Algorithm
Problem Name: 1224. Maximum Equal Frequency
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 <= 1051 <= nums[i] <= 105
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const maxEqualFreq = function (nums) {
const freqCnt = {}, cnt = {}, { max } = Math
let res = 0, maxF = 0, i = 0
for(const e of nums) {
if(cnt[e] == null) cnt[e] = 0
cnt[e]++
const f = cnt[e]
if(freqCnt[f - 1] == null) freqCnt[f - 1] = 0
if(freqCnt[f] == null) freqCnt[f] = 0
if(freqCnt[f - 1] > 0) freqCnt[f - 1]--
freqCnt[f]++
maxF = max(maxF, f)
if(
maxF === 1 ||
maxF * freqCnt[maxF] === i ||
(maxF - 1) * (freqCnt[maxF - 1] + 1) === i
) {
res = i + 1
}
i++
}
return res
}
Input
nums = [2,2,1,1,5,3,3,5]
Output
7
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxEqualFreq(self, nums: List[int]) -> int:
def okay():
if len(dic) == 1 and (1 in dic or 1 in dic.values()):
return True
if len(dic) == 2:
c1, c2 = sorted(dic.keys())
if c2 - c1 == 1 and dic[c2] == 1 or (c1 == 1 and dic[1] == 1):
return True
cnt = collections.Counter(nums)
dic = collections.Counter(cnt.values())
l = len(nums)
for num in nums[::-1]:
if okay():
return l
dic[cnt[num]] -= 1
if not dic[cnt[num]]:
dic.pop(cnt[num])
cnt[num] -= 1
if cnt[num]:
dic[cnt[num]] += 1
l -= 1
if okay():
return l
Input
nums = [2,2,1,1,5,3,3,5]
Output
7