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

## Description

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

## Solutions

Solution 1: Greedy + Enumeration

We use the variable $s$ to represent the sum of the currently selected integers, and the variable $ans$ to represent the number of currently selected integers. We convert the array banned into a hash table for easy determination of whether a certain integer is not selectable.

Next, we start enumerating the integer $i$ from $1$. If $s + i \leq maxSum$ and $i$ is not in banned, then we can select the integer $i$, and add $i$ and $1$ to $s$ and $ans$ respectively.

Finally, we return $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the given integer.

Solution 2: Greedy + Binary Search

If $n$ is very large, the enumeration in Method One will time out.

We can add $0$ and $n + 1$ to the array banned, deduplicate the array banned, remove elements greater than $n+1$, and then sort it.

Next, we enumerate every two adjacent elements $i$ and $j$ in the array banned. The range of selectable integers is $[i + 1, j - 1]$. We use binary search to enumerate the number of elements we can select in this range, find the maximum number of selectable elements, and then add it to $ans$. At the same time, we subtract the sum of these elements from maxSum. If maxSum is less than $0$, we break the loop. Return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array banned.

• class Solution {
public int maxCount(int[] banned, int n, int maxSum) {
Set<Integer> ban = new HashSet<>(banned.length);
for (int x : banned) {
}
int ans = 0, s = 0;
for (int i = 1; i <= n && s + i <= maxSum; ++i) {
if (!ban.contains(i)) {
++ans;
s += i;
}
}
return ans;
}
}

• class Solution {
public:
int maxCount(vector<int>& banned, int n, int maxSum) {
unordered_set<int> ban(banned.begin(), banned.end());
int ans = 0, s = 0;
for (int i = 1; i <= n && s + i <= maxSum; ++i) {
if (!ban.count(i)) {
++ans;
s += i;
}
}
return ans;
}
};

• class Solution:
def maxCount(self, banned: List[int], n: int, maxSum: int) -> int:
ans = s = 0
ban = set(banned)
for i in range(1, n + 1):
if s + i > maxSum:
break
if i not in ban:
ans += 1
s += i
return ans


• func maxCount(banned []int, n int, maxSum int) (ans int) {
ban := map[int]bool{}
for _, x := range banned {
ban[x] = true
}
s := 0
for i := 1; i <= n && s+i <= maxSum; i++ {
if !ban[i] {
ans++
s += i
}
}
return
}

• function maxCount(banned: number[], n: number, maxSum: number): number {
const set = new Set(banned);
let sum = 0;
let ans = 0;
for (let i = 1; i <= n; i++) {
if (i + sum > maxSum) {
break;
}
if (set.has(i)) {
continue;
}
sum += i;
ans++;
}
return ans;
}


• use std::collections::HashSet;
impl Solution {
pub fn max_count(banned: Vec<i32>, n: i32, max_sum: i32) -> i32 {
let mut set = banned.into_iter().collect::<HashSet<i32>>();
let mut sum = 0;
let mut ans = 0;
for i in 1..=n {
if sum + i > max_sum {
break;
}
if set.contains(&i) {
continue;
}
sum += i;
ans += 1;
}
ans
}
}