Is Maximize Fixed Points After Deletions easy or hard?

The problem is labeled Hard because recognizing the subsequence constraint is not immediately obvious. Once you observe that the final array must contain the ordered sequence 1..k, the solution simplifies to a linear greedy scan.

Maximize Fixed Points After Deletions Python/Java solution

The implementation scans the array and increments a target counter whenever the current value equals the expected position. This logic is identical across Python, Java, and C++. Each language simply iterates once through the array and returns target - 1 as the number of fixed points.

How to solve Maximize Fixed Points After Deletions in O(n)?

Maintain a variable called target initialized to 1. Iterate through the array, and whenever an element equals target, include it in the resulting subsequence and increment target. Every accepted element becomes a fixed point after reindexing because its value matches its new position. The final answer is target minus one.

What is the best approach for Maximize Fixed Points After Deletions?

The optimal approach is a greedy subsequence scan. Traverse the array while tracking the next value you want in the final array (1, 2, 3, ...). Whenever the current element equals the target value, keep it and advance the target. This produces the longest subsequence that forms the sequence 1..k, which guarantees k fixed points after deletions. The complexity is O(n) time and O(1) space.

Is Maximize Fixed Points After Deletions asked at Google/Amazon/Meta?

Problems involving fixed points, subsequence construction, and greedy ordering frequently appear in interviews at companies like Google, Amazon, and Meta. Variants typically test whether you recognize that deletions transform the task into building a constrained subsequence.

What data structure is used in Maximize Fixed Points After Deletions?

The optimal solution mainly uses simple variables and sequential array traversal. Conceptually it relates to subsequence matching, which appears in greedy and dynamic programming problems, but no complex data structure is required.

What is the time complexity of Maximize Fixed Points After Deletions?

The optimal solution runs in O(n) time because the array is scanned once while maintaining a single counter for the next required value. Space complexity is O(1) since no auxiliary data structures are needed beyond a few variables.

Description Hints 3 Video 55:26

3920. Maximize Fixed Points After Deletions

Hard19.9% acceptance1 company

You are given an integer array nums.

A position i is called a fixed point if nums[i] == i.

You are allowed to delete any number of elements (including zero) from the array. After each deletion, the remaining elements shift left, and indices are reassigned starting from 0.

Return an integer denoting the maximum number of fixed points that can be achieved after performing any number of deletions.

Example 1:

Input: nums = [0,2,1]

Output: 2

Explanation:

Delete nums[1] = 2. The array becomes [0, 1].
Now, nums[0] = 0 and nums[1] = 1, so both indices are fixed points.
Thus, the answer is 2.

Example 2:

Input: nums = [3,1,2]

Output: 2

Explanation:

Do not delete any elements. The array remains [3, 1, 2].
Here, nums[1] = 1 and nums[2] = 2, so these indices are fixed points.
Thus, the answer is 2.

Example 3:

Input: nums = [1,0,1,2]

Output: 3

Explanation:

Delete nums[0] = 1. The array becomes [0, 1, 2].
Now, nums[0] = 0, nums[1] = 1, and nums[2] = 2, so all indices are fixed points.
Thus, the answer is 3.

Constraints:

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

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

An element can become a fixed point only if i >= nums[i].

Hint 2

Two indices i and j can both be fixed points if they satisfy the first condition and i - nums[i] < j - nums[j].

Hint 3

Find the longest possible increasing subsequence from those indices.

LanguageLoading editor…

1class Solution:
2    def maxFixedPoints(self, nums: list[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[0,2,1]

Input

[3,1,2]

Input

[1,0,1,2]

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