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Leetcode #2229: Check if an Array Is Consecutive

In this guide, we solve Leetcode #2229 Check if an Array Is Consecutive 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

Given an integer array nums, return true if nums is consecutive, otherwise return false. An array is consecutive if it contains every number in the range [x, x + n - 1] (inclusive), where x is the minimum number in the array and n is the length of the array.

Quick Facts

  • Difficulty: Easy
  • Premium: Yes
  • Tags: Array, Hash Table, Sorting

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 = [1,3,4,2] Output: true Explanation: The minimum value is 1 and the length of nums is 4. All of the values in the range [x, x + n - 1] = [1, 1 + 4 - 1] = [1, 4] = (1, 2, 3, 4) occur in nums. Therefore, nums is consecutive.

Python Solution

class Solution: def isConsecutive(self, nums: List[int]) -> bool: mi, mx = min(nums), max(nums) return len(set(nums)) == mx - mi + 1 == len(nums)

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

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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