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 <= 1041 <= banned[i], n <= 1041 <= 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.
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| 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
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| class Solution {
public int maxCount(int[] banned, int n, int maxSum) {
Set<Integer> ban = new HashSet<>(banned.length);
for (int x : banned) {
ban.add(x);
}
int ans = 0, s = 0;
for (int i = 1; i <= n && s + i <= maxSum; ++i) {
if (!ban.contains(i)) {
++ans;
s += i;
}
}
return ans;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
|
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| 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
}
}
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| int cmp(const void* a, const void* b) {
return *(int*) a - *(int*) b;
}
int maxCount(int* banned, int bannedSize, int n, int maxSum) {
qsort(banned, bannedSize, sizeof(int), cmp);
int sum = 0;
int ans = 0;
for (int i = 1, j = 0; i <= n; i++) {
if (sum + i > maxSum) {
break;
}
if (j < bannedSize && i == banned[j]) {
while (j < bannedSize && i == banned[j]) {
j++;
}
} else {
sum += i;
ans++;
}
}
return ans;
}
|
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.
Similar problems:
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| class Solution:
def maxCount(self, banned: List[int], n: int, maxSum: int) -> int:
banned.extend([0, n + 1])
ban = sorted(x for x in set(banned) if x < n + 2)
ans = 0
for i, j in pairwise(ban):
left, right = 0, j - i - 1
while left < right:
mid = (left + right + 1) >> 1
if (i + 1 + i + mid) * mid // 2 <= maxSum:
left = mid
else:
right = mid - 1
ans += left
maxSum -= (i + 1 + i + left) * left // 2
if maxSum <= 0:
break
return ans
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| class Solution {
public int maxCount(int[] banned, int n, int maxSum) {
Set<Integer> black = new HashSet<>();
black.add(0);
black.add(n + 1);
for (int x : banned) {
if (x < n + 2) {
black.add(x);
}
}
List<Integer> ban = new ArrayList<>(black);
Collections.sort(ban);
int ans = 0;
for (int k = 1; k < ban.size(); ++k) {
int i = ban.get(k - 1), j = ban.get(k);
int left = 0, right = j - i - 1;
while (left < right) {
int mid = (left + right + 1) >>> 1;
if ((i + 1 + i + mid) * 1L * mid / 2 <= maxSum) {
left = mid;
} else {
right = mid - 1;
}
}
ans += left;
maxSum -= (i + 1 + i + left) * 1L * left / 2;
if (maxSum <= 0) {
break;
}
}
return ans;
}
}
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| class Solution {
public:
int maxCount(vector<int>& banned, int n, int maxSum) {
banned.push_back(0);
banned.push_back(n + 1);
sort(banned.begin(), banned.end());
banned.erase(unique(banned.begin(), banned.end()), banned.end());
banned.erase(remove_if(banned.begin(), banned.end(), [&](int x) { return x > n + 1; }), banned.end());
int ans = 0;
for (int k = 1; k < banned.size(); ++k) {
int i = banned[k - 1], j = banned[k];
int left = 0, right = j - i - 1;
while (left < right) {
int mid = left + ((right - left + 1) / 2);
if ((i + 1 + i + mid) * 1LL * mid / 2 <= maxSum) {
left = mid;
} else {
right = mid - 1;
}
}
ans += left;
maxSum -= (i + 1 + i + left) * 1LL * left / 2;
if (maxSum <= 0) {
break;
}
}
return ans;
}
};
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| func maxCount(banned []int, n int, maxSum int) (ans int) {
banned = append(banned, []int{0, n + 1}...)
sort.Ints(banned)
ban := []int{}
for i, x := range banned {
if (i > 0 && x == banned[i-1]) || x > n+1 {
continue
}
ban = append(ban, x)
}
for k := 1; k < len(ban); k++ {
i, j := ban[k-1], ban[k]
left, right := 0, j-i-1
for left < right {
mid := (left + right + 1) >> 1
if (i+1+i+mid)*mid/2 <= maxSum {
left = mid
} else {
right = mid - 1
}
}
ans += left
maxSum -= (i + 1 + i + left) * left / 2
if maxSum <= 0 {
break
}
}
return
}
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