to any other person.
Examples:
Input: A[] = {90, 49, 20, 39, 60}, B[] = {14, 24, 82}, X = 15
Output: 3
Explanation:
1st person can pick the 3rd chocolate as the value of the 3rd chocolate ( = 82 ) lies in the range [75 ( = 90 – 15 ), 105 ( = 90 + 15)].
2nd person can’t pick any chocolate because there is no chocolate with value in the range [34 ( = 49 – 15 ), 64 ( = 49 + 15).
3rd person can pick the 1st chocolate as the value of the 1st chocolate lies in the range [5 ( = 20 – 15), 35 ( = 20 – 15)].
4th person can pick the 2nd chocolate because value of the 2nd chocolate lies in the range [ 24 ( = 39 – 15) and 54 ( = 39 – 15)].
5th person can’t pick any chocolate because there is no chocolate with value in the range [45 ( = 60 – 15), 75 ( = 60 – 15)].
Therefore, the total number of people receiving a chocolate is 3, which is the maximum possible.
Input: A[] = {2, 4, 6, 40, 50}, B[] = {38, 36}, X=13
Output: 2
Approach: This problem can be solved using Greedy Technique/a> and Searching. The key observation here is for any ith person, assign the chocolate with smallest possible value that lies in the range, if possible. Otherwise, exclude that person from the result. Follow the steps below:
- Sort the given arrays A[] and B[] in non-decreasing order.
- Initialize a multiset to store the elements of the array B[].
- Initialize a variable count = 0, to store the count of persons receiving a chocolate.
- Traverse the array A[] and find the smallest possible chocolate value that can be assigned to every ith person using Binary Search or lower_bound(). Check if that value lies in the range [ai – X, ai + X] or not.
- If found to be true, then increment count and remove the value of this chocolate from the multiset.
- After completing the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countMaxPersons(int* A, int n, int* B,
int m, int x)
{
int count = 0;
sort(A, A + n);
sort(B, B + m);
multiset<int> s;
for (int i = 0; i < m; i++)
s.insert(B[i]);
for (int i = 0; i < n; i++) {
int val = A[i] - x;
auto it = s.lower_bound(val);
if (it != s.end()
&& *it <= A[i] + x) {
count++;
s.erase(it);
}
}
return count;
}
int main()
{
int A[] = { 90, 49, 20, 39, 49 };
int B[] = { 14, 24, 82 };
int X = 15;
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
cout << countMaxPersons(A, N, B, M, X);
}
|
Java
import java.util.*;
public class Main {
public static int countMaxPersons(int[] A, int n, int[] B,
int m, int x) {
int count = 0;
Arrays.sort(A);
Arrays.sort(B);
TreeSet<Integer> s = new TreeSet<>();
for (int i = 0; i < m; i++)
s.add(B[i]);
for (int i = 0; i < n; i++) {
int val = A[i] - x;
Integer it = s.ceiling(val);
if (it != null && it <= A[i] + x) {
count++;
s.remove(it);
}
}
return count;
}
public static void main(String[] args) {
int[] A = {90, 49, 20, 39, 49};
int[] B = {14, 24, 82};
int X = 15;
int N = A.length;
int M = B.length;
System.out.println(countMaxPersons(A, N, B, M, X));
}
}
|
Python3
from bisect import bisect_left
def countMaxPersons(A, n, B, m, x):
count = 0
A = sorted(A)
B = sorted(B)
s = []
for i in range(m):
s.append(B[i])
for i in range(n):
val = A[i] - x
it = bisect_left(s,val)
if (it != len(s) and it <= A[i] + x):
count += 1
del s[it]
return count
if __name__ == '__main__':
A = [90, 49, 20, 39, 49]
B = [14, 24, 82]
X = 15
N = len(A)
M = len(B)
print(countMaxPersons(A, N, B, M, X))
|
C#
using System;
using System.Collections.Generic;
public class Program
{
public static int CountMaxPersons(int[] A, int n, int[] B,
int m, int x)
{
int count = 0;
Array.Sort(A);
Array.Sort(B);
SortedSet<int> s = new SortedSet<int>();
for (int i = 0; i < m; i++)
s.Add(B[i]);
for (int i = 0; i < n; i++)
{
int val = A[i] - x;
var it = s.GetViewBetween(val, A[i] + x).GetEnumerator();
if (it.MoveNext())
{
count++;
s.Remove(it.Current);
}
}
return count;
}
public static void Main(string[] args)
{
int[] A = { 90, 49, 20, 39, 49 };
int[] B = { 14, 24, 82 };
int X = 15;
int N = A.Length;
int M = B.Length;
Console.WriteLine(CountMaxPersons(A, N, B, M, X));
}
}
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Javascript
function countMaxPersons(A, n, B, m, x) {
let count = 0;
A = A.sort((a, b) => a - b);
B = B.sort((a, b) => a - b);
let s = new Set(B);
for (let i = 0; i < n; i++) {
let val = A[i] - x;
let it = Array.from(s).findIndex(elem => elem >= val);
if (it !== -1 && Array.from(s)[it] <= A[i] + x) {
count += 1;
s.delete(Array.from(s)[it]);
}
}
return count;
}
let A = [90, 49, 20, 39, 49];
let B = [14, 24, 82];
let X = 15;
let N = A.length;
let M = B.length;
console.log(countMaxPersons(A, N, B, M, X));
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Time Complexity: O(N*log N)
Auxiliary Space: O(M)
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