Prefix with Equivalent Frequencies - Amazon Top Interview Questions
Problem Statement :
You are given a list of integers nums. Return the length of the longest prefix of nums such that after we remove one element in the prefix, each number occurs the same number of times. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [5, 5, 3, 7, 3, 9] Output 5 Explanation If we pick the prefix [5, 5, 3, 7, 3] and remove 7 then every number would occur twice.
Solution :
Solution in C++ :
int solve(vector<int>& nums) {
// we use two hashmaps one to keep the track of the frequecny of elements and one to keep the
// track of the of the frequecny of frequencies.
map<int, int> f, fc;
int ans = 0;
for (int i = 0; i < nums.size(); i++) {
// if we have already inserted the element inside our prefix window
// we first remove old frequency contribution
// if there is only one element with this frequency than after the frequency change there
// will be no element remaining with this frequency so we remove it's existance
if (f.find(nums[i]) != f.end() and fc[f[nums[i]]]-- == 1) {
fc.erase(f[nums[i]]);
}
// now we increase the frequency of the element and also the frequency count of f[nums[i]]
// afterwards.
fc[++f[nums[i]]]++;
// now checking if the window so far is valid.
if (fc.size() == 2) {
if (fc.find(1) != fc.end()) {
if (fc[1] == 1) ans = max(ans, i + 1);
}
if (fc.begin()->first == fc.rbegin()->first - 1) {
if (fc.rbegin()->second == 1) ans = max(ans, i + 1);
}
} else if (fc.size() == 1 and (fc.begin()->first == 1 or f.size() == 1)) {
ans = max(ans, i + 1);
}
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
// 1234567 remove one of them
// 1111111 remove one of them
// 111 222 333 4 remove 4
// 111 222 333 4444 remove one of 4
Map<Integer, List<Integer>> window = new HashMap<>();
Map<Integer, Integer> freq = new HashMap<>();
int len = nums.length;
if (len == 2) {
return 2;
}
int res = 0;
for (int i = 0; i < len; i++) {
int n = nums[i];
freq.put(n, freq.getOrDefault(n, 0) + 1);
int f = freq.get(n);
if (window.containsKey(f - 1)) {
window.get(f - 1).remove(new Integer(n));
if (window.get(f - 1).size() == 0) {
window.remove(f - 1);
}
}
window.putIfAbsent(f, new ArrayList<Integer>());
window.get(f).add(n);
// if frequency of all the elements is 1
if (window.size() == 1 && window.containsKey(1)) {
res = i + 1;
// if all the elements are same inside the prefix window
} else if (window.size() == 1) {
for (int key : window.keySet()) {
if (window.get(key).size() == 1) {
res = i + 1;
}
}
} else if (window.size() == 2) {
int first = -1;
int second = -1;
for (int key : window.keySet()) {
if (first == -1) {
first = key;
} else {
second = key;
}
}
if (first + 1 == second && window.get(second).size() == 1) {
res = i + 1;
} else if (second + 1 == first && window.get(first).size() == 1) {
res = i + 1;
} else if (window.containsKey(1) && window.get(1).size() == 1) {
res = i + 1;
}
} else {
continue;
}
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums):
freq = dict() # stores the amount of different counts
counts = dict() # stores the count of each number
for i, num in enumerate(nums):
old_c, new_c = counts.get(num, 0), counts.get(num, 0) + 1
if old_c in freq: # remove previous frequency
freq[old_c] -= 1
if freq[old_c] == 0:
del freq[old_c]
counts[num] = new_c # add the new frequency
freq[new_c] = freq.get(new_c, 0) + 1
if len(freq) == 1:
for k, v in freq.items():
if v == 1 or k == 1:
res = i + 1
if len(freq) == 2:
tmp = [(k, v) for k, v in freq.items()]
(f1, c1), (f2, c2) = tmp[0], tmp[1]
if (f1 == 1 and c1 == 1) or (f2 == 1 and c2 == 1):
res = i + 1
if c1 == 1 and f1 - 1 == f2 or c2 == 1 and f2 - 1 == f1:
res = i + 1
return res
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