In this blog post, we’re tackling a fun and challenging problem: selecting integers while respecting some tricky rules. It’s like solving a puzzle where every piece must fit perfectly. 🤓 Let’s dive in and break it down step by step! 💪

📝 Problem Statement 📋

You are given an integer array banned, and two integers n and maxSum. Your task is to choose integers from the range [1, n] such that:

1. No integers are from the banned list.
2. Each integer can be chosen at most once.
3. The sum of the chosen integers must not exceed maxSum.

Your goal is to maximize the number of integers you can choose while following the above rules.

🔍 Examples 🔗

Example 1:

Input: banned = [1,6,5], n = 5, maxSum = 6
Output: 2
Explanation: You can choose integers 2 and 4. Their sum is 6, which meets the constraints.

Example 2:

Input: banned = [1,2,3,4,5,6,7], n = 8, maxSum = 1
Output: 0
Explanation: You cannot choose any integer within the constraints.

Example 3:

Input: banned = [11], n = 7, maxSum = 50
Output: 7
Explanation: All integers from 1 to 7 can be chosen, as their sum is 28, which is within `maxSum`.

🛠️ Constraints 📐

• \[1 \leq \text{banned.length} \leq 10^4\]
• \[1 \leq \text{banned[i]}, n \leq 10^4\]
• \[1 \leq \text{maxSum} \leq 10^9\]

🧠 Approach and Solutions 🧩

Understanding the problem and its nuances is key to crafting an efficient solution. Here’s a more detailed breakdown of how we approach this problem and arrive at the optimal solution.

🧩 Step 1: Analyze the Problem

We aim to maximize the count of integers selected from the range [1, n] while adhering to three constraints:

1. No banned numbers: Numbers in the banned list are off-limits.
2. Unique selection: Each number can only be chosen once.
3. Sum constraint: The cumulative sum of selected numbers cannot exceed maxSum.

Key Observations:

• The problem revolves around efficiently managing constraints while maximizing the selection.
• We need to identify valid numbers, starting from the smallest ones, to maximize the count before hitting the maxSum threshold.
• The order of numbers matters: prioritizing smaller numbers ensures we stay within the maxSum limit longer.

🛠️ Step 2: Outline the Solution

1. Filter the Banned Numbers:
   • Use a set to store banned numbers. This ensures constant-time lookups, critical for keeping our solution efficient.
2. Iterate Through the Range [1, n]:
   • Start with the smallest number in the range. This is a greedy choice since smaller numbers contribute less to the cumulative sum, allowing us to maximize the count.
   • For each number:
     • Skip if it’s banned.
     • Add it to the cumulative sum if it doesn’t exceed maxSum.
3. Stop Early:
   • Once adding another number would exceed maxSum, break the loop. There’s no need to evaluate further numbers.
4. Return the Count:
   • The final count represents the maximum number of integers we can choose.

📋 Pseudocode for Clarity

Here’s the pseudocode that captures the above logic:

1. Initialize banned_set = set(banned)
2. Initialize total_sum = 0, count = 0
3. For each number i in range 1 to n:
      a. If i is in banned_set, skip to the next number
      b. If total_sum + i > maxSum, break the loop
      c. Otherwise, add i to total_sum and increment count
4. Return count

💡 Why Greedy Works Here

The greedy approach works because:

• Smaller numbers: Selecting numbers from smallest to largest ensures we can pick the maximum count without breaching maxSum.
• Banned set optimization: Using a set to track banned numbers minimizes overhead, allowing us to focus on valid selections.

🔍 Alternative Approaches: Why Not Use Them?

It’s worth exploring other potential solutions to understand why the greedy approach is optimal.

1. Sort and Filter First

One approach could involve:

• Filtering out banned numbers from the range [1, n].
• Sorting the remaining numbers.
• Iterating through the sorted list and summing them until reaching maxSum.

Downside:

• Sorting adds unnecessary overhead of \(O(n \log n)\), while the greedy approach avoids sorting altogether by iterating directly.

2. Dynamic Programming

We could think about this as a subset-sum problem and try solving it with dynamic programming (DP). Each number could either be included or excluded, and we’d track the maximum count of selected integers.

Downside:

• DP requires additional memory and computational complexity (\(O(n \cdot \text{maxSum})\)), which isn’t practical for larger inputs.

🔑 Detailed Steps in Greedy Approach

Let’s break down each part of the greedy approach for clarity.

1. Using a Set for Banned Numbers

banned_set = set(banned)

• The set ensures \(O(1)\) lookups for each number, keeping our algorithm efficient.
• This step preprocesses the banned numbers, which is necessary to quickly check validity during iteration.

2. Iterate and Skip Banned Numbers

for i in range(1, n + 1):
    if i in banned_set:
        continue

• By iterating from 1 to n, we evaluate all possible candidates.
• The continue statement skips numbers that are banned.

3. Check Sum Constraint

if total_sum + i > maxSum:
    break

• This is a critical check. Once the cumulative sum exceeds maxSum, we stop processing further numbers.

4. Update Total and Count

total_sum += i
count += 1

• For valid numbers, we update the cumulative sum and increment the count.

🛠️ Space and Time Complexity Analysis

🕒 Time Complexity

1. Creating the Banned Set:
   • Takes \(O(b)\), where \(b\) is the length of the banned array.
2. Iterating Through Range:
   • Takes \(O(n)\), where \(n\) is the upper limit of the range.

