LeetCode #2554 — MEDIUM

Maximum Number of Integers to Choose From a Range I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Greedy

Example 1

[1,6,5]
5
6

Example 2

[1,2,3,4,5,6,7]
8
1

Example 3

[11]
7
50

Core Insight

What unlocks the optimal approach

Keep the banned numbers that are less than or equal to n in a set.
Loop over the numbers from 1 to n and if the number is not banned, use it.
Keep adding numbers while they are not banned, and their sum is less than or equal to k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2554: Maximum Number of Integers to Choose From a Range I
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;
    }
}

// Accepted solution for LeetCode #2554: Maximum Number of Integers to Choose From a Range I
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
}

# Accepted solution for LeetCode #2554: Maximum Number of Integers to Choose From a Range I
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

// Accepted solution for LeetCode #2554: Maximum Number of Integers to Choose From a Range I
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
    }
}

// Accepted solution for LeetCode #2554: Maximum Number of Integers to Choose From a Range I
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.