Leetcode #1224: Maximum Equal Frequency
In this guide, we solve Leetcode #1224 Maximum Equal Frequency in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an array nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences. If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [2,2,1,1,5,3,3,5]
Output: 7
Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.
Python Solution
class Solution:
def maxEqualFreq(self, nums: List[int]) -> int:
cnt = Counter()
ccnt = Counter()
ans = mx = 0
for i, v in enumerate(nums, 1):
if v in cnt:
ccnt[cnt[v]] -= 1
cnt[v] += 1
mx = max(mx, cnt[v])
ccnt[cnt[v]] += 1
if mx == 1:
ans = i
elif ccnt[mx] * mx + ccnt[mx - 1] * (mx - 1) == i and ccnt[mx] == 1:
ans = i
elif ccnt[mx] * mx + 1 == i and ccnt[1] == 1:
ans = i
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.