Maximum Equal Frequency

Maximum Equal Frequency - Problem

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 occurrences (0).

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,2,1,1,5,3,3,5]

› Output: 7

💡 Note: For prefix [2,2,1,1,5,3,3] of length 7, we can remove one occurrence of element 5, leaving frequencies: 2 appears 2 times, 1 appears 2 times, 3 appears 2 times - all equal

Example 2 — All Same Elements

$ Input: nums = [1,1,1,1]

› Output: 4

💡 Note: All elements are the same, so we can remove any one element and all remaining elements have the same frequency (3)

Example 3 — Remove to Empty

$ Input: nums = [1]

› Output: 1

💡 Note: Single element array - removing it leaves empty array where all elements have same frequency (0)

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁵

Visualization

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Asked in

G Google 25 f Facebook 18

The key insight is to recognize valid frequency patterns mathematically rather than trying all removals. We track element frequencies and frequency counts, then check if the current state matches patterns where removing one element makes all frequencies equal. Best approach uses optimized frequency tracking in O(n) time.

Common Approaches

✓ Bit Manipulation

⏱️ Time: N/A Space: N/A

Optimized

⏱️ Time: N/A Space: N/A

Brute Force - Try All Prefixes

⏱️ Time: O(n³) Space: O(n)

For every possible prefix length, create frequency map, then try removing each unique element and check if remaining elements have equal frequencies. Keep track of maximum valid prefix length.

Optimized Frequency Tracking

⏱️ Time: O(n) Space: O(n)

Instead of trying all removals, identify the mathematical conditions when removing one element can make all frequencies equal. Track count of each frequency value and check valid patterns.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    if (n == 0) return 0;
    
    // Find the largest power of 2 that is <= n
    // This can be done by finding the highest bit position and using it
    int highestBit = 0;
    int temp = n;
    while (temp >>= 1) highestBit++;
    
    return 1 << highestBit;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~35 min Avg. Time

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Smart Actions

💡 Explanation

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