Overall: \(O(n + b)\), which is highly efficient.

💾 Space Complexity

1. Banned Set:
   • Requires \(O(b)\) additional space for storing banned numbers.

Overall: \(O(b)\).

🔍 Edge Cases 🧩

Here are additional edge cases to consider:

Edge Case 1: n is very large, but maxSum is very small

Input: banned = [], n = 10^4, maxSum = 1
Output: 0

Explanation: Even though the range is large, the small maxSum prevents selecting any numbers.

Edge Case 2: All numbers are banned

Input: banned = [1, 2, 3, 4, 5], n = 5, maxSum = 15
Output: 0

Explanation: No numbers can be selected because they are all banned.

Edge Case 3: maxSum is extremely large

Input: banned = [], n = 100, maxSum = 10^9
Output: 100

Explanation: The sum constraint is effectively irrelevant, so we can select all numbers.

Edge Case 4: Large banned list with sparse valid numbers

Input: banned = [1, 3, 5, 7, 9], n = 10, maxSum = 15
Output: 4

Explanation: The valid numbers are \(2, 4, 6, 8\), and their sum \(2 + 4 + 6 + 8 = 20\) exceeds maxSum. We only take \(2, 4, 6\).

🔑 Key Insights

1. Filter banned numbers: Create a set of banned numbers for quick lookup.
2. Iterate through valid numbers: Loop through numbers \(1\) to \(n\) and skip the banned ones.
3. Maintain sum constraint: Track the cumulative sum and stop adding when exceeding maxSum.
4. Maximize count: Count how many numbers we can include.

🛠️ Implementation: One Optimal Solution 🚀

Since both the basic and optimized solutions have the same time complexity, we’ll focus on the single optimal approach.

Optimal Solution 💡

Steps:

1. Add all banned numbers to a set for \(O(1)\) lookup.
2. Iterate through numbers \(1\) to \(n\):
   • Skip the number if it’s in the banned set.
   • Add the number to the cumulative sum if it doesn’t exceed maxSum.
   • Stop if adding another number would exceed maxSum.
3. Return the total count of selected numbers.

Time Complexity: \(O(n + b)\), where \(n\) is the upper limit of the range, and \(b\) is the length of the banned array.

🐍 Python Code 🖥️

def maxCount(banned, n, maxSum):
    # Step 1: Create a set of banned numbers
    banned_set = set(banned)
    total_sum = 0
    count = 0

    # Step 2: Iterate over the range [1, n]
    for i in range(1, n + 1):
        if i in banned_set:
            continue  # Skip banned numbers
        if total_sum + i > maxSum:
            break  # Stop if adding i exceeds maxSum
        total_sum += i
        count += 1

    return count

📊 Example Walkthrough: Large Case 🔍

Let’s walk through a larger example step-by-step to solidify the understanding:

Input:

banned = [5, 10, 15]
n = 20
maxSum = 50

Process:

1. Initialize:
   • banned_set = {5, 10, 15}
   • total_sum = 0, count = 0
2. Iterate from 1 to 20:
   • \(i = 1\): Not banned, \(total\_sum = 1\), \(count = 1\)
   • \(i = 2\): Not banned, \(total\_sum = 3\), \(count = 2\)
   • \(i = 3\): Not banned, \(total\_sum = 6\), \(count = 3\)
   • \(i = 4\): Not banned, \(total\_sum = 10\), \(count = 4\)
   • \(i = 5\): Banned. Skip.
   • \(i = 6\): Not banned, \(total\_sum = 16\), \(count = 5\)
   • \(i = 7\): Not banned, \(total\_sum = 23\), \(count = 6\)
   • \(i = 8\): Not banned, \(total\_sum = 31\), \(count = 7\)
   • \(i = 9\): Not banned, \(total\_sum = 40\), \(count = 8\)
   • \(i = 10\): Banned. Skip.
   • \(i = 11\): Adding exceeds maxSum. Stop.

Output:

8

🔍 Edge Cases 🧩

1. All numbers are banned:

   Input: banned = [1,2,3,4,5], n = 5, maxSum = 10
   Output: 0
   

   Explanation: No valid numbers to choose.

2. High maxSum:

   Input: banned = [], n = 10, maxSum = 1000
   Output: 10
   

   Explanation: All numbers can be chosen since the sum constraint is easily satisfied.

3. Small maxSum:

   Input: banned = [1], n = 10, maxSum = 1
   Output: 0
   

   Explanation: The sum constraint makes it impossible to choose any number.

4. Sparse banned numbers:

   Input: banned = [3, 8], n = 10, maxSum = 25
   Output: 6
   

   Explanation: Chosen numbers are \(1, 2, 4, 5, 6, 7\), with a sum of \(25\).

🏁 Conclusion 🎯

This problem demonstrates the power of a greedy approach and the importance of efficiently managing constraints. By leveraging sets for fast lookups and iterating systematically, we maximize the number of integers while respecting all the rules.

⭐ Final Thoughts 🌟

Problems like this are great for honing your skills in algorithmic thinking and constraint management. Whether you’re prepping for interviews or sharpening your problem-solving abilities, tackling these challenges will make you a better programmer. 🚀

Ready for the next challenge? Let’s keep coding! 💻