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Leetcode #2598: Smallest Missing Non-negative Integer After Operations

In this guide, we solve Leetcode #2598 Smallest Missing Non-negative Integer After Operations in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Greedy, Array, Hash Table, Math

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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.

Python Solution

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

Complexity

The time complexity is O(n)O(n)O(n) and the space complexity is O(value)O(\textit{value})O(value), where nnn is the length of array nums\textit{nums}nums. The space complexity is O(value)O(\textit{value})O(value), where nnn is the length of array nums\textit{nums}nums.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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