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, one pass, no additional memory | slowest-key | 0 | 1 | ```\nclass Solution:\n def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:\n max_dur = releaseTimes[0]\n max_key = keysPressed[0]\n \n for i in range(1, len(releaseTimes)):\n if releaseTimes[i] - releaseTimes[i-1] > max_dur:\n max_dur = releas... | 9 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
Python, one pass, no additional memory | slowest-key | 0 | 1 | ```\nclass Solution:\n def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:\n max_dur = releaseTimes[0]\n max_key = keysPressed[0]\n \n for i in range(1, len(releaseTimes)):\n if releaseTimes[i] - releaseTimes[i-1] > max_dur:\n max_dur = releas... | 9 | You are given an array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
use dictionary | slowest-key | 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 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
use dictionary | slowest-key | 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 array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
My Solution :) BEATS 79.66 % RUNTIME | slowest-key | 0 | 1 | # Intuition\n1. Initialize a dictionary prev to keep track of the previous release time for each key. The first entry in the dictionary is initialized with the first key and its corresponding release time from keysPressed and releaseTimes.\n\n2. Initialize a variable maximumTime with the first release time from release... | 0 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
My Solution :) BEATS 79.66 % RUNTIME | slowest-key | 0 | 1 | # Intuition\n1. Initialize a dictionary prev to keep track of the previous release time for each key. The first entry in the dictionary is initialized with the first key and its corresponding release time from keysPressed and releaseTimes.\n\n2. Initialize a variable maximumTime with the first release time from release... | 0 | You are given an array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
Python Simple Soluiton - Hashmap! | slowest-key | 0 | 1 | \n# Code\n```\nclass Solution:\n def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:\n\n hashmap: dict = { keysPressed[0]:releaseTimes[0] }\n for index in range(1, len(releaseTimes)):\n \n difference: int = releaseTimes[index] - releaseTimes[index - 1]\n ... | 0 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
Python Simple Soluiton - Hashmap! | slowest-key | 0 | 1 | \n# Code\n```\nclass Solution:\n def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:\n\n hashmap: dict = { keysPressed[0]:releaseTimes[0] }\n for index in range(1, len(releaseTimes)):\n \n difference: int = releaseTimes[index] - releaseTimes[index - 1]\n ... | 0 | You are given an array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
python beats 93% | slowest-key | 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 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
python beats 93% | slowest-key | 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 array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
Easy python Solution. Beats 90%. Dictionary. | slowest-key | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n## *Use a python dictionary to calculate the hold time for each key.*\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Store the largest the hold time for each key using a dictionary.\n2. Find the key that was pres... | 0 | A newly designed keypad was tested, where a tester pressed a sequence of `n` keys, one at a time.
You are given a string `keysPressed` of length `n`, where `keysPressed[i]` was the `ith` key pressed in the testing sequence, and a sorted list `releaseTimes`, where `releaseTimes[i]` was the time the `ith` key was releas... | We want to make the smaller digits the most significant digits in the number. For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly. |
Easy python Solution. Beats 90%. Dictionary. | slowest-key | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n## *Use a python dictionary to calculate the hold time for each key.*\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Store the largest the hold time for each key using a dictionary.\n2. Find the key that was pres... | 0 | You are given an array of `events` where `events[i] = [startDayi, endDayi, valuei]`. The `ith` event starts at `startDayi` and ends at `endDayi`, and if you attend this event, you will receive a value of `valuei`. You are also given an integer `k` which represents the maximum number of events you can attend.
