LeetCode #2598 — MEDIUM

Smallest Missing Non-negative Integer After Operations

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

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed integer array nums and an integer value.

In one operation, you can add or subtract value from any element of nums.

For example, if nums = [1,2,3] and value = 2, you can choose to subtract value from nums[0] to make nums = [-1,2,3].

The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it.

For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2.

Return the maximum MEX of nums after applying the mentioned operation any number of times.

Example 1:

Input: nums = [1,-10,7,13,6,8], value = 5
Output: 4
Explanation: One can achieve this result by applying the following operations:
- Add value to nums[1] twice to make nums = [1,0,7,13,6,8]
- Subtract value from nums[2] once to make nums = [1,0,2,13,6,8]
- Subtract value from nums[3] twice to make nums = [1,0,2,3,6,8]
The MEX of nums is 4. It can be shown that 4 is the maximum MEX we can achieve.

Example 2:

Input: nums = [1,-10,7,13,6,8], value = 7
Output: 2
Explanation: One can achieve this result by applying the following operation:
- subtract value from nums[2] once to make nums = [1,-10,0,13,6,8]
The MEX of nums is 2. It can be shown that 2 is the maximum MEX we can achieve.

Constraints:

1 <= nums.length, value <= 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums and an integer value. In one operation, you can add or subtract value from any element of nums. For example, if nums = [1,2,3] and value = 2, you can choose to subtract value from nums[0] to make nums = [-1,2,3]. The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it. For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2. Return the maximum MEX of nums after applying the mentioned operation any number of times.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Greedy

Example 1

[1,-10,7,13,6,8]
5

Example 2

[1,-10,7,13,6,8]
7

Core Insight

What unlocks the optimal approach

Think about using modular arithmetic.
if x = nums[i] (mod value), then we can make nums[i] equal to x after some number of operations
How does finding the frequency of (nums[i] mod value) help?

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 #2598: Smallest Missing Non-negative Integer After Operations
class Solution {
    public int findSmallestInteger(int[] nums, int value) {
        int[] cnt = new int[value];
        for (int x : nums) {
            ++cnt[(x % value + value) % value];
        }
        for (int i = 0;; ++i) {
            if (cnt[i % value]-- == 0) {
                return i;
            }
        }
    }
}

// Accepted solution for LeetCode #2598: Smallest Missing Non-negative Integer After Operations
func findSmallestInteger(nums []int, value int) int {
	cnt := make([]int, value)
	for _, x := range nums {
		cnt[(x%value+value)%value]++
	}
	for i := 0; ; i++ {
		if cnt[i%value] == 0 {
			return i
		}
		cnt[i%value]--
	}
}

# Accepted solution for LeetCode #2598: Smallest Missing Non-negative Integer After Operations
class Solution:
    def findSmallestInteger(self, nums: List[int], value: int) -> int:
        cnt = Counter(x % value for x in nums)
        for i in range(len(nums) + 1):
            if cnt[i % value] == 0:
                return i
            cnt[i % value] -= 1

// Accepted solution for LeetCode #2598: Smallest Missing Non-negative Integer After Operations
impl Solution {
    pub fn find_smallest_integer(nums: Vec<i32>, value: i32) -> i32 {
        let mut cnt = vec![0; value as usize];
        for &x in &nums {
            let idx = ((x % value + value) % value) as usize;
            cnt[idx] += 1;
        }

        let mut i = 0;
        loop {
            let idx = (i % value) as usize;
            if cnt[idx] == 0 {
                return i;
            }
            cnt[idx] -= 1;
            i += 1;
        }
    }
}

// Accepted solution for LeetCode #2598: Smallest Missing Non-negative Integer After Operations
function findSmallestInteger(nums: number[], value: number): number {
    const cnt: number[] = new Array(value).fill(0);
    for (const x of nums) {
        ++cnt[((x % value) + value) % value];
    }
    for (let i = 0; ; ++i) {
        if (cnt[i % value]-- === 0) {
            return i;
        }
    }
}

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(\textitvalue)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.