Here, We see All One Data Structure LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
List of all LeetCode Solution
All One Data Structure LeetCode Solution
Table of Contents
Problem Statement
Design a data structure to store the strings’ count with the ability to return the strings with minimum and maximum counts.
Implement the AllOne class:
AllOne()Initializes the object of the data structure.inc(String key)Increments the count of the stringkeyby1. Ifkeydoes not exist in the data structure, insert it with count1.dec(String key)Decrements the count of the stringkeyby1. If the count ofkeyis0after the decrement, remove it from the data structure. It is guaranteed thatkeyexists in the data structure before the decrement.getMaxKey()Returns one of the keys with the maximal count. If no element exists, return an empty string"".getMinKey()Returns one of the keys with the minimum count. If no element exists, return an empty string"".
Note that each function must run in O(1) average time complexity.
Example 1:
Input [“AllOne”, “inc”, “inc”, “getMaxKey”, “getMinKey”, “inc”, “getMaxKey”, “getMinKey”] [[], [“hello”], [“hello”], [], [], [“leet”], [], []]
Output [null, null, null, “hello”, “hello”, null, “hello”, “leet”]
Explanation
AllOne allOne = new AllOne();
allOne.inc(“hello”);
allOne.inc(“hello”);
allOne.getMaxKey(); // return “hello”
allOne.getMinKey(); // return “hello”
allOne.inc(“leet”);
allOne.getMaxKey(); // return “hello”
allOne.getMinKey(); // return “leet”
All One Data Structure LeetCode Solution C++
class Entry {
public:
int cnt;
unordered_set <string> keys;
Entry(int _cnt, string &key) {
cnt = _cnt;
keys.insert(key);
}
void add(string &key) {
keys.insert(key);
}
void remove(string &key) {
keys.erase(key);
}
bool empty() {
return keys.empty();
}
string getAnyKey() {
return *(keys.begin());
}
};
class AllOne {
public:
list <Entry> data;
unordered_map <string, list<Entry>::iterator> pos;
AllOne() {
}
void inc(string key) {
if (data.empty()) {
data.push_back(Entry(1, key));
pos[key] = data.begin();
return;
}
if (pos.find(key) == pos.end()) {
if (data.begin()->cnt == 1) {
data.begin()->add(key);
pos[key] = data.begin();
}
else {
pos[key] = data.insert(data.begin(), Entry(1, key));
}
}
else {
auto entry = pos[key];
if (entry == prev(data.end())) {
pos[key] = data.insert(data.end(), Entry(entry->cnt+1, key));
}
else if (entry->cnt+1 == next(entry)->cnt) {
next(entry)->add(key);
pos[key] = next(entry);
}
else {
pos[key] = data.insert(next(entry), Entry(entry->cnt+1, key));
}
entry->remove(key);
if (entry->empty()) data.erase(entry);
}
}
void dec(string key) {
auto entry = pos[key];
if (entry->cnt > 1) {
if (entry == data.begin()) {
pos[key] = data.insert(data.begin(), Entry(entry->cnt-1, key));
}
else if (entry->cnt-1 == prev(entry)->cnt) {
prev(entry)->add(key);
pos[key] = prev(entry);
}
else {
pos[key] = data.insert(entry, Entry(entry->cnt-1, key));
}
}
else pos.erase(key);
entry->remove(key);
if (entry->empty()) data.erase(entry);
}
string getMaxKey() {
if (data.empty()) return "";
return prev(data.end())->getAnyKey();
}
string getMinKey() {
if (data.empty()) return "";
return data.begin()->getAnyKey();
}
};Code language: PHP (php)All One Data Structure LeetCode Solution Java
class AllOne {
Map<String, Integer> map;
Map<Integer, Set<String>> valueMap;
LinkedList<Integer> minMax;
public AllOne() {
map = new HashMap<>();
valueMap = new HashMap<>();
minMax = new LinkedList<Integer>();
}
public void inc(String key) {
if(!map.containsKey(key)) {
map.put(key, 1);
putInValueMap(1, key);
} else {
int val = map.get(key);
removeFromValueMap(val, key);
map.put(key, val + 1);
putInValueMap(val + 1, key);
}
}
public void dec(String key) {
if(!map.containsKey(key)) {
return;
}
int val = map.get(key);
if(val == 1) {
map.remove(key);
removeFromValueMap(1, key);
} else {
removeFromValueMap(val, key);
map.put(key, val - 1);
putInValueMap(val - 1, key);
}
}
public String getMaxKey() {
if(minMax.isEmpty()) {
return "";
}
return valueMap.get(minMax.getFirst()).iterator().next();
}
public String getMinKey() {
if(minMax.isEmpty()) {
return "";
}
return valueMap.get(minMax.getLast()).iterator().next();
}
private void putInValueMap(int count, String node) {
if(!valueMap.containsKey(count)) {
valueMap.put(count, new HashSet<String>());
}
valueMap.get(count).add(node);
if(minMax.isEmpty() || minMax.getFirst() < count) {
minMax.addFirst(count);
}
if(!minMax.isEmpty() && minMax.getLast() > count) {
