Problem Description

You are given a 0-indexed array of integers nums.

A sequential prefix is defined as a prefix nums[0..i] where each element (starting from index 1) is exactly 1 more than the previous element. In other words, for all positions 1 <= j <= i, we must have nums[j] = nums[j - 1] + 1. The prefix containing only nums[0] is always considered sequential.

Your task is to:

1. Find the longest sequential prefix starting from nums[0]
2. Calculate the sum of all elements in this longest sequential prefix
3. Find the smallest integer x that is not present in the array nums and is greater than or equal to this sum

For example, if nums = [1, 2, 3, 5, 6], the longest sequential prefix is [1, 2, 3] (since 4 is missing, the sequence breaks). The sum of this prefix is 1 + 2 + 3 = 6. Starting from 6, we need to find the smallest integer not in nums. Since 6 is in nums, we check 7, which is not in nums, so the answer is 7.

The solution works by first iterating through the array to find where the sequential pattern breaks, accumulating the sum along the way. Then it uses a hash set to store all elements from nums for quick lookup, and starting from the calculated sum, it finds the first integer not present in the set.

Intuition

The problem asks us to find a missing integer that's at least as large as the sum of a special prefix. Let's think about this step by step.

First, we need to identify the longest sequential prefix. Since it must start from nums[0] and each subsequent element must be exactly 1 more than the previous, we can simply walk through the array from the beginning and stop as soon as we find an element that doesn't follow the pattern nums[j] = nums[j-1] + 1.

As we traverse this sequential prefix, we can accumulate the sum along the way. This is efficient because we're already checking each element to verify the sequential property.

Now comes the key insight: once we have the sum s of the longest sequential prefix, we need to find the smallest integer >= s that's not in the array. The straightforward approach would be to start from s and check s, s+1, s+2, ... until we find a number not in nums.

To make the checking efficient, we can preprocess all elements of nums into a hash set. This gives us O(1) lookups when checking if a number exists in the array. Without this optimization, we'd need to scan through the entire array for each candidate number, which would be inefficient.

The beauty of this approach is that it breaks the problem into two independent parts: finding the sequential prefix sum, and then finding the first missing number from that point onwards. The hash set bridges these two parts by providing fast membership testing for the second phase.

Learn more about Sorting patterns.

Solution Approach

The solution uses simulation combined with a hash table to efficiently solve the problem in two phases.

Phase 1: Calculate the longest sequential prefix sum

We start by initializing s with nums[0] (the first element) and j with 1 (the index to check next). We then iterate through the array using a while loop:

s, j = nums[0], 1
while j < len(nums) and nums[j] == nums[j - 1] + 1:
    s += nums[j]
    j += 1

The loop continues as long as:

• We haven't reached the end of the array (j < len(nums))
• The current element is exactly 1 more than the previous element (nums[j] == nums[j - 1] + 1)

During each iteration, we add the current element to our sum s and move to the next index. The loop stops when we encounter the first element that breaks the sequential pattern or reach the end of the array.

Phase 2: Find the smallest missing integer

First, we create a hash set from all elements in nums for O(1) lookup:

vis = set(nums)

Then, we use Python's count() function from the itertools module to generate integers starting from s:

for x in count(s):
    if x not in vis:
        return x

The count(s) generates an infinite sequence: s, s+1, s+2, .... For each generated integer x, we check if it exists in our hash set vis. The first integer not found in the set is our answer, and we immediately return it.

The time complexity is O(n + m) where n is the length of the array and m is the difference between the answer and the prefix sum. The space complexity is O(n) for storing the hash set.

Example Walkthrough

Let's walk through the solution with nums = [3, 4, 5, 1, 12, 14, 13].

Phase 1: Finding the longest sequential prefix and its sum

1. Start with s = nums[0] = 3 and j = 1
2. Check nums[1] = 4. Is it equal to nums[0] + 1 = 3 + 1 = 4? Yes!
   • Add to sum: s = 3 + 4 = 7
   • Move to next: j = 2
3. Check nums[2] = 5. Is it equal to nums[1] + 1 = 4 + 1 = 5? Yes!
   • Add to sum: s = 7 + 5 = 12
   • Move to next: j = 3
4. Check nums[3] = 1. Is it equal to nums[2] + 1 = 5 + 1 = 6? No! (1 ≠ 6)
   • The sequential prefix breaks here, so we stop

The longest sequential prefix is [3, 4, 5] with sum s = 12.