You can o... | Get for each press its key and amount of time taken. Iterate on the presses, maintaining the answer so far. The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer. |
1️⃣-liner_brute_force.py😁 | arithmetic-subarrays | 0 | 1 | # Code\n```js\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n return[all(arr[i+1]-arr[i]==arr[1]-arr[0]for i in range(len(arr)-1))for arr in[sorted(nums[a:b+1])for a,b in zip(l,r)]] \n```\n\n is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
1️⃣-liner_brute_force.py😁 | arithmetic-subarrays | 0 | 1 | # Code\n```js\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n return[all(arr[i+1]-arr[i]==arr[1]-arr[0]for i in range(len(arr)-1))for arr in[sorted(nums[a:b+1])for a,b in zip(l,r)]] \n```\n\n_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | arithmetic-subarrays | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(With Sorting)***\n1. **check function:** Sorts the given array and checks if the elements form an arithmetic sequence by verifying if the difference between consecutive elements remains constant.\n1. **checkAri... | 29 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | arithmetic-subarrays | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(With Sorting)***\n1. **check function:** Sorts the given array and checks if the elements form an arithmetic sequence by verifying if the difference between consecutive elements remains constant.\n1. **checkAri... | 29 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
📌📌C++/JAVA/Python3- Most intuitive, concise, and interview friendly code for beginners | arithmetic-subarrays | 1 | 1 | # **Intuition**\n\n* For each query, bring out the required vector to check.\n* sort the array for easy AP check\n* If the size is less than two, then it is surely following AP.\n* else we\'ll check for all the values. If at any place it does not follow the progression, then it is not an AP following the sequence.\n\n-... | 4 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
📌📌C++/JAVA/Python3- Most intuitive, concise, and interview friendly code for beginners | arithmetic-subarrays | 1 | 1 | # **Intuition**\n\n* For each query, bring out the required vector to check.\n* sort the array for easy AP check\n* If the size is less than two, then it is surely following AP.\n* else we\'ll check for all the values. If at any place it does not follow the progression, then it is not an AP following the sequence.\n\n-... | 4 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Simple solution for noobs brute-force | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Simple solution for noobs brute-force | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Easy python sorting array | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Easy python sorting array | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Disgusting one-liner (Python) | arithmetic-subarrays | 0 | 1 | Quite possibly the worst code I\'ve ever written.\nI wanted to see if it was possible.\n\n# Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n return [all([k-(nums2[1]-nums2[0])==nums2[index] for index,k in enumerate(nums2[1:])]) for e... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Disgusting one-liner (Python) | arithmetic-subarrays | 0 | 1 | Quite possibly the worst code I\'ve ever written.\nI wanted to see if it was possible.\n\n# Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n return [all([k-(nums2[1]-nums2[0])==nums2[index] for index,k in enumerate(nums2[1:])]) for e... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
SImple python solution using sorting | arithmetic-subarrays | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Check if each subarry can be rearranged into a arithmetic sequence.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. #### Extract the subarray \n2. #### Check Arithmetic sequence \n3. #### Store the result\... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
SImple python solution using sorting | arithmetic-subarrays | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Check if each subarry can be rearranged into a arithmetic sequence.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. #### Extract the subarray \n2. #### Check Arithmetic sequence \n3. #### Store the result\... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Set Approach Runtime 165ms | arithmetic-subarrays | 0 | 1 | # Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n n=len(l)\n ans=[True for i in range(n)]\n for i in range(n):\n lf,rg=l[i],r[i]\n t=nums[lf:rg+1]\n mini,maxi,s=min(t),max(t),set(t)\n ... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Set Approach Runtime 165ms | arithmetic-subarrays | 0 | 1 | # Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n n=len(l)\n ans=[True for i in range(n)]\n for i in range(n):\n lf,rg=l[i],r[i]\n t=nums[lf:rg+1]\n mini,maxi,s=min(t),max(t),set(t)\n ... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Optimized solution O(MlogN) — SegmentTree + Cumsum + Cumsum of squares | arithmetic-subarrays | 0 | 1 | # Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. In the original solution, the author offered to find minimal and maximal elements just passing through the array for $$O(N)$$. We can optimize the complexity to $$O(logN)$$ using two segment trees.\n2. To check if the subarray is the ar... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Optimized solution O(MlogN) — SegmentTree + Cumsum + Cumsum of squares | arithmetic-subarrays | 0 | 1 | # Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. In the original solution, the author offered to find minimal and maximal elements just passing through the array for $$O(N)$$. We can optimize the complexity to $$O(logN)$$ using two segment trees.\n2. To check if the subarray is the ar... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Simple python solution with explanation | arithmetic-subarrays | 0 | 1 | \n# Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n #variable to store the result\n n=[]\n for i in range(len(l)):\n #taking the subarray which is need to be checked and sorted\n a=nums[l[i]:r[i]+... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Simple python solution with explanation | arithmetic-subarrays | 0 | 1 | \n# Code\n```\nclass Solution:\n def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:\n #variable to store the result\n n=[]\n for i in range(len(l)):\n #taking the subarray which is need to be checked and sorted\n a=nums[l[i]:r[i]+... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Easy Solutions in python code 👍🙏 | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Easy Solutions in python code 👍🙏 | arithmetic-subarrays | 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. $$O(n)$$ --... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Python3 Solution | arithmetic-subarrays | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirstly, we realize we have to check whether the subarray is arithmetic or not. For this condition to be true, the differences between consecutive elements must be the same. \n# Approach\n<!-- Describe your approach to solving the problem... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Python3 Solution | arithmetic-subarrays | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirstly, we realize we have to check whether the subarray is arithmetic or not. For this condition to be true, the differences between consecutive elements must be the same. \n# Approach\n<!-- Describe your approach to solving the problem... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
Simple Brute-Force + Sorting approach | arithmetic-subarrays | 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* Since \'n\' ranges from [2, 500], we can apply a brute-force approach. \n* Here we created a query list store all the queries.(to improve the readability) \n* check ... | 1 | A sequence of numbers is called **arithmetic** if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence `s` is arithmetic if and only if `s[i+1] - s[i] == s[1] - s[0]` for all valid `i`.