minMax.addLast(count);
}
}
private void removeFromValueMap(int count, String node) {
if(!valueMap.containsKey(count)) {
return;
}
valueMap.get(count).remove(node);
if(valueMap.get(count).size() == 0) {
valueMap.remove(count);
if(!minMax.isEmpty() && minMax.getFirst() == count) {
minMax.removeFirst();
}
if(!minMax.isEmpty() && minMax.getLast() == count) {
minMax.removeLast();
}
}
}
}Code language: JavaScript (javascript)All One Data Structure LeetCode Solution JavaScript
var AllOne = function () {
this.minCount = 0;
this.maxCount = 0;
this.keyFreqMap = new Map();
this.freqSetMap = new Map();
};
AllOne.prototype.inc = function (key) {
let count = this.keyFreqMap.get(key);
let newMaxCount = count;
let newMinCount = this.minCount;
if (count) {
let existing_set = this.freqSetMap.get(count);
existing_set.delete(key);
if (existing_set.size == 0) {
this.freqSetMap.delete(count);
} else {
this.freqSetMap.set(count, existing_set);
}
if (existing_set.size == 0 && this.minCount == count)
newMinCount = newMinCount + 1;
let set = this.freqSetMap.get(count + 1) || new Set();
set.add(key);
this.freqSetMap.set(count + 1, set);
this.keyFreqMap.set(key, count + 1);
newMaxCount = count + 1;
} else {
this.keyFreqMap.set(key, 1);
let existing_set = this.freqSetMap.get(1) || new Set();
existing_set.add(key);
this.freqSetMap.set(1, existing_set);
newMinCount = 1;
newMaxCount = 1;
}
this.minCount = newMinCount;
this.maxCount = Math.max(this.maxCount, newMaxCount);
return null;
};
AllOne.prototype.dec = function (key) {
let count = this.keyFreqMap.get(key);
if (count) {
let existing_set = this.freqSetMap.get(count);
existing_set.delete(key);
if (existing_set.size == 0) {
this.freqSetMap.delete(count);
} else {
this.freqSetMap.set(count, existing_set);
}
if (count == 1) {
this.keyFreqMap.delete(key);
} else {
let set = this.freqSetMap.get(count - 1) || new Set();
set.add(key);
this.freqSetMap.set(count - 1, set);
this.keyFreqMap.set(key, count - 1);
}
if (existing_set.size == 0 && this.maxCount == count) this.maxCount -= 1;
if (existing_set.size == 0 && this.minCount == count) {
let min = this.maxCount;
for (let key of this.freqSetMap.keys()) {
min = Math.min(min, key);
}
if (this.freqSetMap.size > 0) this.minCount = min;
}
}
return null;
};
AllOne.prototype.getMaxKey = function () {
let set = this.freqSetMap.get(this.maxCount) || new Set();
let firstElement = "";
for (let val of set) {
firstElement = val;
break;
}
return firstElement;
};
AllOne.prototype.getMinKey = function () {
let set = this.freqSetMap.get(this.minCount) || new Set();
let firstElement = "";
for (let val of set) {
firstElement = val;
break;
}
return firstElement;
};Code language: JavaScript (javascript)All One Data Structure LeetCode Solution Python
class Block(object):
def __init__(self, val=0):
self.val = val
self.keys = set()
self.before = None
self.after = None
def remove(self):
self.before.after = self.after
self.after.before = self.before
self.before, self.after = None, None
def insert_after(self, new_block):
old_after = self.after
self.after = new_block
new_block.before = self
new_block.after = old_after
old_after.before = new_block
class AllOne(object):
def __init__(self):
self.begin = Block()
self.end = Block()
self.begin.after = self.end
self.end.before = self.begin
self.mapping = {}
def inc(self, key):
if not key in self.mapping:
current_block = self.begin
else:
current_block = self.mapping[key]
current_block.keys.remove(key)
if current_block.val + 1 != current_block.after.val:
new_block = Block(current_block.val + 1)
current_block.insert_after(new_block)
else:
new_block = current_block.after
new_block.keys.add(key)
self.mapping[key] = new_block
if not current_block.keys and current_block.val != 0:
current_block.remove()
def dec(self, key):
if not key in self.mapping:
return
current_block = self.mapping[key]
del self.mapping[key]
current_block.keys.remove(key)
if current_block.val != 1:
if current_block.val - 1 != current_block.before.val:
new_block = Block(current_block.val - 1)
current_block.before.insert_after(new_block)
else:
new_block = current_block.before
new_block.keys.add(key)
self.mapping[key] = new_block
if not current_block.keys:
current_block.remove()
def getMaxKey(self):
if self.end.before.val == 0:
return ""
key = self.end.before.keys.pop()
self.end.before.keys.add(key)
return key
def getMinKey(self):
if self.begin.after.val == 0:
return ""
key = self.begin.after.keys.pop()
self.begin.after.keys.add(key)
return key