Phase 2: Finding the smallest missing integer ≥ 12

1. Create a hash set: vis = {3, 4, 5, 1, 12, 14, 13}
2. Start checking from x = 12:
   • Is 12 in vis? Yes (12 is in the array)
   • Is 13 in vis? Yes (13 is in the array)
   • Is 14 in vis? Yes (14 is in the array)
   • Is 15 in vis? No!

The answer is 15 - the smallest integer not in nums that is at least as large as the prefix sum of 12.

Solution Implementation

Time and Space Complexity

The time complexity is O(n) in the best case but can be unbounded in the worst case. The space complexity is O(n), where n is the length of the array nums.

Time Complexity Analysis:

• The first while loop iterates through the consecutive sequence at the beginning of the array, taking O(n) time in the worst case when all elements form a consecutive sequence.
• Creating the set vis from nums takes O(n) time.
• The count(s) generator with the membership check x not in vis could theoretically run indefinitely if all integers starting from s are present in vis. However, since vis contains at most n elements, the loop must terminate within O(s + n) iterations, where s is the sum of the prefix sequence. In the worst case where the prefix is the entire array of consecutive integers starting from 1, s could be O(n²), making the overall time complexity potentially O(n²).

Space Complexity Analysis:

• The set vis stores all elements from nums, requiring O(n) space.
• The variables s, j, and x use O(1) space.
• Overall space complexity is O(n).

Note: The reference answer stating O(n) time complexity appears to be considering only the practical case where the missing integer is found quickly, not the theoretical worst case.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Sequential Prefix Definition

A common mistake is thinking that any increasing sequence counts as a sequential prefix. The problem specifically requires each element to be exactly 1 more than the previous element, starting from index 0.

Incorrect assumption:

# Wrong: This accepts any increasing sequence
while index < len(nums) and nums[index] > nums[index - 1]:
    consecutive_sum += nums[index]
    index += 1

Correct approach:

# Right: Each element must be exactly 1 more than previous
while index < len(nums) and nums[index] == nums[index - 1] + 1:
    consecutive_sum += nums[index]
    index += 1

2. Starting the Search from the Wrong Value

Another pitfall is starting the search for the missing integer from 1 or 0 instead of from the calculated prefix sum.

Incorrect:

# Wrong: Starting from 1
for candidate in count(1):
    if candidate not in seen_numbers:
        return candidate

Correct:

# Right: Starting from the prefix sum
for candidate in count(consecutive_sum):
    if candidate not in seen_numbers:
        return candidate

3. Off-by-One Error in Index Handling

Be careful with index initialization and bounds checking. Starting index at 0 instead of 1 would cause an out-of-bounds error when checking nums[index - 1].

Incorrect:

# Wrong: Starting index at 0 causes nums[-1] access
consecutive_sum = nums[0]
index = 0  # Should be 1!
while index < len(nums) and nums[index] == nums[index - 1] + 1:
    # This will check nums[0] == nums[-1] + 1 on first iteration
    consecutive_sum += nums[index]
    index += 1

4. Not Handling Single Element Arrays

The code handles this correctly, but it's worth noting that for an array with just one element like [5], the sequential prefix is just [5] with sum 5, and we need to find the smallest integer ≥ 5 not in the array.

5. Performance Issue with Large Gaps

While the current solution is correct, if the prefix sum is very small and all numbers in the array are very large, the loop might iterate many times. For example, if nums = [1, 100000], the prefix sum is 1, but we'd need to check numbers 1 through 99999 before finding 2. In practice, this is handled efficiently with the hash set lookup, but for extremely large gaps, consider adding an optimization:

# Optimization for large gaps
min_in_array = min(seen_numbers)
if consecutive_sum < min_in_array:
    return consecutive_sum