Question

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

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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.

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

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;
  }
};

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
  }
}

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:

• 2557. Maximum Number of Integers to Choose From a Range II

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

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;
  }
}

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;
  }
};

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
}