Welcome to Subscribe On Youtube

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) {
            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;
    }
}

  • 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
    }
}

All Problems

All Solutions