How do you solve Maximum Number of Integers to Choose From a Range I on LeetCode?

Maximum Number of Integers to Choose From a Range I (LeetCode #2554) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set for banned numbers allows for quick lookups.

What is the time complexity of Maximum Number of Integers to Choose From a Range I?

The optimal solution for Maximum Number of Integers to Choose From a Range I runs in O(n) time and uses O(n) space. The time complexity is O(n) since we only loop through the range once, and the space complexity is O(n) for storing the banned set.

What is the brute force solution for Maximum Number of Integers to Choose From a Range I?

The brute force approach for Maximum Number of Integers to Choose From a Range I is Brute Force, with O(n²) time complexity and O(n) space complexity. We can simply check each integer from 1 to n and see if it can be included in our selection. This approach is straightforward but inefficient as it checks each number individually against the banned list and the maxSum. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Integers to Choose From a Range I?

Maximum Number of Integers to Choose From a Range I uses the following concepts: Array, Hash Table, Binary Search, Greedy, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Integers to Choose From a Range I?

Always consider edge cases, such as when all numbers are banned. Explain your thought process clearly while coding. Test your code with different inputs to ensure correctness.

What are common mistakes when solving Maximum Number of Integers to Choose From a Range I?

Not using a set for banned numbers, leading to O(n) lookups. Failing to break the loop when maxSum is exceeded.

#2554

Maximum Number of Integers to Choose From a Range I

Medium

ArrayHash TableBinary SearchGreedySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of checking each integer individually, we can iterate through the numbers from 1 to n, skipping the banned numbers and keeping track of the sum. This way, we can efficiently determine how many integers we can select without exceeding maxSum.

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Algorithm

5 steps

1Step 1: Create a set from the banned array for O(1) lookups.
2Step 2: Initialize a variable to keep track of the sum of chosen integers and a counter for the number of integers chosen.
3Step 3: Loop through each integer from 1 to n, check if it's not in the banned set.
4Step 4: If it's not banned and adding it to the sum does not exceed maxSum, add it to the sum and increment the counter.
5Step 5: If adding the next integer exceeds maxSum, break the loop.

solution.py12 lines

1def maxIntegers(banned, n, maxSum):
2    banned_set = set(banned)
3    total_sum = 0
4    count = 0
5    for i in range(1, n + 1):
6        if i not in banned_set:
7            if total_sum + i <= maxSum:
8                total_sum += i
9                count += 1
10            else:
11                break
12    return count

ℹ

Complexity note: The time complexity is O(n) since we only loop through the range once, and the space complexity is O(n) for storing the banned set.

1Using a set for banned numbers allows for quick lookups.
2Iterating through the range and checking conditions in a single pass is efficient.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.