Medium
Distant Barcodes — C++
Full explanation · Time O(n) · Space O(k)
// Time: O(n), k is the number of distinct barcode
// Space: O(k)
class Solution {
public:
vector<int> rearrangeBarcodes(vector<int>& barcodes) {
static const int k = 2;
unordered_map<int, int> cnts;
for (const auto& c : barcodes) {
++cnts[c];
}
const int bucket_cnt = max_element(cbegin(cnts), cend(cnts), [](const auto& a, const auto& b) {
return a.second < b.second;
})->second;
vector<int> partial_sorted_cnts;
for (const auto& [c, v] : cnts) {
if (v == bucket_cnt) {
partial_sorted_cnts.emplace_back(c);
}
}
for (const auto& [c, v] : cnts) {
if (v != bucket_cnt) {
partial_sorted_cnts.emplace_back(c);
}
}
vector<int> result(size(barcodes));
int i = (size(barcodes) - 1) % k;
for (const auto& c : partial_sorted_cnts) {
for (int _ = 0; _ < cnts[c]; ++_) {
result[i] = c;
i += k;
if (i >= size(result)) {
i = (i - 1) % k;
}
}
}
return result;
}
};
// Time: O(n + klogk), k is the number of distinct barcode
// Space: O(k)
class Solution2 {
public:
vector<int> rearrangeBarcodes(vector<int>& barcodes) {
unordered_map<int, int> cnts;
for (const auto& barcode : barcodes) {
++cnts[barcode];
}
vector<pair<int, int>> sorted_cnts;
for (const auto& kvp : cnts) {
sorted_cnts.emplace_back(kvp.second, kvp.first);
}
sort(sorted_cnts.begin(), sorted_cnts.end(),
greater<pair<int, int>>());
int i = 0;
for (const auto& kvp : sorted_cnts) {
for (int j = 0; j < kvp.first; ++j) {
barcodes[i] = kvp.second;
i += 2;
if (i >= barcodes.size()) {
i = 1;
}
}
}
return barcodes;
}
};