Maximum Number of Integers to Choose From a Range I

2554. Maximum Number of Integers to Choose From a Range I

Difficulty: Medium

Topics: Array, Hash Table, Binary Search, Greedy, Sorting

You are given an integer array banned and two integers n and maxSum. You are choosing some number of integers following the below rules:

• The chosen integers have to be in the range [1, n].
• Each integer can be chosen at most once.
• The chosen integers should not be in the array banned.
• The sum of the chosen integers should not exceed maxSum.

Return the maximum number of integers you can choose following the mentioned rules.

Example 1:

• Input: banned = [1,6,5], n = 5, maxSum = 6
• Output: 2
• Explanation: You can choose the integers 2 and 4.
  • 2 and 4 are from the range [1, 5], both did not appear in banned, and their sum is 6, which did not exceed maxSum.

Example 2:

• Input: banned = [1,2,3,4,5,6,7], n = 8, maxSum = 1
• Output: 0
• Explanation: You cannot choose any integer while following the mentioned conditions.

Example 3:

• Input: banned = [11], n = 7, maxSum = 50
• Output: 7
• Explanation: You can choose the integers 1, 2, 3, 4, 5, 6, and 7.
  • They are from the range [1, 7], all did not appear in banned, and their sum is 28, which did not exceed maxSum.

Constraints:

• 1 <= banned.length <= 10⁴
• 1 <= banned[i], n <= 10⁴
• 1 <= maxSum <= 10⁹

Hint:

1. Keep the banned numbers that are less than n in a set.
2. Loop over the numbers from 1 to n and if the number is not banned, use it.
3. Keep adding numbers while they are not banned, and their sum is less than k.

Solution:

We can use a greedy approach where we iterate over the numbers from 1 to n, skipping the banned numbers, and keep adding the valid numbers (those not in the banned array) to a running sum until we reach the maxSum.

Here's the step-by-step solution:

Steps:

1. Convert banned array to a set for quick lookups: Using array_flip() can convert the banned array into a set for O(1) average-time complexity lookups.
2. Iterate from 1 to n: Check each number, if it's not in the banned set and if adding it doesn't exceed maxSum, add it to the sum and increase the count.
3. Stop once adding the next number exceeds maxSum: Since the goal is to maximize the number of chosen integers without exceeding the sum, this greedy approach ensures we take the smallest available numbers first.

Approach:

1. Exclude Banned Numbers: We'll keep track of the banned numbers in a set (or an associative array) for fast lookups.
2. Greedy Selection: Start selecting numbers from 1 to n in ascending order, as this will allow us to maximize the number of integers selected. Each time we select a number, we'll add it to the sum and check if it exceeds maxSum. If it does, stop.
3. Efficiency Considerations: Since we are iterating over numbers from 1 to n, and checking if each is in the banned set (which can be done in constant time), the approach runs in linear time relative to n and the size of the banned list.

Let's implement this solution in PHP: 2554. Maximum Number of Integers to Choose From a Range I

  
  
  Explanation:



Convert banned array to set:


We use array_flip($banned) to create a set from the banned array, which allows for O(1) lookups to check if a number is banned.

Iterate from 1 to n:


We iterate through numbers from 1 to n. For each number:


If the number is not in the banned set (checked using isset($bannedSet[$i])),
And if adding the number to the sum does not exceed maxSum,
We include that number and update the sum and count.


Return the count:


After the loop, we return the number of integers selected ($count).
$bannedSet = array_flip($banned);: This converts the banned list into a set (associative array) for fast lookups.
for ($i = 1; $i <= $n; $i  ): We iterate over all integers from 1 to n.
if (isset($bannedSet[$i])) { continue; }: This checks if the current number is in the banned set. If it is, we skip it.
if ($sum   $i > $maxSum) { break; }: If adding the current number exceeds maxSum, we stop the process.
$sum  = $i; $count  ;: If the number is valid and adding it doesn't exceed maxSum, we include it in our sum and increase the count.



  
  
  Time Complexity:



The creation of the banned set (array_flip) is O(b), where b is the length of the banned array.
The loop iterates n times (for numbers from 1 to n), and each lookup into the banned set takes O(1) time. So, the time complexity of the loop is O(n).
Thus, the overall time complexity is O(n   b), which is efficient given the problem constraints.



  
  
  Example Walkthrough:


For the input:



Input 1: banned = [1, 6, 5], n = 5, maxSum = 6


We create the banned set: {1, 5, 6}.
We iterate through numbers 1 to 5:


1 is banned, skip it.
2 is not banned, add it to sum (sum = 2, count = 1).
3 is not banned, add it to sum (sum = 5, count = 2).
4 is not banned, but adding it to the sum would exceed maxSum (5   4 = 9), so skip it.





The result is 2.







Input 2: banned = [1, 2, 3, 4, 5, 6, 7], n = 8, maxSum = 1


All numbers from 1 to 7 are banned, so no valid numbers can be chosen.
The result is 0.







Input 3: banned = [11], n = 7, maxSum = 50


The only banned number is 11, which is outside the range 1 to 7.
We can select all numbers from 1 to 7, and their sum is 28, which is less than maxSum.
The result is 7.







This solution efficiently handles the problem within the given constraints.

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