LeetCode #1224 — HARD

Maximum Equal Frequency

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

The Problem

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).

Example 1:

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.

Example 2:

Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
Output: 13

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[2,2,1,1,5,3,3,5]

Example 2

[1,1,1,2,2,2,3,3,3,4,4,4,5]

Core Insight

What unlocks the optimal approach

Keep track of the min and max frequencies.
The number to be eliminated must have a frequency of 1, same as the others or the same +1.

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 #1224: Maximum Equal Frequency
class Solution {
    private static int[] cnt = new int[100010];
    private static int[] ccnt = new int[100010];

    public int maxEqualFreq(int[] nums) {
        Arrays.fill(cnt, 0);
        Arrays.fill(ccnt, 0);
        int ans = 0;
        int mx = 0;
        for (int i = 1; i <= nums.length; ++i) {
            int v = nums[i - 1];
            if (cnt[v] > 0) {
                --ccnt[cnt[v]];
            }
            ++cnt[v];
            mx = Math.max(mx, cnt[v]);
            ++ccnt[cnt[v]];
            if (mx == 1) {
                ans = i;
            } else if (ccnt[mx] * mx + ccnt[mx - 1] * (mx - 1) == i && ccnt[mx] == 1) {
                ans = i;
            } else if (ccnt[mx] * mx + 1 == i && ccnt[1] == 1) {
                ans = i;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1224: Maximum Equal Frequency
func maxEqualFreq(nums []int) int {
	cnt := map[int]int{}
	ccnt := map[int]int{}
	ans, mx := 0, 0
	for i, v := range nums {
		i++
		if cnt[v] > 0 {
			ccnt[cnt[v]]--
		}
		cnt[v]++
		mx = max(mx, cnt[v])
		ccnt[cnt[v]]++
		if mx == 1 {
			ans = i
		} else if ccnt[mx]*mx+ccnt[mx-1]*(mx-1) == i && ccnt[mx] == 1 {
			ans = i
		} else if ccnt[mx]*mx+1 == i && ccnt[1] == 1 {
			ans = i
		}
	}
	return ans
}

# Accepted solution for LeetCode #1224: Maximum Equal Frequency
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

// Accepted solution for LeetCode #1224: Maximum Equal Frequency
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn max_equal_freq(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut res = 0;
        let mut count: HashMap<i32, usize> = HashMap::new();
        let mut freq: HashMap<usize, usize> = HashMap::new();
        let mut max_freq = 0;
        for i in 0..n {
            let f = count.entry(nums[i]).or_default();
            if *f != 0 {
                *freq.entry(*f).or_default() -= 1;
            }
            *f += 1;
            *freq.entry(*f).or_default() += 1;
            max_freq = max_freq.max(*f);
            if max_freq == 1
                || *freq.entry(max_freq).or_default() * max_freq == i
                || (*freq.entry(max_freq - 1).or_default() + 1) * (max_freq - 1) == i
            {
                res = res.max(i + 1);
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![2, 2, 1, 1, 5, 3, 3, 5];
    let res = 7;
    assert_eq!(Solution::max_equal_freq(nums), res);
    let nums = vec![1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5];
    let res = 13;
    assert_eq!(Solution::max_equal_freq(nums), res);
    let nums = vec![1, 1, 1, 2, 2, 2];
    let res = 5;
    assert_eq!(Solution::max_equal_freq(nums), res);
}

// Accepted solution for LeetCode #1224: Maximum Equal Frequency
function maxEqualFreq(nums: number[]): number {
    const n = nums.length;
    const map = new Map();
    for (const num of nums) {
        map.set(num, (map.get(num) ?? 0) + 1);
    }

    for (let i = n - 1; i > 0; i--) {
        for (const k of map.keys()) {
            map.set(k, map.get(k) - 1);
            let num = 0;
            for (const v of map.values()) {
                if (v !== 0) {
                    num = v;
                    break;
                }
            }
            let isOk = true;
            let sum = 1;
            for (const v of map.values()) {
                if (v !== 0 && v !== num) {
                    isOk = false;
                    break;
                }
                sum += v;
            }
            if (isOk) {
                return sum;
            }
            map.set(k, map.get(k) + 1);
        }
        map.set(nums[i], map.get(nums[i]) - 1);
    }
    return 1;
}

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.