#2996

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Smallest Missing Integer Greater Than Sequential Prefix Sum

Find the smallest missing integer greater than the sum of the longest sequential prefix in an array.

Problem Statement

You are given a 0-indexed array of integers nums. A prefix nums[0..i] is sequential if, for all 1 <= j <= i, nums[j] = nums[j - 1] + 1. In particular, the prefix consisting only of nums[0] is sequential.

Return the smallest integer x missing from nums such that x is greater than or equal to the sum of the longest sequential prefix.

Examples

Example 1

Input: nums = [1,2,3,2,5]

Output: 6

The longest sequential prefix of nums is [1,2,3] with a sum of 6. 6 is not in the array, therefore 6 is the smallest missing integer greater than or equal to the sum of the longest sequential prefix.

Example 2

Input: nums = [3,4,5,1,12,14,13]

Output: 15

The longest sequential prefix of nums is [3,4,5] with a sum of 12. 12, 13, and 14 belong to the array while 15 does not. Therefore 15 is the smallest missing integer greater than or equal to the sum of the longest sequential prefix.

Constraints

1 <= nums.length <= 50
1 <= nums[i] <= 50

Solution Approach

Identify the longest sequential prefix

To solve the problem, first identify the longest sequential prefix in the array. This can be done by iterating over the array and checking for consecutive numbers (i.e., nums[i] = nums[i-1] + 1).

Calculate the sum of the prefix

Once the longest sequential prefix is identified, compute its sum. This sum is used to determine the smallest missing integer that is greater than or equal to it.

Find the smallest missing integer

With the sum of the longest sequential prefix in hand, scan through the array or use a hash table to identify the smallest integer greater than or equal to this sum that is not present in the array.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of the solution depends on how the array is scanned and the approach for identifying missing integers. If a hash table is used, the space complexity could be O(n), and the time complexity could be O(n) as well, where n is the length of the array.

What Interviewers Usually Probe

The candidate correctly identifies the sequential prefix and calculates its sum.
The candidate demonstrates a clear approach to finding missing integers using efficient data structures.
The candidate optimizes the problem-solving process by considering edge cases and handling array traversal effectively.

Common Pitfalls or Variants

Common pitfalls

Incorrectly identifying the longest sequential prefix due to faulty array traversal.
Failing to calculate the sum of the sequential prefix properly.
Not considering how to find missing integers efficiently, either by skipping hash table usage or miscalculating the sum.

Follow-up variants

Consider implementing the solution with sorting and scanning instead of using hash tables.
Test the problem with an array of maximum size (50) to evaluate the solution's efficiency.
Investigate edge cases with minimal input (i.e., a single element array).

FAQ

How do I identify the longest sequential prefix?

To identify the longest sequential prefix, iterate through the array and check if nums[i] = nums[i-1] + 1. Stop when this condition is violated.

What is the significance of the sum of the sequential prefix?

The sum of the longest sequential prefix is used to determine the smallest integer that is greater than or equal to this sum but not present in the array.

What are the challenges with finding missing integers?

The main challenge is efficiently finding missing integers, especially when the array can have elements that span a large range.

Can this problem be solved without using a hash table?

Yes, it's possible to solve the problem using sorting and array scanning without a hash table. However, a hash table is usually more efficient for lookups.

How does GhostInterview assist with this problem?

GhostInterview offers tailored hints on array scanning, hash lookups, and efficient ways to identify missing integers, making the problem-solving process smoother.

terminal

Solution

Solution 1: Simulation + Hash Table

First, we calculate the longest prefix sum $s$ of the array $nums$. Then, starting from $s$, we enumerate the integer $x$. If $x$ is not in the array $nums$, then $x$ is the answer. Here, we can use a hash table to quickly determine whether an integer is in the array $nums$.

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class Solution:
    def missingInteger(self, nums: List[int]) -> int:
        s, j = nums[0], 1
        while j < len(nums) and nums[j] == nums[j - 1] + 1:
            s += nums[j]
            j += 1
        vis = set(nums)
        for x in count(s):
            if x not in vis:
                return x

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