Maximize Fixed Points After Deletions solution and explanation
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The problem
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] = 0andnums[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] = 1andnums[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, andnums[2] = 2, so all indices are fixed points. - Thus, the answer is 3.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 105
Approach overview
Problem Overview: You are given an array and can delete any elements. After deletions, the remaining elements are reindexed starting from 1. A fixed point occurs when a[i] == i. The goal is to delete elements so the resulting array contains the maximum possible number of fixed points.
Approach 1: Brute Force Subsequence Enumeration (O(2^n) time, O(n) space)
Generate every possible subsequence and check whether elements satisfy the fixed-point condition after reindexing. For each subsequence of length k, verify that the element placed at index j equals j. This approach uses recursion or bitmasks to explore all subsets. It demonstrates the problem mechanics but quickly becomes infeasible once n grows because the search space doubles with each element.
Approach 2: Dynamic Programming on Value Matching (O(n^2) time, O(n) space)
Observe that if the final array has k fixed points, its values must be exactly 1, 2, ..., k in that order. Any valid solution is therefore a subsequence whose values strictly follow this sequence. Define dp[v] as the maximum subsequence length ending with value v. Iterate through the array and attempt to extend subsequences where the previous value is v-1. This ensures the subsequence forms the exact increasing sequence required for fixed points. The DP formulation is straightforward but requires scanning states repeatedly, which leads to quadratic behavior.
Approach 3: Greedy Target Matching (O(n) time, O(1) space)
The optimal observation is that the final array must contain the sequence 1,2,3,...,k in order. While scanning the array from left to right, maintain the next required value target. Whenever a[i] == target, keep that element and increment target. Otherwise, delete the element. Because deletions shift indices left, selecting values in this exact order guarantees each chosen element becomes a fixed point after reindexing. This works because picking the earliest valid match always preserves the maximum remaining opportunities for larger targets. The result is target - 1 fixed points.
This greedy strategy mirrors subsequence matching patterns used in problems involving ordered constraints. The idea also appears in variations of greedy algorithms and subsequence construction problems related to arrays and dynamic programming.
Recommended for interviews: Start by explaining that the final array must equal the sequence 1..k. Mentioning the brute force or DP interpretation shows understanding of the constraint. Interviewers typically expect the greedy scan because it reduces the problem to a simple subsequence match with O(n) time and constant space.
The written solution for this problem is still being prepared. Try it yourself in the meantime.
Try solving it yourselfComplexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Subsequence Enumeration | O(2^n) | O(n) | Only for conceptual understanding or very small inputs |
| Dynamic Programming on Value Matching | O(n^2) | O(n) | When deriving the subsequence constraint formally |
| Greedy Target Matching | O(n) | O(1) | Optimal approach for interviews and production solutions |
Frequently asked questions
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.