LeetCode #3041 — HARD

Maximize Consecutive Elements in an Array After Modification

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Initially, you can increase the value of any element in the array by at most 1.

After that, you need to select one or more elements from the final array such that those elements are consecutive when sorted in increasing order. For example, the elements [3, 4, 5] are consecutive while [3, 4, 6] and [1, 1, 2, 3] are not.

Return the maximum number of elements that you can select.

Example 1:

Input: nums = [2,1,5,1,1]
Output: 3
Explanation: We can increase the elements at indices 0 and 3. The resulting array is nums = [3,1,5,2,1].
We select the elements [3,1,5,2,1] and we sort them to obtain [1,2,3], which are consecutive.
It can be shown that we cannot select more than 3 consecutive elements.

Example 2:

Input: nums = [1,4,7,10]
Output: 1
Explanation: The maximum consecutive elements that we can select is 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums consisting of positive integers. Initially, you can increase the value of any element in the array by at most 1. After that, you need to select one or more elements from the final array such that those elements are consecutive when sorted in increasing order. For example, the elements [3, 4, 5] are consecutive while [3, 4, 6] and [1, 1, 2, 3] are not. Return the maximum number of elements that you can select.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[2,1,5,1,1]

Example 2

[1,4,7,10]

Step 02

Core Insight

What unlocks the optimal approach

Sort the array and try using dynamic programming.
Let <code>dp[i]</code> be the length of the longest consecutive elements ending at element at index <code>i</code> in the sorted array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
class Solution {
  public int maxSelectedElements(int[] nums) {
    int ans = 0;
    // {num: the length of the longest consecutive elements ending in num}
    HashMap<Integer, Integer> dp = new HashMap<>();

    Arrays.sort(nums);

    for (final int num : nums) {
      dp.put(num + 1, dp.getOrDefault(num, 0) + 1);
      dp.put(num, dp.getOrDefault(num - 1, 0) + 1);
      ans = Math.max(ans, Math.max(dp.get(num), dp.get(num + 1)));
    }

    return ans;
  }
}

// Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
package main

import (
	"slices"
)

// https://space.bilibili.com/206214
func maxSelectedElements(nums []int) (ans int) {
	slices.Sort(nums)
	f := map[int]int{}
	for _, x := range nums {
		f[x+1] = f[x] + 1
		f[x] = f[x-1] + 1
	}
	for _, res := range f {
		ans = max(ans, res)
	}
	return
}

# Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
class Solution:
  def maxSelectedElements(self, nums: list[int]) -> int:
    ans = 0
    # {num: the length of the longest consecutive elements ending in num}
    dp = {}

    for num in sorted(nums):
      dp[num + 1] = dp.get(num, 0) + 1
      dp[num] = dp.get(num - 1, 0) + 1
      ans = max(ans, dp[num], dp[num + 1])

    return ans

// Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
// class Solution {
//   public int maxSelectedElements(int[] nums) {
//     int ans = 0;
//     // {num: the length of the longest consecutive elements ending in num}
//     HashMap<Integer, Integer> dp = new HashMap<>();
// 
//     Arrays.sort(nums);
// 
//     for (final int num : nums) {
//       dp.put(num + 1, dp.getOrDefault(num, 0) + 1);
//       dp.put(num, dp.getOrDefault(num - 1, 0) + 1);
//       ans = Math.max(ans, Math.max(dp.get(num), dp.get(num + 1)));
//     }
// 
//     return ans;
//   }
// }

// Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3041: Maximize Consecutive Elements in an Array After Modification
// class Solution {
//   public int maxSelectedElements(int[] nums) {
//     int ans = 0;
//     // {num: the length of the longest consecutive elements ending in num}
//     HashMap<Integer, Integer> dp = new HashMap<>();
// 
//     Arrays.sort(nums);
// 
//     for (final int num : nums) {
//       dp.put(num + 1, dp.getOrDefault(num, 0) + 1);
//       dp.put(num, dp.getOrDefault(num - 1, 0) + 1);
//       ans = Math.max(ans, Math.max(dp.get(num), dp.get(num + 1)));
//     }
// 
//     return ans;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.