LeetCode 3041 - Maximize Consecutive Elements in an Array After Modification

The problem is asking us to maximize the number of consecutive integers we can select from an array after we are allowed to increase any element by at most 1.

leetcodehardarraydynamic-programmingsorting

LeetCode Problem 3041

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Sorting

Solution

Problem Understanding

The problem is asking us to maximize the number of consecutive integers we can select from an array after we are allowed to increase any element by at most 1. The input array nums contains positive integers and can have up to 10^5 elements, with each element ranging from 1 to 10^6. The output should be the largest size of a subset of elements that, when sorted, form a sequence of consecutive integers.

In simpler terms, we are allowed to tweak each number slightly (by adding 1 at most) and then choose elements that can form a consecutive sequence. We need to figure out the largest such sequence. Important constraints are that the array can be large, so brute force checking of all subsets will be too slow. Also, some elements may be duplicates or far apart, which affects whether they can be consecutive after modification.

Important edge cases include arrays where all elements are the same, arrays with elements already consecutive, arrays where elements are very far apart, and arrays with a single element.

Approaches

A brute-force approach would try all possible subsets of the array, apply all possible increment operations, and then check if the subset forms a consecutive sequence. This is correct but computationally infeasible because the number of subsets is 2^n and for each subset, we need sorting and consecutive checking.

The optimal approach relies on the insight that after increasing each number by at most 1, every element can either stay the same or increase by 1. Therefore, to find the longest consecutive subsequence, we can first sort the array and then use a sliding window or two-pointer technique. The key observation is that for a sequence to be consecutive, the difference between the largest and smallest elements in a window should not exceed the window length, because each element can be incremented by at most 1. This allows us to efficiently find the maximum window size that satisfies this property.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * n log n) O(n) Check all subsets and validate consecutiveness after modification.
Optimal O(n log n) O(1) Sort and use a two-pointer sliding window to find the largest valid consecutive sequence.

Algorithm Walkthrough

  1. Sort the input array nums in ascending order. Sorting helps us easily check consecutive sequences and handle duplicates.
  2. Initialize two pointers, left and right, both starting at the beginning of the sorted array.
  3. Iterate with right pointer through the array. At each step, check if the current window [nums[left], nums[right]] can form a consecutive sequence. Specifically, check if nums[right] - nums[left] <= right - left + 1. This works because the difference between the maximum and minimum in a valid window cannot exceed the window length (each element can increase by at most 1 to help close small gaps).
  4. If the condition fails, increment left to shrink the window until the condition holds again.
  5. Keep track of the maximum window length encountered, as this represents the maximum number of consecutive elements possible.
  6. Return the maximum length after processing the entire array.

Why it works: Sorting ensures elements are in order, and the sliding window ensures we are always considering the largest possible group of consecutive elements at each position. By enforcing the difference condition, we account for the fact that each element can be incremented by at most 1, and no larger gaps can be closed.

Python Solution

from typing import List

class Solution:
    def maxSelectedElements(self, nums: List[int]) -> int:
        nums.sort()
        max_len = 0
        left = 0
        
        for right in range(len(nums)):
            while nums[right] - nums[left] > right - left + 1:
                left += 1
            max_len = max(max_len, right - left + 1)
        
        return max_len

The Python implementation first sorts the array to enable easy consecutive checking. The left pointer is used to maintain a valid window while the right pointer expands the window. Whenever the difference between nums[right] and nums[left] exceeds the allowable difference, we increment left to shrink the window. The max_len variable keeps track of the maximum valid window length seen so far.

Go Solution

import "sort"

func maxSelectedElements(nums []int) int {
    sort.Ints(nums)
    maxLen := 0
    left := 0
    
    for right := 0; right < len(nums); right++ {
        for nums[right]-nums[left] > right-left+1 {
            left++
        }
        if right-left+1 > maxLen {
            maxLen = right - left + 1
        }
    }
    
    return maxLen
}

In Go, we also sort the array using sort.Ints. We use a for loop to iterate with right and a nested for loop to increment left when the window is invalid. The window length is calculated and used to update maxLen. Slice indexing and difference calculation in Go is analogous to Python, so the logic directly translates.

Worked Examples

Example 1: nums = [2,1,5,1,1]

After sorting: [1,1,1,2,5]

Initialize left = 0, max_len = 0.

right nums[right] window condition max_len
0 1 [1] 1-1<=1 1
1 1 [1,1] 1-1<=2 2
2 1 [1,1,1] 1-1<=3 3
3 2 [1,1,1,2] 2-1<=4 4
4 5 [1,1,1,2,5] 5-1>5? shrink left 3

Maximum length is 3.

Example 2: nums = [1,4,7,10]

After sorting: [1,4,7,10]

Window lengths never satisfy the difference condition beyond length 1, so the maximum length is 1.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting takes O(n log n), and the two-pointer sliding window iterates over the array in O(n).
Space O(1) Only pointers and counters are used; no extra data structures proportional to n are required.

Sorting dominates the time complexity. The sliding window ensures that we only pass over the array once, making it efficient.

Test Cases

# Provided examples
assert Solution().maxSelectedElements([2,1,5,1,1]) == 3  # max consecutive after +1
assert Solution().maxSelectedElements([1,4,7,10]) == 1    # elements too far apart

# Edge cases
assert Solution().maxSelectedElements([1]) == 1            # single element
assert Solution().maxSelectedElements([5,5,5,5]) == 4     # all same elements
assert Solution().maxSelectedElements([1,2,2,3,4]) == 5   # duplicates and consecutive
assert Solution().maxSelectedElements([1,3,5,7]) == 1     # all gaps >1
assert Solution().maxSelectedElements([1,2,3,4,5,6]) == 6 # already consecutive
Test Why
[2,1,5,1,1] Checks if increments are correctly used to maximize consecutive sequence
[1,4,7,10] Checks behavior when gaps are too large
[1] Minimal input edge case
[5,5,5,5] Duplicate elements and maximum consecutive selection
[1,2,2,3,4] Mix of duplicates and consecutive numbers
[1,3,5,7] Sparse array, no consecutive elements possible
[1,2,3,4,5,6] Already consecutive sequence

Edge Cases

  1. Single-element array: This is the smallest possible input. The maximum consecutive sequence is always 1. The implementation handles it naturally as the sliding window immediately returns length 1.
  2. All identical elements: For arrays like [5,5,5,5], each element can increase by 1. The algorithm ensures that the window grows to include all identical elements because the difference condition is satisfied for all elements in the sorted array.
  3. Large gaps between elements: Arrays like [1,4,7,10] have elements too far apart to form a sequence even after incrementing by 1. The sliding window will shrink continuously to maintain the difference condition, resulting in a maximum consecutive length of 1.

This approach handles all edge cases efficiently and correctly while maintaining optimal performance.