For example, these are **arithmetic** sequences:
1... | If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same. If the range (high - low + 1) is odd, the solution will depend on the parity of high and low. |
Simple Brute-Force + Sorting approach | arithmetic-subarrays | 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* Since \'n\' ranges from [2, 500], we can apply a brute-force approach. \n* Here we created a query list store all the queries.(to improve the readability) \n* check ... | 1 | Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
There may be **duplicates** in the original array.
**Note:** An array `A` rotated by `x` positions results in an array `B` of the sa... | To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same. If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arith... |
【Video】Simple Solution with Min Heap and BFS - Python, JavaScript, Java and C++ | path-with-minimum-effort | 1 | 1 | # Intuition\r\nUsing min heap and BFS.\r\n\r\n---\r\n\r\n# Solution Video\r\n\r\nhttps://youtu.be/yQtfAG1wI48\r\n\r\nIn the video, the steps of approach below are visualized using diagrams and drawings. I\'m sure you understand the solution easily!\r\n\r\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my ch... | 8 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
【Video】Simple Solution with Min Heap and BFS - Python, JavaScript, Java and C++ | path-with-minimum-effort | 1 | 1 | # Intuition\r\nUsing min heap and BFS.\r\n\r\n---\r\n\r\n# Solution Video\r\n\r\nhttps://youtu.be/yQtfAG1wI48\r\n\r\nIn the video, the steps of approach below are visualized using diagrams and drawings. I\'m sure you understand the solution easily!\r\n\r\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my ch... | 8 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
Rust/Python BFS with explanation | path-with-minimum-effort | 0 | 1 | # Intuition\nThere are at least two possible solutions here. One involves dijkstra where you maintain the min-heap of positions with smallest effor seen so far. But you can also adjust a standard BFS to solve the problem.\n\nTo do this, create a frontier which starts with `(0, 0)` and maintain your current knowledge of... | 1 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
Rust/Python BFS with explanation | path-with-minimum-effort | 0 | 1 | # Intuition\nThere are at least two possible solutions here. One involves dijkstra where you maintain the min-heap of positions with smallest effor seen so far. But you can also adjust a standard BFS to solve the problem.\n\nTo do this, create a frontier which starts with `(0, 0)` and maintain your current knowledge of... | 1 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
✅ 97.67% Optimal Dijkstra with Heap | path-with-minimum-effort | 1 | 1 | # Comprehensive Guide to Solving "Path With Minimum Effort": Conquering the Terrain Efficiently\r\n\r\n## Introduction & Problem Statement\r\n\r\nGreetings, algorithmic adventurers! Today\'s quest leads us through a matrix of heights, representing terrains that a hiker needs to cross. The task is to find the path that ... | 95 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
✅ 97.67% Optimal Dijkstra with Heap | path-with-minimum-effort | 1 | 1 | # Comprehensive Guide to Solving "Path With Minimum Effort": Conquering the Terrain Efficiently\r\n\r\n## Introduction & Problem Statement\r\n\r\nGreetings, algorithmic adventurers! Today\'s quest leads us through a matrix of heights, representing terrains that a hiker needs to cross. The task is to find the path that ... | 95 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
Python3 Solution | path-with-minimum-effort | 0 | 1 | \r\n```\r\nclass Solution:\r\n def minimumEffortPath(self, heights: List[List[int]]) -> int:\r\n m=len(heights)\r\n n=len(heights[0])\r\n queue={(0,0):0}\r\n seen={(0,0):0}\r\n ans=inf\r\n while queue:\r\n newqueue={}\r\n for (i,j),h in queue.items():\r... | 1 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
