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 = 15Output:3Explanation:

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.

A[] = {2, 4, 6, 40, 50}, B[] = {38, 36}, X=13Input:Output:2

**Approach:** This problem can be solved using Greedy Technique/a> and Searching. The key observation here is for any **i ^{th} 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 i
^{th}person using Binary Search or lower_bound(). Check if that value lies in the range**[a**or not._{i }– X, a_{i }+ X] - 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++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the maximum number` `// of persons receiving a chocolate` `int` `countMaxPersons(` `int` `* A, ` `int` `n, ` `int` `* B,` ` ` `int` `m, ` `int` `x)` `{` ` ` `// Initialize count as 0` ` ` `int` `count = 0;` ` ` `// Sort the given arrays` ` ` `sort(A, A + n);` ` ` `sort(B, B + m);` ` ` `// Initialize a multiset` ` ` `multiset<` `int` `> s;` ` ` `// Insert B[] array values into` ` ` `// the multiset` ` ` `for` `(` `int` `i = 0; i < m; i++)` ` ` `s.insert(B[i]);` ` ` `// Traverse elements in array A[]` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `int` `val = A[i] - x;` ` ` `// Search for the lowest value in B[]` ` ` `auto` `it = s.lower_bound(val);` ` ` `// If found, increment count,` ` ` `// and delete from set` ` ` `if` `(it != s.end()` ` ` `&& *it <= A[i] + x) {` ` ` `count++;` ` ` `s.erase(it);` ` ` `}` ` ` `}` ` ` `// Return the number of people` ` ` `return` `count;` `}` `// Driver Code` `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]);` ` ` `// Function Call` ` ` `cout << countMaxPersons(A, N, B, M, X);` `}` |

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## Python3

`# Python3 program for the above approach` `from` `bisect ` `import` `bisect_left` `# Function to count the maximum number` `# of persons receiving a chocolate` `def` `countMaxPersons(A, n, B, m, x):` ` ` `# Initialize count as 0` ` ` `count ` `=` `0` ` ` `# Sort the given arrays` ` ` `A ` `=` `sorted` `(A)` ` ` `B ` `=` `sorted` `(B)` ` ` `# Initialize a multiset` ` ` `s ` `=` `[]` ` ` `# Insert B[] array values into` ` ` `# the multiset` ` ` `for` `i ` `in` `range` `(m):` ` ` `s.append(B[i])` ` ` `# Traverse elements in array A[]` ` ` `for` `i ` `in` `range` `(n):` ` ` `val ` `=` `A[i] ` `-` `x` ` ` `# Search for the lowest value in B[]` ` ` `it ` `=` `bisect_left(s,val)` ` ` `# If found, increment count,` ` ` `# and delete from set` ` ` `if` `(it !` `=` `len` `(s) ` `and` `it <` `=` `A[i] ` `+` `x):` ` ` `count ` `+` `=` `1` ` ` `del` `s[it]` ` ` `# Return the number of people` ` ` `return` `count` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `A ` `=` `[` `90` `, ` `49` `, ` `20` `, ` `39` `, ` `49` `]` ` ` `B ` `=` `[` `14` `, ` `24` `, ` `82` `]` ` ` `X ` `=` `15` ` ` `N ` `=` `len` `(A)` ` ` `M ` `=` `len` `(B)` ` ` `# Function Call` ` ` `print` `(countMaxPersons(A, N, B, M, X))` ` ` `# This code is contributed by mohit kumar 29` |

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**Output:**

3

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

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