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Nick Camilleri

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Design And Implement An LFU Cache Problem Statement

Least Frequently Used (LFU) is one of the cache algorithms (also called cache replacement algorithms or policies). When the cache is full and a new element needs to be added, this algorithm evicts/purges the least frequently used element to free up room for adding the new one.

Cache has two operations:

• `put(key, value)` does not return anything. It inserts or updates (if one already present) the value for the given key in the cache. If the cache is at capacity and a key-value pair needs to be inserted, the least frequently used key-value gets evicted before adding a new one. If there are multiple LFU keys, the least recently used one of them gets evicted.

• `get(key)` returns the value currently associated with the given key in the cache. If there is no value for that key, it returns -1.

Calling any one of these operations for a key increments that key's usage count by one.

Once a key-value pair is evicted from the cache, the key's usage count is reset/forgotten as if it was never in the cache.

Given a capacity and a set of operations to perform on a LFU cache, return results of those operations.

Each operation will be given as either two or three numbers:

• `[0, key, value]` means `put(key, value)`, and
• `[1, key]` means `get(key)`.

Example

``````{
"capacity": 2,
"operations": [
[0, 1, 2],
[0, 3, 6],
[1, 1],
[0, 4, 8],
[1, 3],
[1, 4],
[1, 1]
]
}
``````

Output:

``````[2, -1, 8, 2]
``````

Output contains results of all `get` operations, in order:

• 2 was the value in the cache corresponding to key 1 at the time when `get(1)` was called for the first time,
• -1 was returned because by the time `get(3)` was called, the cache did not contain a value for key 3 (it was evicted, as the least frequently used one, while processing `put(4, 8)`),
• 8 was the value for key 4, and
• 2 was the value for key 1.

Notes

• It is a good idea to implement the cache as a class (or object or struct - depending on the language you use). In function `implement_lfu_cache` you would then create one instance of that class/object/struct and call its methods/functions to process the operations.
• Structure of the class may look like this: \`\`\` class LFUCache { LFUCache(int capacity) { // This is a constructor. // Initialize the data structures of the cache. }

int get(int key) { // ... }

void put(int key, int value) { // ... } }

Constraints

• 0 <= cache capacity <= 104
• 0 <= a key in the cache <= 105
• 0 <= a value in the cache <= 109
• 1 <= number of operations <= 105

Code For Design And Implement An Lfu Cache Solution: Optimal

``````/*
Asymptotic complexity in terms of number of operations \`q\`, complexity of \`get\` operation \`n1\`, complexity of \`put\` operation \`n2\`,
cache capacity \`c\`, key in the cache \`n\`:
* Time: O(q * (n1 + n2)), where O(n1) = O(1) and O(n2) = O(1).
* Auxiliary space: O(n + c).
* Total space: O(n + c).
*/

class LFUCache {
public:
const static int N = 1e5 + 5;

int cache_value[N];

int use_counter[N];

list<int>::iterator position[N]; // This will store the position of a key in the deque.

list<int> cache_deque[N]; // This will store all the key in least frequently order for each use_counter value.

int min_use_count = 1, cache_size = 0;

int capacity;

LFUCache(int capacity_) {
capacity = capacity_;
memset(cache_value, -1, sizeof(cache_value));
}

int get(int key) {
if (!capacity || cache_value[key] == -1) {
return -1;
}

return cache_value[key];
}

/* This function will update the use_counter corresponding to the passed \`key\` and will
also make the necessary updates in \`cache_deque\`.
*/
if (use_counter[key]) {
cache_deque[use_counter[key]].erase(position[key]);

if (cache_deque[min_use_count].empty()) {
min_use_count++;
}
}

use_counter[key]++;

if (use_counter[key] == 1) {
min_use_count = 1;
}

cache_deque[use_counter[key]].push_back(key);
position[key] = cache_deque[use_counter[key]].end();
position[key]--;
}

void put(int key, int value) {
if (capacity == 0) {
return;
}

if (cache_size == capacity && use_counter[key] == 0) {
// Cache is full and this is a new key.
invalidate();
}

cache_value[key] = value;

if (use_counter[key] == 1) {
// This key is inserted for the first time.
cache_size++;
}
}

/*
This function will purge the Least Frequently Used key from the cache.
*/
void invalidate() {
int key = cache_deque[min_use_count].front();

use_counter[key] = 0;
cache_value[key] = -1;

cache_deque[min_use_count].pop_front();

cache_size--;
}
};

vector<int> implement_lfu_cache(int capacity, vector<vector<int>> operations) {
LFUCache lfu(capacity);

vector<int> result;

for (auto operation : operations) {
if (operation[0] == 0) { // put
int key = operation[1];
int value = operation[2];
lfu.put(key, value);
} else { // get
int key = operation[1];
result.push_back(lfu.get(key));
}
}

return result;
}
``````

We hope that these solutions to the 4 sum problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart’s FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

Try yourself in the Editor

Note: Input and Output will already be taken care of.

