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Leetcode #2113: Elements in Array After Removing and Replacing Elements

In this guide, we solve Leetcode #2113 Elements in Array After Removing and Replacing Elements 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. Initially on minute 0, the array is unchanged.

Quick Facts

  • Difficulty: Medium
  • Premium: Yes
  • Tags: Array

Intuition

The constraints allow a direct scan that keeps only the essential state.

By translating the requirements into a clean loop, the logic stays easy to reason about.

Approach

Iterate through the data once, updating the state needed to compute the answer.

Return the final state after the traversal is complete.

Steps:

  • Parse the input.
  • Iterate and update state.
  • Return the computed answer.

Example

Input: nums = [0,1,2], queries = [[0,2],[2,0],[3,2],[5,0]] Output: [2,2,-1,0] Explanation: Minute 0: [0,1,2] - All elements are in the nums. Minute 1: [1,2] - The leftmost element, 0, is removed. Minute 2: [2] - The leftmost element, 1, is removed. Minute 3: [] - The leftmost element, 2, is removed. Minute 4: [0] - 0 is added to the end of nums. Minute 5: [0,1] - 1 is added to the end of nums. At minute 0, nums[2] is 2. At minute 2, nums[0] is 2. At minute 3, nums[2] does not exist. At minute 5, nums[0] is 0.

Python Solution

class Solution: def elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]: n, m = len(nums), len(queries) ans = [-1] * m for j, (t, i) in enumerate(queries): t %= 2 * n if t < n and i < n - t: ans[j] = nums[i + t] elif t > n and i < t - n: ans[j] = nums[i] return ans

Complexity

The time complexity is O(m)O(m)O(m), where mmm is the length of the array queriesqueriesqueries. The space complexity is O(1)O(1)O(1).

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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