title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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Python 3 || 9 lines, w/ explanation || T/M: 89% / 49% | number-of-orders-in-the-backlog | 0 | 1 | Here\'s the plan:\n1. Establish maxheap `buy` for buy orders and minheap `sell` for sell orders.\n2. Iterate through `orders`. Push each element onto the appropriate heap.\n3. During each iteration, peak at the heads of each list and determine whether they allow a transaction. If so, pop both heads, and if one has a po... | 3 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python3] priority queue | number-of-orders-in-the-backlog | 0 | 1 | \n```\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n ans = 0\n buy, sell = [], [] # max-heap & min-heap \n \n for p, q, t in orders: \n ans += q\n if t: # sell order\n while q and buy and -buy[0][0] >= p: # mat... | 6 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python3] priority queue | number-of-orders-in-the-backlog | 0 | 1 | \n```\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n ans = 0\n buy, sell = [], [] # max-heap & min-heap \n \n for p, q, t in orders: \n ans += q\n if t: # sell order\n while q and buy and -buy[0][0] >= p: # mat... | 6 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Heap/PQ solution with logic [Python] | number-of-orders-in-the-backlog | 0 | 1 | Logic is:\n* Since we need to do a lot of **find min/max** operations: use 2 heaps:\n\t* max-heap `b` for buy, min-heap `s` for sell\n\t* So, max buy offer is on top of heap `b`\n\t* Min sell offer is on top of heap `s`\n* Each element of heap is an array: `[price, amount]`\n* *For* each buy/sell order:\n\t* Check for ... | 9 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Heap/PQ solution with logic [Python] | number-of-orders-in-the-backlog | 0 | 1 | Logic is:\n* Since we need to do a lot of **find min/max** operations: use 2 heaps:\n\t* max-heap `b` for buy, min-heap `s` for sell\n\t* So, max buy offer is on top of heap `b`\n\t* Min sell offer is on top of heap `s`\n* Each element of heap is an array: `[price, amount]`\n* *For* each buy/sell order:\n\t* Check for ... | 9 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python] Concise Heap Implementation | number-of-orders-in-the-backlog | 0 | 1 | ### Introduction\n\nWe need to find the total amount of orders that remain un-executed in the backlog of orders after a series of buy and sell orders have passed. A buy order is executed if there exists a sell order in the backlog that has a selling price lower than or equal to the buying price, and a sell order is exe... | 4 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python] Concise Heap Implementation | number-of-orders-in-the-backlog | 0 | 1 | ### Introduction\n\nWe need to find the total amount of orders that remain un-executed in the backlog of orders after a series of buy and sell orders have passed. A buy order is executed if there exists a sell order in the backlog that has a selling price lower than or equal to the buying price, and a sell order is exe... | 4 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python3] OOD | Interview Friendly Solution | number-of-orders-in-the-backlog | 0 | 1 | # Complexity\nTime: $$O(nlogn)$$\nSpace: $$O(n)$$\n\n\n# Code\n```\nimport abc\nimport heapq\nfrom enum import Enum\nfrom abc import ABC, abstractmethod\nfrom typing import Optional, List\n\nMOD = int(1e9+7)\nclass OrderType(Enum):\n BUY = 0\n SELL = 1\n\n @classmethod\n def get(cls, order_type_id: int):\n ... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python3] OOD | Interview Friendly Solution | number-of-orders-in-the-backlog | 0 | 1 | # Complexity\nTime: $$O(nlogn)$$\nSpace: $$O(n)$$\n\n\n# Code\n```\nimport abc\nimport heapq\nfrom enum import Enum\nfrom abc import ABC, abstractmethod\nfrom typing import Optional, List\n\nMOD = int(1e9+7)\nclass OrderType(Enum):\n BUY = 0\n SELL = 1\n\n @classmethod\n def get(cls, order_type_id: int):\n ... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python - MaxHeap for Buy Backlog & MinHeap for Sell Backlog ✅ | number-of-orders-in-the-backlog | 0 | 1 | As the problem statement says, if there is a Buy order, we want to quickly get the sell order with the smallest price currently in the backlog, if it exists. If the price of that sell order is <= the price of buy order, we can execute the order. Again, it is not necessary that all the buy orders get executed with the s... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python - MaxHeap for Buy Backlog & MinHeap for Sell Backlog ✅ | number-of-orders-in-the-backlog | 0 | 1 | As the problem statement says, if there is a Buy order, we want to quickly get the sell order with the smallest price currently in the backlog, if it exists. If the price of that sell order is <= the price of buy order, we can execute the order. Again, it is not necessary that all the buy orders get executed with the s... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Priority Queue] with Intuition and Approach Explained | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\nWe want to find a way to match the best sell orders with the best buy orders, and vice versa. Since this matching requires us to think about the "minimum" price of the sell orders, and the "maximum" price of the buy orders in the backlog, we know we should use some form of heap or priority queue in order t... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Priority Queue] with Intuition and Approach Explained | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\nWe want to find a way to match the best sell orders with the best buy orders, and vice versa. Since this matching requires us to think about the "minimum" price of the sell orders, and the "maximum" price of the buy orders in the backlog, we know we should use some form of heap or priority queue in order t... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python3] Two Heap | number-of-orders-in-the-backlog | 0 | 1 | # Code\n```\nimport heapq\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int: \n buyQ = []\n sellQ = []\n\n for o in orders:\n price = o[0]\n amount = o[1]\n if o[2]==0:\n # buy\n while amount > 0 a... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python3] Two Heap | number-of-orders-in-the-backlog | 0 | 1 | # Code\n```\nimport heapq\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int: \n buyQ = []\n sellQ = []\n\n for o in orders:\n price = o[0]\n amount = o[1]\n if o[2]==0:\n # buy\n while amount > 0 a... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Just clean and readable code. | number-of-orders-in-the-backlog | 0 | 1 | # Approach\nWe represent the backlog as tuple of two heaps (or alternatively sorted arrays or priority queues) - one for buy orders (reversed) and one for sell orders. This way if any orders can match they will be at the top. This way we get all $$m$$ matches in the backlog in $$O(m)$$ steps.\n\n# Complexity\n$$n = len... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Just clean and readable code. | number-of-orders-in-the-backlog | 0 | 1 | # Approach\nWe represent the backlog as tuple of two heaps (or alternatively sorted arrays or priority queues) - one for buy orders (reversed) and one for sell orders. This way if any orders can match they will be at the top. This way we get all $$m$$ matches in the backlog in $$O(m)$$ steps.\n\n# Complexity\n$$n = len... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Heap Python3 Solution | number-of-orders-in-the-backlog | 0 | 1 | ```\nclass Solution:\n \n # O(nlogn) time,\n # O(n) space,\n # Approach: heap, \n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n MOD = 10**9 + 7\n sell_log = []\n buy_log = []\n in_log = 0\n \n for order in orders:\n price, amoun... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Heap Python3 Solution | number-of-orders-in-the-backlog | 0 | 1 | ```\nclass Solution:\n \n # O(nlogn) time,\n # O(n) space,\n # Approach: heap, \n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n MOD = 10**9 + 7\n sell_log = []\n buy_log = []\n in_log = 0\n \n for order in orders:\n price, amoun... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python sortedcontainers | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python sortedcontainers | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python] - Using heap T: O(nlogn) S: O(n) | number-of-orders-in-the-backlog | 0 | 1 | \n# Complexity\n- Time complexity: O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n \n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n MOD = 10**9 + 7\n... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python] - Using heap T: O(nlogn) S: O(n) | number-of-orders-in-the-backlog | 0 | 1 | \n# Complexity\n- Time complexity: O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n \n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n MOD = 10**9 + 7\n... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python (Simple Heap) | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python (Simple Heap) | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
PYTHON FASTER SOLUTION - Easy To Understand | number-of-orders-in-the-backlog | 0 | 1 | # Please Upvote If It\'s Helpful :)\n\n# Intuition\nWe keep min heap for selling because we look for the smallest price when there is buying input. And, we also keep max heap for buying because we look for the biggest price when there is selling input. If it is buying, while length of min heap bigger than 0 and amount ... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
PYTHON FASTER SOLUTION - Easy To Understand | number-of-orders-in-the-backlog | 0 | 1 | # Please Upvote If It\'s Helpful :)\n\n# Intuition\nWe keep min heap for selling because we look for the smallest price when there is buying input. And, we also keep max heap for buying because we look for the biggest price when there is selling input. If it is buying, while length of min heap bigger than 0 and amount ... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python3 solution beats 97%. exPlained | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\nMy first thought was to use a greedy approach. For each order, we can check if the order can be fulfilled by a previous order. If not, we can add it to a heap to track the backlog orders.\n\n# Approach\nWe can use two heaps to track all the buy and sell orders. We iterate through the orders and check if th... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python3 solution beats 97%. exPlained | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\nMy first thought was to use a greedy approach. For each order, we can check if the order can be fulfilled by a previous order. If not, we can add it to a heap to track the backlog orders.\n\n# Approach\nWe can use two heaps to track all the buy and sell orders. We iterate through the orders and check if th... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python easy understand | number-of-orders-in-the-backlog | 0 | 1 | \n\n# Complexity\n- Time complexity:\nO(nlogn)\n\n- Space complexity:\nO(n)\n\n\n# Code\n```\nimport heapq\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n buy, sell = [], []\n for p, a, t in orders:\n if t == 0:\n while a and sell and s... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python easy understand | number-of-orders-in-the-backlog | 0 | 1 | \n\n# Complexity\n- Time complexity:\nO(nlogn)\n\n- Space complexity:\nO(n)\n\n\n# Code\n```\nimport heapq\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n buy, sell = [], []\n for p, a, t in orders:\n if t == 0:\n while a and sell and s... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
[Python3] - Commented SortedList - Better Space efficiency? | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe definetely need some kind of data structure where we can quickly access the minimum/maximum sales price of the orders.\n\nTherefore we could use Min/Maxheaps or a SortedList.\nMy first intuition is to to use a SortedList as we then can... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
[Python3] - Commented SortedList - Better Space efficiency? | number-of-orders-in-the-backlog | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe definetely need some kind of data structure where we can quickly access the minimum/maximum sales price of the orders.\n\nTherefore we could use Min/Maxheaps or a SortedList.\nMy first intuition is to to use a SortedList as we then can... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python | Heap | O(nlogn) | number-of-orders-in-the-backlog | 0 | 1 | # Code\n```\nfrom heapq import heappush, heappop\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n buyBacklog = []\n sellBacklog = []\n for p, amount, t in orders:\n if t:\n while amount and buyBacklog and -buyBacklog[0][0] >= p:\n... | 0 | You are given a 2D integer array `orders`, where each `orders[i] = [pricei, amounti, orderTypei]` denotes that `amounti` orders have been placed of type `orderTypei` at the price `pricei`. The `orderTypei` is:
* `0` if it is a batch of `buy` orders, or
* `1` if it is a batch of `sell` orders.
Note that `orders[i]... | null |
Python | Heap | O(nlogn) | number-of-orders-in-the-backlog | 0 | 1 | # Code\n```\nfrom heapq import heappush, heappop\nclass Solution:\n def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:\n buyBacklog = []\n sellBacklog = []\n for p, amount, t in orders:\n if t:\n while amount and buyBacklog and -buyBacklog[0][0] >= p:\n... | 0 | There is a country of `n` cities numbered from `0` to `n - 1` where **all the cities are connected** by bi-directional roads. The roads are represented as a 2D integer array `edges` where `edges[i] = [xi, yi, timei]` denotes a road between cities `xi` and `yi` that takes `timei` minutes to travel. There may be multiple... | Store the backlog buy and sell orders in two heaps, the buy orders in a max heap by price and the sell orders in a min heap by price. Store the orders in batches and update the fields according to new incoming orders. Each batch should only take 1 "slot" in the heap. |
Python🔥Java🔥C++🔥Simple Solution🔥Easy to Understand | maximum-value-at-a-given-index-in-a-bounded-array | 1 | 1 | # !! BIG ANNOUNCEMENT !!\nI am currently Giving away my premium content well-structured assignments and study materials to clear interviews at top companies related to computer science and data science to my current Subscribers. This is only for the first 10,000 Subscribers. **DON\'T FORGET** to Subscribe.\n\n\n# Searc... | 11 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
Python🔥Java🔥C++🔥Simple Solution🔥Easy to Understand | maximum-value-at-a-given-index-in-a-bounded-array | 1 | 1 | # !! BIG ANNOUNCEMENT !!\nI am currently Giving away my premium content well-structured assignments and study materials to clear interviews at top companies related to computer science and data science to my current Subscribers. This is only for the first 10,000 Subscribers. **DON\'T FORGET** to Subscribe.\n\n\n# Searc... | 11 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
Simple Python solution using Binary Search | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n maxSum -= n\n def check(x):\n y = max(0, x - index)\n ans = (x+y) * (x-y+1) // 2\n y = max(0, x - ((n-1) - index))\n ans += (x+y) * (x-y+1) // 2\n retu... | 3 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
Simple Python solution using Binary Search | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n maxSum -= n\n def check(x):\n y = max(0, x - index)\n ans = (x+y) * (x-y+1) // 2\n y = max(0, x - ((n-1) - index))\n ans += (x+y) * (x-y+1) // 2\n retu... | 3 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
Growing the mountain | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn my mind, I visualized `nums` as an array of heights. Maximizing `nums[index]` while adhering to the rules would create a mountain shape. This would be a direct calculation, except that the boundaries of the index could truncate either ... | 2 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
Growing the mountain | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn my mind, I visualized `nums` as an array of heights. Maximizing `nums[index]` while adhering to the rules would create a mountain shape. This would be a direct calculation, except that the boundaries of the index could truncate either ... | 2 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
Easy, Binary Search Python3 || Beats 63% 🔥 | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | Easy, Intuitive and a bit mathematical solution with commented code.\nSince, we are using binary search, and constant extra space:\n**Time complexity =** $$O(logn)$$\n**Space complexity =** $$O(1)$$\n\n# Code\n```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n r = n - index -... | 2 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
Easy, Binary Search Python3 || Beats 63% 🔥 | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | Easy, Intuitive and a bit mathematical solution with commented code.\nSince, we are using binary search, and constant extra space:\n**Time complexity =** $$O(logn)$$\n**Space complexity =** $$O(1)$$\n\n# Code\n```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n r = n - index -... | 2 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
python 3 - binary search + greedy + triangle number | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | (This one is quite hard)\n\n# Intuition\nEvery array element has to be 1 in the beginning. At most you case raise nums[i] for (ms - 1), which is the upper boundary. Lower boundary = 0 for sure. -> This is a binary search pattern.\n\nYou will definitely try to raise nums[i] for value x. Value will not be randomly distri... | 1 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
python 3 - binary search + greedy + triangle number | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | (This one is quite hard)\n\n# Intuition\nEvery array element has to be 1 in the beginning. At most you case raise nums[i] for (ms - 1), which is the upper boundary. Lower boundary = 0 for sure. -> This is a binary search pattern.\n\nYou will definitely try to raise nums[i] for value x. Value will not be randomly distri... | 1 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
Python3 || Tc: O(logn), SC: O(1) || Binary Search || Commented code | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | ```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n low, high = 1, maxSum\n while low <= high: \n mid = (low + high) // 2\n if self.helper(mid, index, maxSum, n):\n ans = mid \n low = mid + 1\n else:\n ... | 1 | You are given three positive integers: `n`, `index`, and `maxSum`. You want to construct an array `nums` (**0-indexed**) that satisfies the following conditions:
* `nums.length == n`
* `nums[i]` is a **positive** integer where `0 <= i < n`.
* `abs(nums[i] - nums[i+1]) <= 1` where `0 <= i < n-1`.
* The sum of a... | Simulate the given in the statement Calculate those who will eat instead of those who will not. |
Python3 || Tc: O(logn), SC: O(1) || Binary Search || Commented code | maximum-value-at-a-given-index-in-a-bounded-array | 0 | 1 | ```\nclass Solution:\n def maxValue(self, n: int, index: int, maxSum: int) -> int:\n low, high = 1, maxSum\n while low <= high: \n mid = (low + high) // 2\n if self.helper(mid, index, maxSum, n):\n ans = mid \n low = mid + 1\n else:\n ... | 1 | Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` (**0-indexed**).
Specifically, `ans` is the **concatenation** of two `nums` arrays.
Return _the array_ `ans`.
**Example 1:**
**Input:** nums = \[1,2,1\... | What if the problem was instead determining if you could generate a valid array with nums[index] == target? To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum. n is... |
[Python3] trie | count-pairs-with-xor-in-a-range | 0 | 1 | \n```\nclass Trie: \n def __init__(self): \n self.root = {}\n \n def insert(self, val): \n node = self.root \n for i in reversed(range(15)):\n bit = (val >> i) & 1\n if bit not in node: node[bit] = {"cnt": 0}\n node = node[bit]\n node["cnt"] ... | 45 | Given a **(0-indexed)** integer array `nums` and two integers `low` and `high`, return _the number of **nice pairs**_.
A **nice pair** is a pair `(i, j)` where `0 <= i < j < nums.length` and `low <= (nums[i] XOR nums[j]) <= high`.
**Example 1:**
**Input:** nums = \[1,4,2,7\], low = 2, high = 6
**Output:** 6
**Explan... | Iterate on the customers, maintaining the time the chef will finish the previous orders. If that time is before the current arrival time, the chef starts immediately. Else, the current customer waits till the chef finishes, and then the chef starts. Update the running time by the time when the chef starts preparing + p... |
[Python] O(N log N) using math (Fourier transform) | count-pairs-with-xor-in-a-range | 0 | 1 | We can use some number theory to calculate all possible XOR values of two numbers in the range (0, max(nums)). Obviously this knowledge isn\'t expected in a normal interview, but it can come in handy.\n\nThe trick, which is used in cryptography/signal processing, is to convert our array into a frequency array, then use... | 16 | Given a **(0-indexed)** integer array `nums` and two integers `low` and `high`, return _the number of **nice pairs**_.
A **nice pair** is a pair `(i, j)` where `0 <= i < j < nums.length` and `low <= (nums[i] XOR nums[j]) <= high`.
**Example 1:**
**Input:** nums = \[1,4,2,7\], low = 2, high = 6
**Output:** 6
**Explan... | Iterate on the customers, maintaining the time the chef will finish the previous orders. If that time is before the current arrival time, the chef starts immediately. Else, the current customer waits till the chef finishes, and then the chef starts. Update the running time by the time when the chef starts preparing + p... |
[C++ || Python || Java] Brief code without explicit Trie | count-pairs-with-xor-in-a-range | 1 | 1 | > **I know almost nothing about English, pointing out the mistakes in my article would be much appreciated.**\n\n> **In addition, I\'m too weak, please be critical of my ideas.**\n---\n# Intuition\n1. For the counting pairs problem, the first intuition is to enumerate one of the pair, and count another. Such as [327. C... | 1 | Given a **(0-indexed)** integer array `nums` and two integers `low` and `high`, return _the number of **nice pairs**_.
A **nice pair** is a pair `(i, j)` where `0 <= i < j < nums.length` and `low <= (nums[i] XOR nums[j]) <= high`.
**Example 1:**
**Input:** nums = \[1,4,2,7\], low = 2, high = 6
**Output:** 6
**Explan... | Iterate on the customers, maintaining the time the chef will finish the previous orders. If that time is before the current arrival time, the chef starts immediately. Else, the current customer waits till the chef finishes, and then the chef starts. Update the running time by the time when the chef starts preparing + p... |
[Python3] Trie with diagram and explanations | count-pairs-with-xor-in-a-range | 0 | 1 | ## Idea\n1. We scan the numbers one by one and use a trie to store the counts of existing nums.\n2. Whenever we have a new number `v`, we first find a path in the trie that ensures the XOR between the new number and numbers in the trie to be less than `high`, then we add this number `v` into the trie and update the cou... | 1 | Given a **(0-indexed)** integer array `nums` and two integers `low` and `high`, return _the number of **nice pairs**_.
A **nice pair** is a pair `(i, j)` where `0 <= i < j < nums.length` and `low <= (nums[i] XOR nums[j]) <= high`.
**Example 1:**
**Input:** nums = \[1,4,2,7\], low = 2, high = 6
**Output:** 6
**Explan... | Iterate on the customers, maintaining the time the chef will finish the previous orders. If that time is before the current arrival time, the chef starts immediately. Else, the current customer waits till the chef finishes, and then the chef starts. Update the running time by the time when the chef starts preparing + p... |
Efficient Python Program Using Map and Regular Expression. | number-of-different-integers-in-a-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(set(map(int,re.findall(r\'\\d+\',word))))\n``` | 1 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Efficient Python Program Using Map and Regular Expression. | number-of-different-integers-in-a-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(set(map(int,re.findall(r\'\\d+\',word))))\n``` | 1 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Efficient Python3 Program Using Regular Expression Method | number-of-different-integers-in-a-string | 0 | 1 | # Intuition:\n\nThe goal is to find and count different integers within the given string, where non-digit characters are replaced with spaces.\n\n# Approach:\n\nImport the re module for regular expressions.\nUse re.findall to extract all the integers from the word string.\nConvert the extracted integers to a set to rem... | 1 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Efficient Python3 Program Using Regular Expression Method | number-of-different-integers-in-a-string | 0 | 1 | # Intuition:\n\nThe goal is to find and count different integers within the given string, where non-digit characters are replaced with spaces.\n\n# Approach:\n\nImport the re module for regular expressions.\nUse re.findall to extract all the integers from the word string.\nConvert the extracted integers to a set to rem... | 1 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Easy method | 3 line code | 80% beats | number-of-different-integers-in-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\n\n\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n ... | 0 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Easy method | 3 line code | 80% beats | number-of-different-integers-in-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\n\n\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n ... | 0 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Simplest solution for beginners in python beat 80% | number-of-different-integers-in-a-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n nums=\'0123456789\'\n\n for i in range(len(word)):\n if word[i] not in nums:\n word= word.replace(word[i],\' \')\n num_list=word.split()\n\n res=[]\n for i in range(len(... | 3 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Simplest solution for beginners in python beat 80% | number-of-different-integers-in-a-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n nums=\'0123456789\'\n\n for i in range(len(word)):\n if word[i] not in nums:\n word= word.replace(word[i],\' \')\n num_list=word.split()\n\n res=[]\n for i in range(len(... | 3 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Python One Line Efficient Solution. | number-of-different-integers-in-a-string | 0 | 1 | # Intuition\n The goal is to find and count different integers within the given string, where non-digit characters are replaced with spaces.\n\n# Approach\n 1. Use regular expressions to extract all the integers from the \'word\' string.\n 2. Convert the extracted integers to a set to remove duplicates.\n 3. Return the... | 1 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Python One Line Efficient Solution. | number-of-different-integers-in-a-string | 0 | 1 | # Intuition\n The goal is to find and count different integers within the given string, where non-digit characters are replaced with spaces.\n\n# Approach\n 1. Use regular expressions to extract all the integers from the \'word\' string.\n 2. Convert the extracted integers to a set to remove duplicates.\n 3. Return the... | 1 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Python | Easy Solution✅ | number-of-different-integers-in-a-string | 0 | 1 | # Code \u2705\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int: # // word = "a123bc34d8ef34"\n digit_index = 0\n digit_found = False\n number_list = []\n for i in range(len(word)):\n if word[i].isdigit() and not digit_found: # // condition will be true w... | 5 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Python | Easy Solution✅ | number-of-different-integers-in-a-string | 0 | 1 | # Code \u2705\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int: # // word = "a123bc34d8ef34"\n digit_index = 0\n digit_found = False\n number_list = []\n for i in range(len(word)):\n if word[i].isdigit() and not digit_found: # // condition will be true w... | 5 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
[Python3] Short and easy to understand | number-of-different-integers-in-a-string | 0 | 1 | We want to return only the part that matches the condition **(\'\\d+)\'** (matches one or more digits), then convert every element to **int** and the length of the **set(nums)** will be the answer.\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n word = re.findall(\'(\\d+)\', word)... | 9 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
[Python3] Short and easy to understand | number-of-different-integers-in-a-string | 0 | 1 | We want to return only the part that matches the condition **(\'\\d+)\'** (matches one or more digits), then convert every element to **int** and the length of the **set(nums)** will be the answer.\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n word = re.findall(\'(\\d+)\', word)... | 9 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Python set comprehension with itertools.groupby | number-of-different-integers-in-a-string | 0 | 1 | ```python []\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(\n set(\n int("".join(grp))\n for k, grp in groupby(word, key=lambda x: x.isdigit())\n if k\n )\n )\n\n``` | 3 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Python set comprehension with itertools.groupby | number-of-different-integers-in-a-string | 0 | 1 | ```python []\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(\n set(\n int("".join(grp))\n for k, grp in groupby(word, key=lambda x: x.isdigit())\n if k\n )\n )\n\n``` | 3 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
Very easy and straightfoward O(n) solution | number-of-different-integers-in-a-string | 0 | 1 | Runtime: 28 ms, faster than 91.62% of Python3 online submissions for Number of Different Integers in a String.\nMemory Usage: 14.1 MB, less than 95.04% of Python3 online submissions for Number of Different Integers in a String.\n\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n out ... | 4 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
Very easy and straightfoward O(n) solution | number-of-different-integers-in-a-string | 0 | 1 | Runtime: 28 ms, faster than 91.62% of Python3 online submissions for Number of Different Integers in a String.\nMemory Usage: 14.1 MB, less than 95.04% of Python3 online submissions for Number of Different Integers in a String.\n\n```\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n out ... | 4 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
One-line simple solution with regex | number-of-different-integers-in-a-string | 0 | 1 | # Code\n```python\nimport re\n\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(set(map(int, re.sub(r\'[a-z]\', \' \', word).split())))\n\n``` | 0 | You are given a string `word` that consists of digits and lowercase English letters.
You will replace every non-digit character with a space. For example, `"a123bc34d8ef34 "` will become `" 123 34 8 34 "`. Notice that you are left with some integers that are separated by at least one space: `"123 "`, `"34 "`, `"8 "`, ... | Choose k 1s and determine how many steps are required to move them into 1 group. Maintain a sliding window of k 1s, and maintain the steps required to group them. When you slide the window across, should you move the group to the right? Once you move the group to the right, it will never need to slide to the left again... |
One-line simple solution with regex | number-of-different-integers-in-a-string | 0 | 1 | # Code\n```python\nimport re\n\nclass Solution:\n def numDifferentIntegers(self, word: str) -> int:\n return len(set(map(int, re.sub(r\'[a-z]\', \' \', word).split())))\n\n``` | 0 | A **value-equal** string is a string where **all** characters are the same.
* For example, `"1111 "` and `"33 "` are value-equal strings.
* In contrast, `"123 "` is not a value-equal string.
Given a digit string `s`, decompose the string into some number of **consecutive value-equal** substrings where **exactly o... | Try to split the string so that each integer is in a different string. Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique. |
[Python3] simulation | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | \n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n ans = 0\n perm = list(range(n))\n while True: \n ans += 1\n perm = [perm[n//2+(i-1)//2] if i&1 else perm[i//2] for i in range(n)]\n if all(perm[i] == i for i in range(n)): return ans\n``... | 4 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
[Python3] simulation | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | \n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n ans = 0\n perm = list(range(n))\n while True: \n ans += 1\n perm = [perm[n//2+(i-1)//2] if i&1 else perm[i//2] for i in range(n)]\n if all(perm[i] == i for i in range(n)): return ans\n``... | 4 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
Very Easy Solution | Python3 | Simple To Understand | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Approach\n**Just Following what the problem description is saying thats it.**\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def reinitializePermutation(self,... | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
Very Easy Solution | Python3 | Simple To Understand | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Approach\n**Just Following what the problem description is saying thats it.**\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def reinitializePermutation(self,... | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
Easy to understand Python3 solution | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
Easy to understand Python3 solution | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
[Python3] Brutal Force | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n orig = [i for i in range(n)]\n perm = [i for i in range(n)]\n arr = [0]*n\n \n cnt = 1\n for i in range(len(arr)):\n if i%2==0:\n arr[i] = perm[i//2]\n els... | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
[Python3] Brutal Force | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n orig = [i for i in range(n)]\n perm = [i for i in range(n)]\n arr = [0]*n\n \n cnt = 1\n for i in range(len(arr)):\n if i%2==0:\n arr[i] = perm[i//2]\n els... | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
Python 3 - 4 liner Beats 99% | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Intuition\nFind the minimum $m$ such that $2^m - 1 \\text{ mod } n-1 = 0.$\n\n# Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n if n == 2: return 1\n m = 1\n while (2 ** m) % (n - 1) != 1: m += 1\n return m\n``` | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
Python 3 - 4 liner Beats 99% | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Intuition\nFind the minimum $m$ such that $2^m - 1 \\text{ mod } n-1 = 0.$\n\n# Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n if n == 2: return 1\n m = 1\n while (2 ** m) % (n - 1) != 1: m += 1\n return m\n``` | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
take care of perm = arr.copy(), you can try change it to perm = arr :P | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | \n# Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n goal = list(range(n))\n perm = list(range(n))\n arr = list(range(n))\n \n step = 0\n while True:\n for i in range(len(perm)):\n if i % 2==0:\n ar... | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
take care of perm = arr.copy(), you can try change it to perm = arr :P | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | \n# Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n goal = list(range(n))\n perm = list(range(n))\n arr = list(range(n))\n \n step = 0\n while True:\n for i in range(len(perm)):\n if i % 2==0:\n ar... | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
minimum-number-of-operations-to-reinitialize-a-permutation | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n b = 0\n l = []\n p = []\n for i in range(n):\n p.append(i)\n l.append(i)\n while True:\n for i in range(len(p)):\n if p[i]%2==0:\n a... | 0 | You are given an **even** integer `n`. You initially have a permutation `perm` of size `n` where `perm[i] == i` **(0-indexed)**.
In one operation, you will create a new array `arr`, and for each `i`:
* If `i % 2 == 0`, then `arr[i] = perm[i / 2]`.
* If `i % 2 == 1`, then `arr[i] = perm[n / 2 + (i - 1... | Simulate the tournament as given in the statement. Be careful when handling odd integers. |
minimum-number-of-operations-to-reinitialize-a-permutation | minimum-number-of-operations-to-reinitialize-a-permutation | 0 | 1 | # Code\n```\nclass Solution:\n def reinitializePermutation(self, n: int) -> int:\n b = 0\n l = []\n p = []\n for i in range(n):\n p.append(i)\n l.append(i)\n while True:\n for i in range(len(p)):\n if p[i]%2==0:\n a... | 0 | There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string `text` of words separated by a single space (no leading or trailing spaces) and a string `brokenLetters` of all **distinct** letter keys that are broken, return _the **number of words** i... | It is safe to assume the number of operations isn't more than n The number is small enough to apply a brute force solution. |
Python Easy to Understand | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def evaluate(self, s: str, knowledge: List[List[str]]) -> str:\n d={}\n for i in knowledge:\n d[i[0]]=i[1]\n temp=\'\'\n res=\'\'\n flag=0\n for i in s:\n if i==\'(\':\n flag=1\n elif i!=\')\'... | 1 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
Python: Regular Expression Solution | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | Runtime: 2292 ms, beats 56.60%\nMemory: 52.4 MB, beats 94.04%\n\n```python\nimport re\n\nclass Solution:\n def evaluate(self, s: str, knowledge: list[list[str]]) -> str:\n\t # Transform the list into a dictonary for quick lookups.\n knowledge = {k[0]: k[1] for k in knowledge}\n\t\t\n\t\t# Lookup function t... | 1 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
✅ Python Simple Logic ✅ | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | # Code\n```\nclass Solution:\n def evaluate(self, s: str, knowledge: List[List[str]]) -> str:\n d = {key:sol for key, sol in knowledge}\n ans = add = \'\'\n\n for char in s:\n if char == \'(\':\n ans += add\n add = \'\'\n elif char == \')\':\n ... | 1 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
Concise Two Pointers with HashMap | Beats 100% | Rust & Python | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)\\ extra\\ memory$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Rust Code\n```rust\nuse std::collections::HashMap;\nimpl Solution {\n pub fn evaluate(s: String, k: Vec<Vec<St... | 0 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
Straight Forward T:O(N)|S:O(N) Approach in Python | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | # Intuition\nSince each opening "(" must be closed by a ")", we can solve in one pass.\n\n# Approach\nConvert 2D list of key value pairs into a hashmap for every lookup. Go through the string and check for opening "(". If found continue untill you see a ")". Note here we are not technically doing an expensive double wh... | 0 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
Python simple-solution | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | # Intuition\nUsing the flag to check whether the subsentence should be substituded or not. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution:\n def evaluate(self, s: str, knowledge: List[List[s... | 0 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
python3 parsing | evaluate-the-bracket-pairs-of-a-string | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def evaluate(self, s: str, knowledge: List[List[str]]) -> str:\n d = defaultdict(lambda : \'?\')\n res = \'\'\n for x,y in knowledge:\n d[x] = y\n i = 0\n while i < len(s):\n if s[i] != \'(\':\n res += s[i]\n ... | 0 | You are given a string `s` that contains some bracket pairs, with each pair containing a **non-empty** key.
* For example, in the string `"(name)is(age)yearsold "`, there are **two** bracket pairs that contain the keys `"name "` and `"age "`.
You know the values of a wide range of keys. This is represented by a 2D ... | Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit. If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input. Thus the answer is equal to the max digit. |
Solution | maximize-number-of-nice-divisors | 0 | 1 | \n# Code\n```\nclass Solution:\n def maxNiceDivisors(self, primeFactors: int) -> int:\n if primeFactors <= 3:\n return primeFactors\n \n MOD = int(1e9 + 7)\n if primeFactors % 3 == 0:\n power = primeFactors // 3\n return self.calculateNiceDivisors(3, power... | 0 | You are given a positive integer `primeFactors`. You are asked to construct a positive integer `n` that satisfies the following conditions:
* The number of prime factors of `n` (not necessarily distinct) is **at most** `primeFactors`.
* The number of nice divisors of `n` is maximized. Note that a divisor of `n` is... | The constraints are small enough for an N^2 solution. Try using dynamic programming. |
Solution | maximize-number-of-nice-divisors | 0 | 1 | \n# Code\n```\nclass Solution:\n def maxNiceDivisors(self, primeFactors: int) -> int:\n if primeFactors <= 3:\n return primeFactors\n \n MOD = int(1e9 + 7)\n if primeFactors % 3 == 0:\n power = primeFactors // 3\n return self.calculateNiceDivisors(3, power... | 0 | You are given a **strictly increasing** integer array `rungs` that represents the **height** of rungs on a ladder. You are currently on the **floor** at height `0`, and you want to reach the last rung.
You are also given an integer `dist`. You can only climb to the next highest rung if the distance between where you a... | The number of nice divisors is equal to the product of the count of each prime factor. Then the problem is reduced to: given n, find a sequence of numbers whose sum equals n and whose product is maximized. This sequence can have no numbers that are larger than 4. Proof: if it contains a number x that is larger than 4, ... |
Python3 solution exPlained | maximize-number-of-nice-divisors | 0 | 1 | # Intuition\nThe goal of this problem is to find the maximum number of nice divisors of a given number, where a nice divisor is a number that is a product of only prime numbers. My first thought is to use a combination of prime factorization and dynamic programming to find the maximum number of nice divisors for a give... | 0 | You are given a positive integer `primeFactors`. You are asked to construct a positive integer `n` that satisfies the following conditions:
* The number of prime factors of `n` (not necessarily distinct) is **at most** `primeFactors`.
* The number of nice divisors of `n` is maximized. Note that a divisor of `n` is... | The constraints are small enough for an N^2 solution. Try using dynamic programming. |
Python3 solution exPlained | maximize-number-of-nice-divisors | 0 | 1 | # Intuition\nThe goal of this problem is to find the maximum number of nice divisors of a given number, where a nice divisor is a number that is a product of only prime numbers. My first thought is to use a combination of prime factorization and dynamic programming to find the maximum number of nice divisors for a give... | 0 | You are given a **strictly increasing** integer array `rungs` that represents the **height** of rungs on a ladder. You are currently on the **floor** at height `0`, and you want to reach the last rung.
You are also given an integer `dist`. You can only climb to the next highest rung if the distance between where you a... | The number of nice divisors is equal to the product of the count of each prime factor. Then the problem is reduced to: given n, find a sequence of numbers whose sum equals n and whose product is maximized. This sequence can have no numbers that are larger than 4. Proof: if it contains a number x that is larger than 4, ... |
Python 3 Solution | Better than 100% at runtime | Better than 91.43% at memory | maximize-number-of-nice-divisors | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxNiceDivisors(self, primeFactors: int) -> int:\n if primeFactors == 1:\n return 1\n Mod = 1_000_000_007\n num2 = -primeFactors%3\... | 0 | You are given a positive integer `primeFactors`. You are asked to construct a positive integer `n` that satisfies the following conditions:
* The number of prime factors of `n` (not necessarily distinct) is **at most** `primeFactors`.
* The number of nice divisors of `n` is maximized. Note that a divisor of `n` is... | The constraints are small enough for an N^2 solution. Try using dynamic programming. |
Python 3 Solution | Better than 100% at runtime | Better than 91.43% at memory | maximize-number-of-nice-divisors | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def maxNiceDivisors(self, primeFactors: int) -> int:\n if primeFactors == 1:\n return 1\n Mod = 1_000_000_007\n num2 = -primeFactors%3\... | 0 | You are given a **strictly increasing** integer array `rungs` that represents the **height** of rungs on a ladder. You are currently on the **floor** at height `0`, and you want to reach the last rung.
You are also given an integer `dist`. You can only climb to the next highest rung if the distance between where you a... | The number of nice divisors is equal to the product of the count of each prime factor. Then the problem is reduced to: given n, find a sequence of numbers whose sum equals n and whose product is maximized. This sequence can have no numbers that are larger than 4. Proof: if it contains a number x that is larger than 4, ... |
Solution | maximize-number-of-nice-divisors | 1 | 1 | ```C++ []\nclass Solution {\n public:\n int maxNiceDivisors(int primeFactors) {\n if (primeFactors <= 3)\n return primeFactors;\n if (primeFactors % 3 == 0)\n return myPow(3, primeFactors / 3) % kMod;\n if (primeFactors % 3 == 1)\n return (4 * myPow(3, (primeFactors - 4) / 3)) % kMod;\n retu... | 0 | You are given a positive integer `primeFactors`. You are asked to construct a positive integer `n` that satisfies the following conditions:
* The number of prime factors of `n` (not necessarily distinct) is **at most** `primeFactors`.
* The number of nice divisors of `n` is maximized. Note that a divisor of `n` is... | The constraints are small enough for an N^2 solution. Try using dynamic programming. |
Solution | maximize-number-of-nice-divisors | 1 | 1 | ```C++ []\nclass Solution {\n public:\n int maxNiceDivisors(int primeFactors) {\n if (primeFactors <= 3)\n return primeFactors;\n if (primeFactors % 3 == 0)\n return myPow(3, primeFactors / 3) % kMod;\n if (primeFactors % 3 == 1)\n return (4 * myPow(3, (primeFactors - 4) / 3)) % kMod;\n retu... | 0 | You are given a **strictly increasing** integer array `rungs` that represents the **height** of rungs on a ladder. You are currently on the **floor** at height `0`, and you want to reach the last rung.
You are also given an integer `dist`. You can only climb to the next highest rung if the distance between where you a... | The number of nice divisors is equal to the product of the count of each prime factor. Then the problem is reduced to: given n, find a sequence of numbers whose sum equals n and whose product is maximized. This sequence can have no numbers that are larger than 4. Proof: if it contains a number x that is larger than 4, ... |
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