Python3 Solution | path-with-minimum-effort | 0 | 1 | \r\n```\r\nclass Solution:\r\n def minimumEffortPath(self, heights: List[List[int]]) -> int:\r\n m=len(heights)\r\n n=len(heights[0])\r\n queue={(0,0):0}\r\n seen={(0,0):0}\r\n ans=inf\r\n while queue:\r\n newqueue={}\r\n for (i,j),h in queue.items():\r... | 1 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
🚀 99.92% || Dijkstra's Algorithm || Binary Search || Commented Code 🚀 | path-with-minimum-effort | 1 | 1 | # Problem Description\r\nYou are a hiker preparing for an upcoming hike and need to plan your route efficiently. Given a 2D array of heights representing the elevation of each cell, your objective is to find the path from the **top-left** cell to the **bottom-right** cell with the **minimum** effort for the whole path.... | 59 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
🚀 99.92% || Dijkstra's Algorithm || Binary Search || Commented Code 🚀 | path-with-minimum-effort | 1 | 1 | # Problem Description\r\nYou are a hiker preparing for an upcoming hike and need to plan your route efficiently. Given a 2D array of heights representing the elevation of each cell, your objective is to find the path from the **top-left** cell to the **bottom-right** cell with the **minimum** effort for the whole path.... | 59 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
Thought Process from Naive BFS to Dijkstra | path-with-minimum-effort | 0 | 1 | **TLDR**: Naive BFS doesn\'t work because you only visit the node once, and therefore you **only consider one of many paths.** Secondly, a naive BFS will give you the path with minimal nodes visited and doesn\'t care about the `effort`. In this problem, though, we can clearly see that `optimal path != path with minimal... | 285 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
Thought Process from Naive BFS to Dijkstra | path-with-minimum-effort | 0 | 1 | **TLDR**: Naive BFS doesn\'t work because you only visit the node once, and therefore you **only consider one of many paths.** Secondly, a naive BFS will give you the path with minimal nodes visited and doesn\'t care about the `effort`. In this problem, though, we can clearly see that `optimal path != path with minimal... | 285 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
12 Line Python | path-with-minimum-effort | 0 | 1 | ```python\nclass Solution:\n def minimumEffortPath(self, heights: List[List[int]]) -> int:\n d = [[float(\'inf\')] * len(heights[0]) for _ in range(len(heights))]\n d[0][0] = 0\n q = [(0, 0, 0)]\n while q:\n e, x, y = heappop(q)\n if (x, y) == (len(heights) - 1, len(heights[0]) - 1): return e\n ... | 0 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
12 Line Python | path-with-minimum-effort | 0 | 1 | ```python\nclass Solution:\n def minimumEffortPath(self, heights: List[List[int]]) -> int:\n d = [[float(\'inf\')] * len(heights[0]) for _ in range(len(heights))]\n d[0][0] = 0\n q = [(0, 0, 0)]\n while q:\n e, x, y = heappop(q)\n if (x, y) == (len(heights) - 1, len(heights[0]) - 1): return e\n ... | 0 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
✔Graph || Binary Search 📈✨|| Medium-->Easy🏆 || Easy to understand 😎|| #Beginner | path-with-minimum-effort | 1 | 1 | # Intuition\r\n- The problem requires finding the minimum effort path from the top-left corner of a 2D grid to the bottom-right corner. \r\n- The effort of a path is defined as the maximum absolute difference in heights between adjacent cells along the path. We need to minimize this maximum effort.\r\n\r\n# Approach\r\... | 16 | You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**... | Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ? if the current accu sum is odd, we care only about previous even accu sums and vice versa. |
✔Graph || Binary Search 📈✨|| Medium-->Easy🏆 || Easy to understand 😎|| #Beginner | path-with-minimum-effort | 1 | 1 | # Intuition\r\n- The problem requires finding the minimum effort path from the top-left corner of a 2D grid to the bottom-right corner. \r\n- The effort of a path is defined as the maximum absolute difference in heights between adjacent cells along the path. We need to minimize this maximum effort.\r\n\r\n# Approach\r\... | 16 | You are playing a solitaire game with **three piles** of stones of sizes `a`, `b`, and `c` respectively. Each turn you choose two **different non-empty** piles, take one stone from each, and add `1` point to your score. The game stops when there are **fewer than two non-empty** piles (meaning there ar... | Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells. If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost. Binary search the k value. |
Sorting - GroupBy - DSU | rank-transform-of-a-matrix | 0 | 1 | # Complexity\n```haskell\nTime complexity: O(nlogn + n * A(n))\nSpace complexity: O(n)\n```\n# Code\n```\nclass Solution:\n def matrixRankTransform(self, M: List[List[int]]) -> List[List[int]]:\n mr = [0 for _ in M]\n mc = [0 for _ in range(len(M[0]))]\n ii = sorted([(i, j) for j in range(len(M[... | 0 | Given an `m x n` `matrix`, return _a new matrix_ `answer` _where_ `answer[row][col]` _is the_ _**rank** of_ `matrix[row][col]`.
The **rank** is an **integer** that represents how large an element is compared to other elements. It is calculated using the following rules:
* The rank is an integer starting from `1`.
*... | Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index. |
Rank Transform of a Matrix | rank-transform-of-a-matrix | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe idea behind this algorithm is to iterate through the elements of the matrix in ascending order, maintaining a data structure (DSU - Disjoint Set Union) to keep track of the connected components based on the values. We also keep track ... | 0 | Given an `m x n` `matrix`, return _a new matrix_ `answer` _where_ `answer[row][col]` _is the_ _**rank** of_ `matrix[row][col]`.
The **rank** is an **integer** that represents how large an element is compared to other elements. It is calculated using the following rules:
* The rank is an integer starting from `1`.
*... | Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index. |
Python3 || DSU Algorithm | rank-transform-of-a-matrix | 0 | 1 | \n# Code\n```\nclass Solution:\n def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:\n\n m = len(matrix)\n n = len(matrix[0])\n\n # Iterate over all elements and create a map with that element along with its parent array to get connected components for all equal values\n\n... | 0 | Given an `m x n` `matrix`, return _a new matrix_ `answer` _where_ `answer[row][col]` _is the_ _**rank** of_ `matrix[row][col]`.
The **rank** is an **integer** that represents how large an element is compared to other elements. It is calculated using the following rules:
* The rank is an integer starting from `1`.
*... | Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index. |
[C++ || Python || Java] Union find with path compression | rank-transform-of-a-matrix | 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> **Vote welcome if this solution helped.**\n---\n\n# Intuition\n- TBD\n\n# Approach\n- TBD\n\n# Complexity\n- Time complexity: $O(mn\\log... | 0 | Given an `m x n` `matrix`, return _a new matrix_ `answer` _where_ `answer[row][col]` _is the_ _**rank** of_ `matrix[row][col]`.
The **rank** is an **integer** that represents how large an element is compared to other elements. It is calculated using the following rules:
* The rank is an integer starting from `1`.
*... | Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index. |
[Python3] UF | rank-transform-of-a-matrix | 0 | 1 | Learned this approach from @lee215 in this [post](https://leetcode.com/problems/rank-transform-of-a-matrix/discuss/909142/Python-Union-Find). \n```\nclass Solution:\n def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:\n m, n = len(matrix), len(matrix[0]) # dimension \n # mapping... | 5 | Given an `m x n` `matrix`, return _a new matrix_ `answer` _where_ `answer[row][col]` _is the_ _**rank** of_ `matrix[row][col]`.
The **rank** is an **integer** that represents how large an element is compared to other elements. It is calculated using the following rules:
* The rank is an integer starting from `1`.
*... | Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index. |
heap | sort-array-by-increasing-frequency | 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. $$O(n)$$ --... | 3 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Python keep it simple | sort-array-by-increasing-frequency | 0 | 1 | # Complexity\n- Time complexity:$$O(nlogn)$$\n- Space complexity:$$O(n)$$\n\n# Code\n```\nclass Solution:\n def frequencySort(self, nums: List[int]) -> List[int]:\n counts_dict = Counter(nums)\n return sorted(nums, key = lambda x: (counts_dict[x], -x))\n``` | 1 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Python 3 solution - with process of thinking and improvement | sort-array-by-increasing-frequency | 0 | 1 | # Part 1\n\nSunday morning, Spotify, coffee, console... Task from the list of tasks to solve later.\nLet\'s go!\n\n```python\n~ \u276F python3 at 10:38:03\nPython 3.9.1 (default, Feb 3 2021, 07:38:02)\n[... | 261 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Python || 94.84% Faster || Max Heap | sort-array-by-increasing-frequency | 0 | 1 | ```\nclass Solution:\n def frequencySort(self, nums: List[int]) -> List[int]:\n c=Counter(nums)\n pq=[]\n for item,freq in c.items():\n heapq.heappush(pq,(-freq,item)) # \'-\' because we are using max heap\n result=[]\n while pq:\n freq, item = heapq.heappop(p... | 1 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Python Easy Solution || Sorting || Hashmap | sort-array-by-increasing-frequency | 0 | 1 | # Code\n```\nclass Solution:\n def sort(self,nums: list)-> list:\n i=0\n while(i<len(nums)):\n swapped=0\n j=0\n while(j<len(nums)-i-1):\n if nums[j]>nums[j+1]:\n \n nums[j+1],nums[j]=nums[j],nums[j+1]\n ... | 1 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
2 lines solution very easy | sort-array-by-increasing-frequency | 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. $$O(n)$$ --... | 2 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Easiest one line solution in python | sort-array-by-increasing-frequency | 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. $$O(n)$$ --... | 11 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Easiest & Shortest readable HashMap/HashTable Solution. | sort-array-by-increasing-frequency | 0 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Consider example:\n - `nums = [3, 4, 2, 1, 2, 1]` \n- First dictionary will store the frequency of the numbers\n - `firstDic = {2: 2, 1: 2, 3: 1, 4: 1}`\n- Second dictionary will store all the values with same frequency, meaning if two numbe... | 1 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
Easy Python Solution with Explanation | sort-array-by-increasing-frequency | 0 | 1 | \tfrom collections import Counter\n\tclass Solution:\n\t\tdef frequencySort(self, nums: List[int]) -> List[int]:\n\t\t\td = Counter(nums)\n\t\t\tdef check(x):\n\t\t\t\treturn d[x]\n\n\t\t\tnums.sort(reverse=True)\n\t\t\tnums.sort(key=check)\n\t\t\treturn nums\n\t\t\t\nSo the basic idea is we use the Python\'s inbuilt s... | 32 | Given an array of integers `nums`, sort the array in **increasing** order based on the frequency of the values. If multiple values have the same frequency, sort them in **decreasing** order.
Return the _sorted array_.
**Example 1:**
**Input:** nums = \[1,1,2,2,2,3\]
**Output:** \[3,1,1,2,2,2\]
**Explanation:** '3' h... | Count number of 1s in each consecutive-1 group. For a group with n consecutive 1s, the total contribution of it to the final answer is (n + 1) * n // 2. |
✅98.74%🔥Easy Solution🔥with explanation🔥 | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | \n\n---\n# Intuition\n##### The problem asks for the widest vertical area between two points on a 2D plane. To find this, we can sort the given points based on their x-coordinates. Af... | 62 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
✅98.74%🔥Easy Solution🔥with explanation🔥 | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | \n\n---\n# Intuition\n##### The problem asks for the widest vertical area between two points on a 2D plane. To find this, we can sort the given points based on their x-coordinates. Af... | 62 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
【Video】Give me 5 minutes - How we think about a solution + Bonus code in Python | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # Intuition\nSort x-coordinates\n\n---\n\n# Solution Video\n\nhttps://youtu.be/q628cVxEFvU\n\n\u25A0 Timeline of the video\n\n`0:06` Explain key points\n`3:16` Coding\n`4:38` Time Complexity and Space Complexity\n`4:57` Step by step algorithm of my solution code\n`5:08` Bonus coding in Python\n`6:19` Time Complexity an... | 49 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
【Video】Give me 5 minutes - How we think about a solution + Bonus code in Python | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # Intuition\nSort x-coordinates\n\n---\n\n# Solution Video\n\nhttps://youtu.be/q628cVxEFvU\n\n\u25A0 Timeline of the video\n\n`0:06` Explain key points\n`3:16` Coding\n`4:38` Time Complexity and Space Complexity\n`4:57` Step by step algorithm of my solution code\n`5:08` Bonus coding in Python\n`6:19` Time Complexity an... | 49 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
✅☑Beats 100% || [C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n#### ***Approach 1(Sorting)***\n1. **Sorting Points:**\n\n- The function begins by sorting the `points` vector based on the x-coordinate of each point. This is done using `sort(points.begin(), points.end())`.\n1. **Calculating Maximu... | 39 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
✅☑Beats 100% || [C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n#### ***Approach 1(Sorting)***\n1. **Sorting Points:**\n\n- The function begins by sorting the `points` vector based on the x-coordinate of each point. This is done using `sort(points.begin(), points.end())`.\n1. **Calculating Maximu... | 39 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
6 methods||Beats 100%||Python liners, C++ set/priority_queue/sort&adjacent_difference | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOnly x-coordinate is important.\n\nUsing sorting is easy; a 2 line code for python\nThe 2nd Python code is also 2 line code which uses the matrix-transpose-like zip & beats 100% in speed.\n\nC++ has 3 solutions, not only uses sort, but al... | 15 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
6 methods||Beats 100%||Python liners, C++ set/priority_queue/sort&adjacent_difference | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nOnly x-coordinate is important.\n\nUsing sorting is easy; a 2 line code for python\nThe 2nd Python code is also 2 line code which uses the matrix-transpose-like zip & beats 100% in speed.\n\nC++ has 3 solutions, not only uses sort, but al... | 15 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Easy solution using sets || c++ || java || python | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nGiven the problem description, our focus is solely on the x coordinates, disregarding the y coordinates. In this context, we can efficiently address the issue by gathering and sorting all x coordinates, utilizing a set to el... | 8 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Easy solution using sets || c++ || java || python | widest-vertical-area-between-two-points-containing-no-points | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nGiven the problem description, our focus is solely on the x coordinates, disregarding the y coordinates. In this context, we can efficiently address the issue by gathering and sorting all x coordinates, utilizing a set to el... | 8 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
✅Python || Beats 100% Runtime | 99.58% Memory | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n# What is the Problem \n- Problem nature: The problem involves finding the maximum width of a vertical area formed by points on a 2D plane.\n- Key insight: The widths of vertical areas are determined solely by the differences in x-coordinates of points.\n- Focus on x-coordinates: We can focus on extracting and sortin... | 3 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
✅Python || Beats 100% Runtime | 99.58% Memory | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n# What is the Problem \n- Problem nature: The problem involves finding the maximum width of a vertical area formed by points on a 2D plane.\n- Key insight: The widths of vertical areas are determined solely by the differences in x-coordinates of points.\n- Focus on x-coordinates: We can focus on extracting and sortin... | 3 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
✅ One Line Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Code #1 - The Most Concise\nTime complexity: $$O(n*log(n))$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n return max(b-a for (a,_),(b,_) in pairwise(sorted(points)))\n```\n\n# Code #2\nTime complexity: $$O(n*log(n))$$. Space complex... | 4 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
✅ One Line Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Code #1 - The Most Concise\nTime complexity: $$O(n*log(n))$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n return max(b-a for (a,_),(b,_) in pairwise(sorted(points)))\n```\n\n# Code #2\nTime complexity: $$O(n*log(n))$$. Space complex... | 4 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Python3 Soltion | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n points.sort()\n n=len(points)\n ans=0\n for i in range(1,n):\n ans=max(ans,(points[i][0]-points[i-1][0]))\n return ans \n \n``` | 3 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Python3 Soltion | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n points.sort()\n n=len(points)\n ans=0\n for i in range(1,n):\n ans=max(ans,(points[i][0]-points[i-1][0]))\n return ans \n \n``` | 3 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
[We💕Simple] One Liner! (Beat 100%) | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | ``` Kotlin []\nclass Solution {\n fun maxWidthOfVerticalArea(points: Array<IntArray>): Int =\n points.map { it[0] }.sorted().zipWithNext { a, b -> b - a }.max()\n}\n```\n.\n``` Python3 []\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n return max([\n ... | 2 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
[We💕Simple] One Liner! (Beat 100%) | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | ``` Kotlin []\nclass Solution {\n fun maxWidthOfVerticalArea(points: Array<IntArray>): Int =\n points.map { it[0] }.sorted().zipWithNext { a, b -> b - a }.max()\n}\n```\n.\n``` Python3 []\nclass Solution:\n def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:\n return max([\n ... | 2 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Normal Approach to problem solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs per given question, whatever the value of y-coordinate we need the maxwidth/max difference between x-coordinates so that no other x-coordinate in between those points.\n# Approach\n<!-- Describe your approach to solving the problem. --... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Normal Approach to problem solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs per given question, whatever the value of y-coordinate we need the maxwidth/max difference between x-coordinates so that no other x-coordinate in between those points.\n# Approach\n<!-- Describe your approach to solving the problem. --... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Easy approach // Beat 100% user | widest-vertical-area-between-two-points-containing-no-points | 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. $$O(n)$$ --... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Easy approach // Beat 100% user | widest-vertical-area-between-two-points-containing-no-points | 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. $$O(n)$$ --... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Fastest Python Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFaster than 100% of solutions. We can reword the question to find the maximum distance between two adjacent points\n\n\n\n\n... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Fastest Python Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFaster than 100% of solutions. We can reword the question to find the maximum distance between two adjacent points\n\n\n\n\n... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
Easy python solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\nto be consistent in leetcode\n\n# Approach\nused sorting,difference and iterative method\n\n# Complexity\n- beats 100% users in python by time\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n60.5 MB\n\n# Code\n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points:... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
Easy python solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | # Intuition\nto be consistent in leetcode\n\n# Approach\nused sorting,difference and iterative method\n\n# Complexity\n- beats 100% users in python by time\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n60.5 MB\n\n# Code\n```\nclass Solution:\n def maxWidthOfVerticalArea(self, points:... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
✅ 100% SPEED SOLUTION IN 1 LINE! 😲😲 | FROM USERS 🐋 TO USERS 🐳 | widest-vertical-area-between-two-points-containing-no-points | 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$$O (n*log(n))$$\n\n- Space complexity:\n<!-- Add your space complexity here... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
✅ 100% SPEED SOLUTION IN 1 LINE! 😲😲 | FROM USERS 🐋 TO USERS 🐳 | widest-vertical-area-between-two-points-containing-no-points | 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$$O (n*log(n))$$\n\n- Space complexity:\n<!-- Add your space complexity here... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
PYTHON3 SOLUTION | RUNTIME BEATS 100% 🐍 | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe just have to find the width of the widest area. Because of the infinite height, the y coordinates can be ignored.\n\n... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
PYTHON3 SOLUTION | RUNTIME BEATS 100% 🐍 | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe just have to find the width of the widest area. Because of the infinite height, the y coordinates can be ignored.\n\n... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
My Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | Given n points on a 2D plane where points[i] = [xi, yi], Return the widest vertical area between two points such that no points are inside the area.\n\nA vertical area is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The widest vertical area is the one with the maximum width.\n\n... | 1 | Given `n` `points` on a 2D plane where `points[i] = [xi, yi]`, Return _the **widest vertical area** between two points such that no points are inside the area._
A **vertical area** is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The **widest vertical area** is the one with the ... | Use dynamic programming. The state of the DP can be the current index and the remaining characters to delete. Having a prefix sum for each character can help you determine for a certain character c in some specific range, how many characters you need to delete to merge all occurrences of c in that range. |
My Solution | widest-vertical-area-between-two-points-containing-no-points | 0 | 1 | Given n points on a 2D plane where points[i] = [xi, yi], Return the widest vertical area between two points such that no points are inside the area.\n\nA vertical area is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The widest vertical area is the one with the maximum width.\n\n... | 1 | You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
Your job at this factory is to put each ball in the box with a number equal to the sum of digits ... | Try sorting the points Think is the y-axis of a point relevant |
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