Design And Implement An LFU Cache Problem Statement

Least Frequently Used (LFU) is one of the cache algorithms (also called cache replacement algorithms or policies). When the cache is full and a new element needs to be added, this algorithm evicts/purges the least frequently used element to free up room for adding the new one.

Cache has two operations:

• `put(key, value)` does not return anything. It inserts or updates (if one already present) the value for the given key in the cache. If the cache is at capacity and a key-value pair needs to be inserted, the least frequently used key-value gets evicted before adding a new one. If there are multiple LFU keys, the least recently used one of them gets evicted.

• `get(key)` returns the value currently associated with the given key in the cache. If there is no value for that key, it returns -1.

Calling any one of these operations for a key increments that key's usage count by one.

Once a key-value pair is evicted from the cache, the key's usage count is reset/forgotten as if it was never in the cache.

Given a capacity and a set of operations to perform on a LFU cache, return results of those operations.

Each operation will be given as either two or three numbers:

• `[0, key, value]` means `put(key, value)`, and
• `[1, key]` means `get(key)`.

Example

``````{
"capacity": 2,
"operations": [
[0, 1, 2],
[0, 3, 6],
[1, 1],
[0, 4, 8],
[1, 3],
[1, 4],
[1, 1]
]
}
``````

Output:

``````[2, -1, 8, 2]
``````

Output contains results of all `get` operations, in order:

• 2 was the value in the cache corresponding to key 1 at the time when `get(1)` was called for the first time,
• -1 was returned because by the time `get(3)` was called, the cache did not contain a value for key 3 (it was evicted, as the least frequently used one, while processing `put(4, 8)`),
• 8 was the value for key 4, and
• 2 was the value for key 1.

Notes

• It is a good idea to implement the cache as a class (or object or struct - depending on the language you use). In function `implement_lfu_cache` you would then create one instance of that class/object/struct and call its methods/functions to process the operations.
• Structure of the class may look like this: \`\`\` class LFUCache { LFUCache(int capacity) { // This is a constructor. // Initialize the data structures of the cache. }

int get(int key) { // ... }

void put(int key, int value) { // ... } }

Constraints

• 0 <= cache capacity <= 104
• 0 <= a key in the cache <= 105
• 0 <= a value in the cache <= 109
• 1 <= number of operations <= 105

Code For Design And Implement An Lfu Cache Solution: Optimal

``````/*
Asymptotic complexity in terms of number of operations \`q\`, complexity of \`get\` operation \`n1\`, complexity of \`put\` operation \`n2\`,
cache capacity \`c\`, key in the cache \`n\`:
* Time: O(q * (n1 + n2)), where O(n1) = O(1) and O(n2) = O(1).
* Auxiliary space: O(n + c).
* Total space: O(n + c).
*/

class LFUCache {
public:
const static int N = 1e5 + 5;

int cache_value[N];

int use_counter[N];

list<int>::iterator position[N]; // This will store the position of a key in the deque.

list<int> cache_deque[N]; // This will store all the key in least frequently order for each use_counter value.

int min_use_count = 1, cache_size = 0;

int capacity;

LFUCache(int capacity_) {
capacity = capacity_;
memset(cache_value, -1, sizeof(cache_value));
}

int get(int key) {
if (!capacity || cache_value[key] == -1) {
return -1;
}

return cache_value[key];
}

/* This function will update the use_counter corresponding to the passed \`key\` and will
also make the necessary updates in \`cache_deque\`.
*/
if (use_counter[key]) {
cache_deque[use_counter[key]].erase(position[key]);

if (cache_deque[min_use_count].empty()) {
min_use_count++;
}
}

use_counter[key]++;

if (use_counter[key] == 1) {
min_use_count = 1;
}

cache_deque[use_counter[key]].push_back(key);
position[key] = cache_deque[use_counter[key]].end();
position[key]--;
}

void put(int key, int value) {
if (capacity == 0) {
return;
}

if (cache_size == capacity && use_counter[key] == 0) {
// Cache is full and this is a new key.
invalidate();
}

cache_value[key] = value;

if (use_counter[key] == 1) {
// This key is inserted for the first time.
cache_size++;
}
}

/*
This function will purge the Least Frequently Used key from the cache.
*/
void invalidate() {
int key = cache_deque[min_use_count].front();

use_counter[key] = 0;
cache_value[key] = -1;

cache_deque[min_use_count].pop_front();

cache_size--;
}
};

vector<int> implement_lfu_cache(int capacity, vector<vector<int>> operations) {
LFUCache lfu(capacity);

vector<int> result;

for (auto operation : operations) {
if (operation[0] == 0) { // put
int key = operation[1];
int value = operation[2];
lfu.put(key, value);
} else { // get
int key = operation[1];
result.push_back(lfu.get(key));
}
}

return result;
}
``````

We hope that these solutions to the 4 sum problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart’s FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as: