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number-of-arithmetic-triplets
✅ A generic Dynamic Programming (A[i] - diff) || [Intuition Updated]
a-generic-dynamic-programming-ai-diff-in-y2fl
This Approach even works if the array is unsorted or elements are negative or diff is negative\n\nSimilar to 300. Longest Increasing Subsequence.\nWe simply sav
xxvvpp
NORMAL
2022-08-07T04:15:10.106293+00:00
2022-08-07T09:09:42.671871+00:00
2,919
false
**This Approach even works if the array is unsorted or elements are negative or diff is negative**\n\n**Similar to [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/).**\nWe simply save the value for every we have seen till now in a array and our count will be `cnt[A[i]]...
23
2
['Dynamic Programming', 'C']
1
number-of-arithmetic-triplets
BRUTE FORCE 3 LOOP's && Optimized O(N)
brute-force-3-loops-optimized-on-by-up15-as5k
Contest Submission\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int ans=0;\n for(int i=0;i<nums.size(
up1512001
NORMAL
2022-08-07T04:02:53.017214+00:00
2022-08-08T13:06:29.695911+00:00
3,699
false
**Contest Submission**\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int ans=0;\n for(int i=0;i<nums.size();i++){\n for(int j=i+1;j<nums.size();j++){\n for(int k=j+1;k<nums.size();k++){\n if((nums[j]-nums[i])==diff...
20
0
['C', 'C++']
1
number-of-arithmetic-triplets
Java / C++ Brute Force to Set
java-c-brute-force-to-set-by-fllght-mwgi
The brute force method is pretty fast here, but using a set is preferable\n\n### Java Brute Force\njava\npublic int arithmeticTriplets(int[] nums, int diff) {\n
FLlGHT
NORMAL
2022-08-07T06:07:06.808460+00:00
2022-08-07T07:27:36.949896+00:00
1,809
false
The brute force method is pretty fast here, but using a set is preferable\n\n### Java Brute Force\n```java\npublic int arithmeticTriplets(int[] nums, int diff) {\n int count = 0;\n\n for (int i = 0; i < nums.length - 2; i++) {\n for (int j = i + 1; j < nums.length - 1; j++) {\n f...
16
0
['C', 'Ordered Set', 'Java']
3
number-of-arithmetic-triplets
easy hashmap solution and got 54ms ✅
easy-hashmap-solution-and-got-54ms-by-mj-xnn4
\nvar arithmeticTriplets = function(nums, diff) {\n \n let hash = new Map();\n let count = 0;\n\n for(let i=0; i<nums.length; i++){\n let tem
MJWHY
NORMAL
2022-08-18T00:11:46.230923+00:00
2022-08-18T17:17:04.109785+00:00
1,179
false
```\nvar arithmeticTriplets = function(nums, diff) {\n \n let hash = new Map();\n let count = 0;\n\n for(let i=0; i<nums.length; i++){\n let temp = nums[i] - diff;\n \n if(hash.has(temp) && hash.has(temp - diff)){\n count++;\n }\n hash.set(nums[i] , "Hard choice...
14
0
['JavaScript']
2
number-of-arithmetic-triplets
✔ C++ using Map with Explanation || Very Easy and Simple to understand Solution
c-using-map-with-explanation-very-easy-a-4x19
Up Vote if you like the solution\n# Check for the nums[i] -diff and nums[i] - 2*diff, if these are present in the map or not. If present count a triplet. Then
kreakEmp
NORMAL
2022-08-07T04:01:32.339003+00:00
2022-08-07T05:18:24.991141+00:00
2,122
false
<b> Up Vote if you like the solution\n# Check for the nums[i] -diff and nums[i] - 2*diff, if these are present in the map or not. If present count a triplet. Then strore the current number in map for future use.\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n unordere...
14
0
[]
0
number-of-arithmetic-triplets
Python | Easy Solution✅
python-easy-solution-by-gmanayath-777z
\ndef arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n ans = 0\n for i in range(len(nums)): # nums = [0,1,4,6,7,10], diff = 3\n
gmanayath
NORMAL
2022-08-12T15:45:44.309663+00:00
2022-12-22T16:50:28.013817+00:00
1,712
false
```\ndef arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n ans = 0\n for i in range(len(nums)): # nums = [0,1,4,6,7,10], diff = 3\n j = nums[i] - diff # 4 - 3 = 1 (when i ==2)\n k = diff + nums[i] # 3 + 4 = 7 (when i ==2)\n if j in nums and k in nums: # 1 ...
13
0
['Python', 'Python3']
4
number-of-arithmetic-triplets
✅Easy Java solution||Straight Forward||Beginner Friendly🔥
easy-java-solutionstraight-forwardbeginn-qfyc
If you really found my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community, if you have some queries
deepVashisth
NORMAL
2022-09-30T19:21:27.691409+00:00
2022-09-30T19:21:27.691448+00:00
1,081
false
**If you really found my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community, if you have some queries or some improvements please feel free to comment and share your views.**\n```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n ...
11
0
['Java']
2
number-of-arithmetic-triplets
( Beats 99%)3 Java Solution | Binary Search | Map | 3 Pointers
beats-993-java-solution-binary-search-ma-z9ot
3 Different JAVA Solutions \n\nSolution Using Binary Search (Time Complexity : O(nlog(n)) , Space Complexity : O(1))\n\nclass Solution {\n public static bool
Aman__Bhardwaj
NORMAL
2022-09-08T17:19:27.645130+00:00
2022-09-08T17:19:27.645171+00:00
740
false
# 3 Different JAVA Solutions \n\n**Solution Using Binary Search (Time Complexity : O(nlog(n)) , Space Complexity : O(1))**\n```\nclass Solution {\n public static boolean BinarySearch(int[] arr, int key){\n int low = 0 , high = arr.length-1;\n while(low <= high){\n int mid = low + (high-low)/...
10
0
['Binary Tree', 'Java']
1
number-of-arithmetic-triplets
✅ Easy Python Solution||Binary Search Solution
easy-python-solutionbinary-search-soluti-zrt1
Intution:-\n PLEASE UPVOTE\uD83D\uDC4D\uD83D\uDC4D IF YOU LIKE THE SOLUTION\n## Binary Search\nUse binary search to find out if nums[i]+diff and nums[i]+2diff i
amandgp
NORMAL
2023-03-26T13:41:24.646179+00:00
2023-03-26T13:41:24.646222+00:00
506
false
# Intution:-\n **PLEASE UPVOTE\uD83D\uDC4D\uD83D\uDC4D IF YOU LIKE THE SOLUTION**\n## Binary Search\nUse binary search to find out if nums[i]+diff and nums[i]+2*diff is present in array or not if both are present in array increment the count by 1 if any one of them is not present in array continue:\n<!-- Describe your ...
8
0
['Array', 'Two Pointers', 'Binary Search', 'Enumeration', 'Python3']
0
number-of-arithmetic-triplets
C++|| Multiple Approach || Bruteforce to optimized || Easy Explanation
c-multiple-approach-bruteforce-to-optimi-hg87
\n\t// Brute force \n\t// we will check condition for k only if nums[j]-nums[i]==diff\n\n\tclass Solution {\n\tpublic:\n\t\tint arithmeticTriplets(vector& nums,
anubhavsingh11
NORMAL
2022-08-09T19:47:57.117384+00:00
2023-01-16T05:38:00.637389+00:00
969
false
\n\t// Brute force \n\t// we will check condition for k only if nums[j]-nums[i]==diff\n\n\tclass Solution {\n\tpublic:\n\t\tint arithmeticTriplets(vector<int>& nums, int diff) {\n\t\t\tint ans=0;\n\t\t\tfor(int i=0;i<nums.size()-2;i++)\n\t\t\t{\n\t\t\t\tfor(int j=i+1;j<nums.size()-1;j++)\n\t\t\t\t{\n\t\t\t\t\tif(nums[j...
8
0
['C', 'Ordered Set']
2
number-of-arithmetic-triplets
Python easy understand solution O(n) space, O(n) time
python-easy-understand-solution-on-space-dzdk
Use a set to memorize each interger is exist or not. \npython\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n s
amikai
NORMAL
2022-08-07T13:49:28.493723+00:00
2022-08-07T14:45:40.890742+00:00
1,517
false
Use a set to memorize each interger is exist or not. \n```python\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n s = set(nums)\n count = 0\n for num in nums:\n if (num + diff) in s and (num + diff + diff) in s:\n count += 1\n ...
8
0
['Ordered Set', 'Python', 'Python3']
5
number-of-arithmetic-triplets
Java Easy solution using HashMap
java-easy-solution-using-hashmap-by-basi-kc53
For each element check element + diff && element + 2 * diff exist in map.\n```\npublic int arithmeticTriplets(int[] nums, int diff) {\n HashMap map = new
basitimam
NORMAL
2022-08-07T04:01:23.409876+00:00
2022-08-07T04:08:56.653024+00:00
963
false
For each element check element + diff && element + 2 * diff exist in map.\n```\npublic int arithmeticTriplets(int[] nums, int diff) {\n HashMap<Integer, Integer> map = new HashMap<>();\n for(int i : nums){\n map.put(i, map.getOrDefault(i, 0) + 1);\n }\n int count = 0;\n for...
7
0
['Java']
3
number-of-arithmetic-triplets
Two (Three) Pointers. Beates 100 %
two-three-pointers-beates-100-by-sberik-gq2s
Approach\nThe main idea of the code is to use three pointers to search for arithmetic triplets in thenumsarray, where the difference between adjacent elements i
SBerik
NORMAL
2023-10-31T04:53:08.569549+00:00
2023-10-31T05:03:58.697126+00:00
185
false
# Approach\nThe main idea of the code is to use three pointers to search for arithmetic triplets in the`nums`array, where the difference between adjacent elements is equal to the given `diff`. The pointers `i`, `j`, and `k` move through the array.\n\n# Complexity\n- Time complexity: $$O(N)$$\n\n- Space complexity: $$O(...
6
0
['Two Pointers', 'Java']
0
number-of-arithmetic-triplets
C++ || Unordered-map || Easy-to-Understand || no of arithmetic tripplet.
c-unordered-map-easy-to-understand-no-of-yuil
# Intuition \n\n\n\n\n\n\n\n\n\n\n\n\n# Code\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int n=nums.size(
m_isha_125
NORMAL
2022-11-10T20:55:26.502303+00:00
2022-11-10T20:55:26.502344+00:00
567
false
<!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n<!-- # Approach -->\n<!-- Describe your approach to solving the problem. -->\n\n<!-- # Complexity -->\n<!-- - Time complexity: -->\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n<!-- - Space complexity: -->\n<!-- Add...
6
0
['Hash Table', 'C++']
0
number-of-arithmetic-triplets
Python O(n) Two Pointers Approach
python-on-two-pointers-approach-by-brent-e45q
We can take advantage of the fact that the array is already sorted to find arithmetic triples quickly.\nThe big idea is that if either nums[j] - nums[i] or nums
brent_pappas
NORMAL
2022-10-08T15:05:42.229714+00:00
2022-10-08T15:05:42.229746+00:00
661
false
We can take advantage of the fact that the array is already sorted to find arithmetic triples quickly.\nThe big idea is that if either `nums[j] - nums[i]` or `nums[k] - nums[j]` is not equal to diff, we increment the appropriate index.\n\n```\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int...
6
0
['Two Pointers', 'Python']
1
number-of-arithmetic-triplets
python 95% fast and 97% memory
python-95-fast-and-97-memory-by-tul-jo37
\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n dic = {} # store nums[i]\n quest = {} # store require nu
TUL
NORMAL
2022-08-29T04:52:21.911732+00:00
2022-08-29T04:52:21.911789+00:00
899
false
```\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n dic = {} # store nums[i]\n quest = {} # store require number you possible find after nums[i]\n count = 0 # count answer\n for i in nums:\n dic[i] = True \n if i in quest: count ...
6
0
['Hash Table', 'Python', 'Python3']
1
number-of-arithmetic-triplets
✅SIMPLE AND EASY C++ SOLUTION USING HASHMAP
simple-and-easy-c-solution-using-hashmap-uq8n
Time Complexity: O(n)\n\nCode:\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n unordered_
ke4e
NORMAL
2022-08-21T07:08:20.426713+00:00
2022-08-21T07:08:20.426759+00:00
458
false
Time Complexity: O(n)\n\nCode:\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n unordered_map<int, bool> data;\n for (int num:nums)\n data[num]=true;\n \n for (int num: nums)\n if (data[num-diff] && data[...
6
0
['C']
0
number-of-arithmetic-triplets
Common Difference A.P||C++
common-difference-apc-by-priyanshu3237-obvv
class Solution {\npublic:\n int arithmeticTriplets(vector& nums, int diff) {\n int ans = 0;\n for(int j = 1;j=0 && k<nums.size()){\n
priyanshu3237
NORMAL
2022-08-07T04:06:04.059362+00:00
2022-08-07T06:49:19.591583+00:00
526
false
class Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int ans = 0;\n for(int j = 1;j<nums.size()-1;j++){\n int i = j-1, k = j+1;\n \n while(i>=0 && k<nums.size()){\n \n if(nums[j]-nums[i] == diff && nums[k]-nu...
6
0
[]
2
number-of-arithmetic-triplets
JAVA || 10000% beats ❤️ || 3 Solution || HashSet || ArrayList || ❤️
java-10000-beats-3-solution-hashset-arra-06tc
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
saurabh_kumar1
NORMAL
2023-08-30T16:22:01.360415+00:00
2023-08-30T16:22:01.360434+00:00
653
false
# 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:0(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:0(1)\n<!-- Add your space complexity here, e.g. $$O...
5
0
['Array', 'Hash Table', 'C++', 'Java']
1
number-of-arithmetic-triplets
my_arithmeticTriplets
my_arithmetictriplets-by-rinatmambetov-4ww3
# Intuition \n\n\n\n\n\n\n\n\n\n\n\n# Code\n\n/**\n * @param {number[]} nums\n * @param {number} diff\n * @return {number}\n */\nvar arithmeticTriplets = func
RinatMambetov
NORMAL
2023-05-14T14:23:07.857122+00:00
2023-05-14T14:23:07.857163+00:00
881
false
<!-- # 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 spa...
5
0
['JavaScript']
0
number-of-arithmetic-triplets
Two simple and fast Map and Array based solutions
two-simple-and-fast-map-and-array-based-wy2n2
Approach\nEach number can have only one potenial triplet and the numbers should go in sequential order. So we just need to check that. If array we can check for
balfin
NORMAL
2023-02-11T09:50:15.872915+00:00
2023-02-12T05:26:56.603744+00:00
804
false
# Approach\nEach number can have only one potenial triplet and the numbers should go in sequential order. So we just need to check that. If array we can check forward values, if map we need to fill it with data and check backwards. Map solution is a bit slower per leetcode tests, looks like map.has is not that optimize...
5
0
['JavaScript']
0
number-of-arithmetic-triplets
using binary_search
using-binary_search-by-bhupendraj9-9zev
\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n for(int i=0;i<nums.size();i++)\n {\n
bhupendraj9
NORMAL
2022-11-08T09:14:00.153022+00:00
2023-11-01T12:20:05.427694+00:00
438
false
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n for(int i=0;i<nums.size();i++)\n {\n if(binary_search(nums.begin(),nums.end(),nums[i]+diff) && binary_search(nums.begin(),nums.end(),nums[i]+2*diff))\n count++;\n ...
5
0
['Binary Search', 'C++']
3
number-of-arithmetic-triplets
Swift One-Liner | Also: Optimized Alternates
swift-one-liner-also-optimized-alternate-weos
One-Liner (accepted answer)\n\nclass Solution {\n func arithmeticTriplets(_ nums: [Int], _ diff: Int) -> Int {\n nums.filter { nums.contains($0+diff)
UpvoteThisPls
NORMAL
2022-08-08T23:19:36.474617+00:00
2022-08-08T23:19:36.474660+00:00
102
false
**One-Liner (accepted answer)**\n```\nclass Solution {\n func arithmeticTriplets(_ nums: [Int], _ diff: Int) -> Int {\n nums.filter { nums.contains($0+diff) && nums.contains($0+diff*2) }.count\n }\n}\n```\n\n**Two-Liner, much faster approach (accepted answer)**\n```\nclass Solution {\n func arithmeticTr...
5
0
[]
1
number-of-arithmetic-triplets
Python easy solution for beginners using slightly optimized brute force
python-easy-solution-for-beginners-using-ova9
This solution is slightly better than pure brute force as it calls the 3rd loop only if a condition is met.\n\n```\nclass Solution:\n def arithmeticTriplets(
elefant1805
NORMAL
2022-08-07T07:08:51.665627+00:00
2022-08-07T07:13:07.895189+00:00
438
false
##### This solution is slightly better than pure brute force as it calls the 3rd loop only if a condition is met.\n\n```\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n res = 0\n for i in range(len(nums)):\n for j in range(i+1, len(nums)):\n ...
5
0
['Python', 'Python3']
0
number-of-arithmetic-triplets
C++ ✅ using SET O(n2) ✅ ✅
c-using-set-on2-by-pratapsingha264-vob1
class Solution {\npublic:\n int arithmeticTriplets(vector& nums, int diff) \n {\n int count=0;\n int n= nums.size();\n set s;\n
Pratapsingha264
NORMAL
2022-08-07T04:17:52.085132+00:00
2022-08-07T04:20:56.378463+00:00
586
false
class Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) \n {\n int count=0;\n int n= nums.size();\n set<int> s;\n \n \n for(int i=0;i<n;i++)\n {\n s.insert(nums[i]);\n for(int j=i+1;j<n;j++)\n {\n ...
5
1
['C', 'Ordered Set']
0
number-of-arithmetic-triplets
SIMPLE SET COMMENTED C++ SOLUTION WITH EXPLANATION
simple-set-commented-c-solution-with-exp-wd9i
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
Jeffrin2005
NORMAL
2024-07-18T12:39:01.981499+00:00
2024-07-18T12:39:01.981523+00:00
107
false
# 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:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:o(n)\n<!-- Add your space complexity here, e.g. $$O...
4
0
['C++']
0
number-of-arithmetic-triplets
Easiest way
easiest-way-by-nurzhansultanov-1wg5
\n\nclass Solution(object):\n def arithmeticTriplets(self, nums, diff):\n result = []\n for i in range(len(nums)-1, 1, -1):\n for j
NurzhanSultanov
NORMAL
2024-01-30T14:55:26.936789+00:00
2024-01-30T14:55:26.936824+00:00
174
false
\n```\nclass Solution(object):\n def arithmeticTriplets(self, nums, diff):\n result = []\n for i in range(len(nums)-1, 1, -1):\n for j in range(i-1, 0, -1):\n for k in range(j-1, -1, -1):\n if nums[i] - nums[j] == nums[j] - nums[k] == diff:\n ...
4
0
['Python']
1
number-of-arithmetic-triplets
TypeScript/JavaScript O(n) Solution
typescriptjavascript-on-solution-by-hals-xjn3
\n# Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n\n# Code\n\n
halshar
NORMAL
2023-09-22T18:48:12.736640+00:00
2023-09-26T14:38:33.860133+00:00
312
false
\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfunction arithmeticTriplets(nums: number[], diff: number): number {\n const hash = new Set<number>();\n let ans = 0;\n\n // what we ...
4
0
['TypeScript', 'JavaScript']
1
number-of-arithmetic-triplets
📍||USING SET ||2 Liner ||🔥🔥🔥🔥|| OPTIMISED SOLUTION
using-set-2-liner-optimised-solution-by-jr0bn
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
Sautramani
NORMAL
2023-08-20T05:54:31.523936+00:00
2023-08-20T05:57:32.374694+00:00
802
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n...
4
0
['Array', 'Ordered Set', 'Python', 'C++', 'Java']
1
number-of-arithmetic-triplets
Go. Two pointers. Time complexity O(n) and Space complexity O(1)
go-two-pointers-time-complexity-on-and-s-trtu
Intuition\n\n##### Analyzing the problem, the following points can be emphasized:\n1. We are given an array of integers sorted in ascending order.\n1. We need t
sergei-m
NORMAL
2023-08-01T15:54:07.222484+00:00
2023-08-05T23:22:40.020599+00:00
264
false
# Intuition\n\n##### Analyzing the problem, the following points can be emphasized:\n1. We are given an array of integers sorted in ascending order.\n1. We need to find pairs of values that will match the condition `y - x = diff && z - y = diff`.\n1. This means that there is no need to look for a pair `z - y = diff` if...
4
0
['Array', 'Two Pointers', 'Go']
0
number-of-arithmetic-triplets
FASTEST solution with JAVA(1ms)
fastest-solution-with-java1ms-by-amin_az-ygj2
Code\n\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n int count = 0; \n Set<Integer> list = new HashSet<>();\n
amin_aziz
NORMAL
2023-04-12T09:05:48.679776+00:00
2023-04-12T18:07:18.237541+00:00
1,893
false
# Code\n```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n int count = 0; \n Set<Integer> list = new HashSet<>();\n for(int a : nums){\n if(list.contains(a-diff) && list.contains(a-diff*2)) count++;\n list.add(a);\n }\n return coun...
4
0
['Java']
3
number-of-arithmetic-triplets
C++ easy solution with loops.
c-easy-solution-with-loops-by-aloneguy-e06b
\n# Complexity\n- Time complexity: 6 ms.Beats 65.9% solutions.\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:8.6 Mb.Beats 88.98% solutions.\
aloneguy
NORMAL
2023-03-23T07:03:21.747706+00:00
2023-03-23T07:03:21.747748+00:00
454
false
\n# Complexity\n- Time complexity: 6 ms.Beats 65.9% solutions.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:8.6 Mb.Beats 88.98% solutions.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int ...
4
0
['C++']
0
number-of-arithmetic-triplets
C# easy solution.
c-easy-solution-by-aloneguy-vux7
\n# Complexity\n- Time complexity: 89 ms.Beats 66.24% of other solutions.\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: 38.1 Mb.Beats 85.99
aloneguy
NORMAL
2023-03-23T06:57:44.832885+00:00
2023-03-23T06:57:44.832925+00:00
373
false
\n# Complexity\n- Time complexity: 89 ms.Beats 66.24% of other solutions.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 38.1 Mb.Beats 85.99% of others.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\npublic class Solution {\n public int count=0;\n public int...
4
0
['Math', 'C#']
0
number-of-arithmetic-triplets
java solution using HashSet !!
java-solution-using-hashset-by-aanshi_07-kg9o
\n# Code\n\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set<Integer> set = new HashSet<>();\n for(int n: nums){
aanshi_07
NORMAL
2022-12-10T10:14:36.778815+00:00
2022-12-10T10:18:10.855162+00:00
797
false
\n# Code\n```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set<Integer> set = new HashSet<>();\n for(int n: nums){\n set.add(n);\n }\n int count =0;\n for(int n:nums){\n if(set.contains(n+diff) && set.contains(n+(2*diff)))\n ...
4
0
['Hash Table', 'Java']
3
number-of-arithmetic-triplets
100% Faster || TypeScript || Easy to understand
100-faster-typescript-easy-to-understand-4btn
Let me know in comments if you have any doubts. I will be happy to answer.\nPlease upvote if you found the solution useful.\n\n\nconst arithmeticTriplets = (num
viniciusteixeiradias
NORMAL
2022-11-14T02:36:28.105429+00:00
2022-11-16T01:48:43.163198+00:00
106
false
Let me know in comments if you have any doubts. I will be happy to answer.\nPlease upvote if you found the solution useful.\n\n```\nconst arithmeticTriplets = (nums: number[], diff: number): number => {\n let sum = 0;\n let hash = new Set();\n \n for(let i = 0; i < nums.length; i++) {\n if (hash.has(...
4
0
['TypeScript']
0
number-of-arithmetic-triplets
C++ Naive Approach Solution
c-naive-approach-solution-by-king_of_kin-ugy0
int arithmeticTriplets(vector& nums, int diff) {\n\n int ans = 0 ;\n for(int i = 0 ; i < nums.size()-2 ; i++)\n {\n for(int j =
King_of_Kings101
NORMAL
2022-09-23T07:03:31.293751+00:00
2022-09-23T07:06:36.055976+00:00
623
false
int arithmeticTriplets(vector<int>& nums, int diff) {\n\n int ans = 0 ;\n for(int i = 0 ; i < nums.size()-2 ; i++)\n {\n for(int j = i+1 ; j < nums.size()-1 ; j++)\n {\n if(nums[i]+diff==nums[j])\n {\n for(int k = j +1; k < nums.si...
4
0
['Array', 'C', 'C++']
0
number-of-arithmetic-triplets
C Sharp Solution
c-sharp-solution-by-fadydev123-3n1c
\nint count = 0;\nHashSet<int> hash_num = new HashSet<int>();\nforeach(int num in nums)\n{\n\tif(hash_num.Contains(num - diff) && hash_num.Contains(num - 2 * di
fadydev123
NORMAL
2022-09-11T07:08:23.635368+00:00
2022-09-11T07:08:39.984015+00:00
271
false
```\nint count = 0;\nHashSet<int> hash_num = new HashSet<int>();\nforeach(int num in nums)\n{\n\tif(hash_num.Contains(num - diff) && hash_num.Contains(num - 2 * diff))\n\t\tcount++;\n\thash_num.Add(num); \n}\nreturn count;\n\n```
4
0
['C#']
0
number-of-arithmetic-triplets
JavaScript | Hashmap solution | O(n)
javascript-hashmap-solution-on-by-zhe1ka-megg
```\nfunction arithmeticTriplets(nums: number[], diff: number): number {\n let sum = 0;\n let hash = new Set();\n \n for (let numIndex = 0; numIndex
Zhe1ka
NORMAL
2022-08-29T06:34:27.142829+00:00
2022-08-29T06:34:27.142854+00:00
533
false
```\nfunction arithmeticTriplets(nums: number[], diff: number): number {\n let sum = 0;\n let hash = new Set();\n \n for (let numIndex = 0; numIndex < nums.length; numIndex++) {\n const currentNum = nums[numIndex];\n \n if (hash.has(currentNum - diff) && hash.has(currentNum - diff - dif...
4
0
['TypeScript', 'JavaScript']
0
number-of-arithmetic-triplets
Three Pointers, one pass O(n) time and O(1) space
three-pointers-one-pass-on-time-and-o1-s-osqq
Intuition"Strictly increasing" is the key.Complexity Time complexity:O(n) Space complexity:O(1) Code
SoulMan
NORMAL
2025-02-09T08:55:22.860188+00:00
2025-02-19T09:10:59.236542+00:00
227
false
# Intuition "Strictly increasing" is the key. <!-- Describe your first thoughts on how to solve this problem. --> # Complexity - Time complexity: $$O(n)$$ <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: $$O(1)$$ <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```cpp [] class S...
3
0
['C++', 'Java']
0
number-of-arithmetic-triplets
💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 💯
easiestfaster-lesser-cpython3javacpython-vnsh
Intuition\n\n\nThese are the almost same questions which are similar to this one which are as follows :-\n\n---\nQ - 1460 --> https://leetcode.com/problems/make
Edwards310
NORMAL
2024-07-26T13:19:02.822977+00:00
2024-07-26T14:15:15.201707+00:00
357
false
# Intuition\n![0ehh83fsnh811.jpg](https://assets.leetcode.com/users/images/402c9963-629f-4a75-91f6-301b8ba012b4_1721999712.5771427.jpeg)\n![Screenshot 2024-07-26 184436.png](https://assets.leetcode.com/users/images/14d3e21d-8b12-4232-8f46-febf0177db00_1721999722.453408.png)\nThese are the almost same questions which ar...
3
0
['Array', 'Hash Table', 'Two Pointers', 'C', 'Enumeration', 'Python', 'C++', 'Java', 'Python3', 'C#']
0
number-of-arithmetic-triplets
Easy to understand - Java
easy-to-understand-java-by-ungureanu_ovi-o5g5
Upvote if you like the solution\n\n# Intuition\n\nThe problem requires finding unique arithmetic triplets in an array, where the difference between consecutive
Ungureanu_Ovidiu
NORMAL
2024-07-02T10:09:52.126395+00:00
2024-07-02T10:09:52.126431+00:00
142
false
##### Upvote if you like the solution\n\n# Intuition\n\nThe problem requires finding unique arithmetic triplets in an array, where the difference between consecutive elements in the triplet is a given value `diff`.\n\n## Approach\n\n1. **Set for Lookup**:\n - Use a `HashSet` called `presentNums` to store all the inte...
3
0
['Java']
0
number-of-arithmetic-triplets
Easy solution java
easy-solution-java-by-abhinav29code-zoz2
\n\n# Code\n\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n if(nums.length < 3) return 0;\n\n int count = 0;\n
abhinav29code
NORMAL
2023-12-26T04:22:39.315944+00:00
2023-12-26T04:22:39.315983+00:00
364
false
\n\n# Code\n```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n if(nums.length < 3) return 0;\n\n int count = 0;\n HashSet<Integer> set = new HashSet<>();\n for(int num : nums) {\n if(set.contains(num - diff) && set.contains(num - diff*2)) {\n ...
3
0
['Java']
0
number-of-arithmetic-triplets
Easy to understand for beginner using "HashMap"
easy-to-understand-for-beginner-using-ha-2u6h
Approach\n\nfirst of all you should be aware about hashmap and its functions.\nOne such function which I have used is "containsKey" it return true if a key is p
Priyansh_max
NORMAL
2023-12-17T15:44:08.631449+00:00
2023-12-17T15:44:08.631470+00:00
387
false
# Approach\n\nfirst of all you should be aware about hashmap and its functions.\nOne such function which I have used is "containsKey" it return true if a key is present in the hashmap that you have inputed.\n\n1 --> creating a hashmap\n2 --> create a <key , value> pair \n\n3 --> check if "diff + current index is presen...
3
0
['Java']
2
number-of-arithmetic-triplets
Better than nested loops.
better-than-nested-loops-by-drugged_monk-c7i3
Intuition\nProblem should be solved in one loop.\n\n# Approach\nFirst I convert array nums to HashSet. Then I traverse array nums to find every triplet nums[i],
Drugged_Monkey
NORMAL
2023-12-01T20:20:53.173227+00:00
2024-06-08T15:09:40.389792+00:00
102
false
# Intuition\nProblem should be solved in one loop.\n\n# Approach\nFirst I convert array `nums` to `HashSet`. Then I traverse array `nums` to find every triplet `nums[i]`, `nums[i] - diff` and `nums[i] + diff`. That\'s it.\n\n# Complexity\n- Time complexity:\n$$O(n)$$. Despite `HashSet.Contains()` implementation isn\'t ...
3
0
['C#']
2
number-of-arithmetic-triplets
✅Iteratively||🔥Simple Solution 🔥|| JAVA || C++|| Python || 🚀Clean Codes🚀 ||Easy to Understand✅✅
iterativelysimple-solution-java-c-python-xpng
\n# Approach\n Describe your approach to solving the problem. \nHere\'s a breakdown of the code:\n\n1. int ans = 0;: This initializes a variable ans to store th
preans1412
NORMAL
2023-08-11T16:47:23.178415+00:00
2023-08-11T16:47:23.178434+00:00
497
false
\n# Approach\n<!-- Describe your approach to solving the problem. -->\nHere\'s a breakdown of the code:\n\n1. int ans = 0;: This initializes a variable ans to store the count of arithmetic triplets found in the vector.\n\n2. The code uses three nested for loops to iterate through all possible combinations of three indi...
3
0
['Array', 'Python', 'C++', 'Java', 'Python3']
1
number-of-arithmetic-triplets
Easy C++ Solution
easy-c-solution-by-adityagarg217-wmqq
class Solution {\npublic:\n\n int arithmeticTriplets(vector& nums, int diff) {\n int ans = 0 ;\n \n for(int i=0 ;i<nums.size()-2 ;i++){\n
adityagarg217
NORMAL
2023-06-04T10:39:36.839663+00:00
2023-06-05T11:49:43.064449+00:00
1,475
false
class Solution {\npublic:\n\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int ans = 0 ;\n \n for(int i=0 ;i<nums.size()-2 ;i++){\n int count = 0 ;\n for(int j=i+1 ; j<nums.size() ;j++){\n \n if(nums[j]-nums[i]==diff || nums[j]-nums[i...
3
0
['Array', 'C']
0
number-of-arithmetic-triplets
[Java] Easy set solution 100% O(n)
java-easy-set-solution-100-on-by-ytchoua-mu5m
java\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set<Integer> set = new HashSet<>();\n int i = 0, k = 1, result
YTchouar
NORMAL
2023-04-01T07:04:11.285197+00:00
2023-04-01T07:04:25.892014+00:00
2,685
false
```java\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set<Integer> set = new HashSet<>();\n int i = 0, k = 1, result = 0;\n\n for(int num : nums)\n set.add(num);\n\n while(k < nums.length) {\n if(nums[k] - nums[i] == 2 * diff && set.cont...
3
0
['Java']
3
number-of-arithmetic-triplets
single pass
single-pass-by-rajsiddi-r2ed
\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int trip = 0;\n for(int i=0;i<nums.size();i++){\n
rajsiddi
NORMAL
2023-03-05T09:35:59.011537+00:00
2023-03-05T09:44:41.552036+00:00
1,060
false
\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int trip = 0;\n for(int i=0;i<nums.size();i++){\n int find = nums[i]+diff;\n if(count(nums.begin(),nums.end(),find)>0){\n int find2 = find+diff;\n if(count(nums.begin(),num...
3
0
['C++']
1
number-of-arithmetic-triplets
Runtime 52 ms Beats 57.18% Memory 13.9 MB Beats 55.96%, beginner easy to understand
runtime-52-ms-beats-5718-memory-139-mb-b-xzlv
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
NEWYORK2009
NORMAL
2023-02-09T17:48:08.180089+00:00
2023-02-09T17:48:08.180144+00:00
697
false
# 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
0
['Python3']
1
number-of-arithmetic-triplets
Beats 97.36% || Number of Arithmetic Triplets
beats-9736-number-of-arithmetic-triplets-9x8p
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
himshrivas
NORMAL
2023-01-29T17:23:05.418261+00:00
2023-01-29T17:23:05.418307+00:00
1,779
false
# 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
0
['Python3']
4
number-of-arithmetic-triplets
O(N) Set + reduce solution
on-set-reduce-solution-by-ods967-xhzm
Code\n\n/**\n * @param {number[]} nums\n * @param {number} diff\n * @return {number}\n */\nvar arithmeticTriplets = function(nums, diff) {\n const set = new
ods967
NORMAL
2023-01-05T06:52:29.348880+00:00
2023-01-05T06:52:29.348926+00:00
417
false
# Code\n```\n/**\n * @param {number[]} nums\n * @param {number} diff\n * @return {number}\n */\nvar arithmeticTriplets = function(nums, diff) {\n const set = new Set();\n\n return nums.reduce((acc, num) => {\n set.add(num);\n\n if (set.has(num - diff) && set.has(num - 2 * diff)) {\n retur...
3
0
['JavaScript']
0
number-of-arithmetic-triplets
3 loops || JAVA || Easy to understand
3-loops-java-easy-to-understand-by-im_ob-0iqq
Intuition\n\nDon\'t forget to upvote if you found it useful\n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity:\
im_obid
NORMAL
2022-11-21T08:58:00.843601+00:00
2022-11-21T08:58:00.843633+00:00
1,012
false
# Intuition\n\nDon\'t forget to upvote if you found it useful\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:\nO(n^3)\n\n# Code\n```\nclass Solution {\n public int arithmeticTriple...
3
0
['Java']
0
number-of-arithmetic-triplets
O(n) Solution | Java
on-solution-java-by-pawan300-7681
```\n public int arithmeticTriplets(int[] nums, int diff) {\n int n = nums.length;\n int res =0;\n boolean arr[] = new boolean[nums[n-1]+di
pawan300
NORMAL
2022-09-30T18:09:30.960908+00:00
2022-09-30T18:09:30.960941+00:00
369
false
```\n public int arithmeticTriplets(int[] nums, int diff) {\n int n = nums.length;\n int res =0;\n boolean arr[] = new boolean[nums[n-1]+diff+1];\n for(int num:nums){\n arr[num]= true;\n }\n for(int i=0;i<n-2;i++){\n int val = nums[i];\n int val...
3
0
['Java']
0
number-of-arithmetic-triplets
First SC - O(1) Solution | TC - O(n) | CPP, C, JAVA
first-sc-o1-solution-tc-on-cpp-c-java-by-plt8
Please upvote If you like\n\nCPP\n\n\tunsigned short int count = 0,\n\ti = 0,\n\tj = i + 1,\n\tk = j + 1;\n\twhile (i < nums.size() && j < nums.size() && k < nu
rushil12
NORMAL
2022-09-13T03:41:59.724695+00:00
2022-09-15T03:59:31.908930+00:00
112
false
**Please upvote If you like**\n\n**CPP**\n\n\tunsigned short int count = 0,\n\ti = 0,\n\tj = i + 1,\n\tk = j + 1;\n\twhile (i < nums.size() && j < nums.size() && k < nums.size()) {\n\n\t\tif (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) {\n\t\t\tcount++;\n\t\t\ti++;\n\t\t\tj++;\n\t\t\tk++;\n\t\t\tcontinue;\n...
3
0
[]
0
number-of-arithmetic-triplets
[Java] Most easiest solution with explanation.
java-most-easiest-solution-with-explanat-trle
//This is a two pointer simple approach. Take a variable count to keep a count of the triplets.\n//initialiaze i=0; and then take a while loop which will run ti
vishady7
NORMAL
2022-08-20T19:53:29.879137+00:00
2022-08-20T19:53:29.879169+00:00
520
false
//This is a two pointer simple approach. Take a variable count to keep a count of the triplets.\n//initialiaze i=0; and then take a while loop which will run till i is less than nums.length-2.\n//inside this while loop take two variable j = i+1 and k = j+1;\n\n\nclass Solution {\n public int arithmeticTriplets(int[]...
3
0
['Two Pointers', 'Java']
0
number-of-arithmetic-triplets
✅✅Faster || Easy To Understand || C++ Code
faster-easy-to-understand-c-code-by-__kr-qbb3
Using Unordered Set\n\n Time Complexity :- O(N)\n\n Space Complexity :- O(N)\n\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int d
__KR_SHANU_IITG
NORMAL
2022-08-17T07:10:16.977521+00:00
2022-08-17T07:10:16.977552+00:00
399
false
* ***Using Unordered Set***\n\n* ***Time Complexity :- O(N)***\n\n* ***Space Complexity :- O(N)***\n\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n \n int n = nums.size();\n \n // insert all the elements into the set\n \n unordered_...
3
0
['C', 'C++']
1
number-of-arithmetic-triplets
C++ 0ms Faster than 100%
c-0ms-faster-than-100-by-husain_267-6i8r
class Solution {\npublic:\n int arithmeticTriplets(vector& nums, int diff) {\n int cnt = 0;\n unordered_setset;\n for(int i=0;i<nums.siz
husain_267
NORMAL
2022-08-13T10:28:36.635983+00:00
2022-08-13T10:28:36.636025+00:00
241
false
class Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int cnt = 0;\n unordered_set<int>set;\n for(int i=0;i<nums.size();i++){\n set.insert(nums[i]);\n }\n for(int i=0;i<nums.size();i++){\n if(set.find(nums[i]+diff)!=set.end() && s...
3
0
['C', 'Ordered Set']
0
number-of-arithmetic-triplets
Python Elegant & Short | Two solutions | O(n) and O(n*log(n)) | Binary search
python-elegant-short-two-solutions-on-an-1ekf
\tclass Solution:\n\t\t"""\n\t\tTime: O(n*log(n))\n\t\tMemory: O(1)\n\t\t"""\n\n\t\tdef arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n\t\t\tco
Kyrylo-Ktl
NORMAL
2022-08-12T10:27:36.955680+00:00
2022-08-12T10:35:41.767537+00:00
381
false
\tclass Solution:\n\t\t"""\n\t\tTime: O(n*log(n))\n\t\tMemory: O(1)\n\t\t"""\n\n\t\tdef arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n\t\t\tcount = 0\n\t\t\tleft, right = 0, len(nums) - 1\n\n\t\t\tfor j, num in enumerate(nums):\n\t\t\t\tif self.binary_search(nums, num - diff, left, j - 1) != -1 and \\...
3
0
['Binary Tree', 'Python', 'Python3']
1
number-of-arithmetic-triplets
O(n^3) Solution for Beginners
on3-solution-for-beginners-by-ssbaviskar-ufvb
\n\nlet arr = []\n for (let i=0;i<=nums.length-3;i++){\n for(let j=i+1;j<=nums.length-2;j++){\n for(let k = j+1;k<=nums.length-1;k++){\n
SSBaviskar
NORMAL
2022-08-09T13:38:29.314767+00:00
2022-09-25T18:03:59.899868+00:00
289
false
```\n\nlet arr = []\n for (let i=0;i<=nums.length-3;i++){\n for(let j=i+1;j<=nums.length-2;j++){\n for(let k = j+1;k<=nums.length-1;k++){\n if(nums[j]-nums[i]==diff && nums[k]-nums[j]==diff){\n arr.push([nums[i],nums[j],nums[k]]);\n }\n }\...
3
0
['JavaScript']
1
number-of-arithmetic-triplets
C++ Simple 2 pointer approach
c-simple-2-pointer-approach-by-hi2406-hnef
\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n for(int i=0;i<nums.size()-2;i++){\n int sta
hi2406
NORMAL
2022-08-07T21:14:47.401703+00:00
2022-08-07T21:14:47.401732+00:00
381
false
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n for(int i=0;i<nums.size()-2;i++){\n int start = i+1,end = nums.size()-1;\n while(start < end){\n if(nums[start]-nums[i] == diff && nums[end]-nums[start] == diff){\n co...
3
0
['Two Pointers', 'C']
0
number-of-arithmetic-triplets
C# Hashset
c-hashset-by-zloytvoy-595a
\tpublic int ArithmeticTriplets(int[] nums, int diff) \n\t\t{\n\t\t\tvar hashset = new HashSet();\n\t\t\tfor(int i = 0; i < nums.Length; ++i)\n\t\t\t{\n\t\t\t\t
zloytvoy
NORMAL
2022-08-07T20:27:41.355796+00:00
2022-08-07T20:27:41.355829+00:00
57
false
\tpublic int ArithmeticTriplets(int[] nums, int diff) \n\t\t{\n\t\t\tvar hashset = new HashSet<int>();\n\t\t\tfor(int i = 0; i < nums.Length; ++i)\n\t\t\t{\n\t\t\t\thashset.Add(nums[i]); \n\t\t\t}\n\n\t\t\tint result = 0;\n\t\t\tfor(int i = 0; i < nums.Length; ++i)\n\t\t\t{\n\t\t\t\tif(hashset.Contains(nums[i] + dif...
3
0
[]
0
number-of-arithmetic-triplets
C++ | O(n) | Map | Easy to understand | Commented & Readable
c-on-map-easy-to-understand-commented-re-p89e
\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n unordered_map<int, int> mp;\n int ans = 0;\n \n//
shreyanshxyz
NORMAL
2022-08-07T19:24:29.844077+00:00
2022-08-07T19:24:29.844117+00:00
60
false
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n unordered_map<int, int> mp;\n int ans = 0;\n \n// we check in the map if the number with the current number\'s difference & its 2x difference exists or not, if it does not, we simply push the current number ...
3
0
['C']
0
number-of-arithmetic-triplets
C++ solution || Using unordered map || 💯Faster
c-solution-using-unordered-map-faster-by-7qxt
\n int cnt = 0;\n unordered_map<int,bool> mp;\n for(int i=0;i<nums.size();i++)\n mp[nums[i]] = true;\n for(int i=0;i<nums.size(
Amanmalviya
NORMAL
2022-08-07T11:14:36.337316+00:00
2022-08-07T11:14:36.337356+00:00
402
false
```\n int cnt = 0;\n unordered_map<int,bool> mp;\n for(int i=0;i<nums.size();i++)\n mp[nums[i]] = true;\n for(int i=0;i<nums.size();i++)\n {\n if(mp[nums[i]-diff] && mp[nums[i]+diff])\n cnt++;\n } \n return cnt;\n```
3
0
['C++']
0
number-of-arithmetic-triplets
Latest Solution || Detailed Explanation || 2 Approaches
latest-solution-detailed-explanation-2-a-eoio
BIG IDEA: SAY YOU ARE GIVEN SINGLE NUMBER 6 AND DIFF = 3 , WHAT OTHER TWO NUMBERS DO YOU NEED TO SATISY CONDITIONS GIVEN IN THE PROBLEM? YOU NEED 6- DIFF AND 6
progressivefailure
NORMAL
2022-08-07T07:36:50.976604+00:00
2022-08-07T07:36:50.976640+00:00
188
false
# BIG IDEA: SAY YOU ARE GIVEN SINGLE NUMBER 6 AND DIFF = 3 , WHAT OTHER TWO NUMBERS DO YOU NEED TO SATISY CONDITIONS GIVEN IN THE PROBLEM? YOU NEED 6- DIFF AND 6 + DIFF. SO WE ARE SIMPLY DOING THAT, THE THING IS YOU ONLY NEED TO IDENTIFY THIS AND THEN THE QUESTION IS A CAKEWALK.\n# FIRST APPROACH \n# NOTE: AS IT IS GIV...
3
0
['Binary Tree', 'Python']
1
number-of-arithmetic-triplets
JAVA | easy to understand | for loop
java-easy-to-understand-for-loop-by-pazu-413s
Assume all number can be the first number of triplet\n2. if map[num - diff] == 1, it means that num is the second number of triplet\n\tor map[num - diff] == 2,
PazuC
NORMAL
2022-08-07T04:13:40.750241+00:00
2022-08-07T04:19:50.375160+00:00
706
false
1. Assume all number can be the first number of triplet\n2. if map[num - diff] == 1, it means that num is the second number of triplet\n\tor map[num - diff] == 2, it means that num is the third number of triplet\n\t...\n3. so, just need to count how many map[num] >= 3 (inside a triplet)\n\nin case 1, 2 triplets are [1,...
3
0
['Java']
1
number-of-arithmetic-triplets
✅Easy C++ || Looop || Basic Approach || No extra space needed!
easy-c-looop-basic-approach-no-extra-spa-jwac
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector& nums, int diff) {\n int n = nums.size();\n int c=0;\n for(int i=0;i<n-2;
gittyAditya
NORMAL
2022-08-07T04:08:17.716726+00:00
2022-08-07T04:08:17.716753+00:00
153
false
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int n = nums.size();\n int c=0;\n for(int i=0;i<n-2;++i)\n for(int j=i+1;j<n-1;++j)\n for(int k=j+1;k<n;++k)\n if(nums[j]-nums[i] == diff && nums[k]-nums[j] == di...
3
0
['C']
0
number-of-arithmetic-triplets
Python Solution
python-solution-by-swanand_joshi-flt9
IntuitionApproachComplexity Time complexity: Space complexity: Code
Swanand_Joshi
NORMAL
2025-02-21T18:32:09.251736+00:00
2025-02-21T18:32:09.251736+00:00
61
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code `...
2
0
['Python3']
0
number-of-arithmetic-triplets
Number of Arithmetic Triplets || Python || Easy Solution
number-of-arithmetic-triplets-python-eas-db42
Step-by-Step Explanation1. Initialize a Counter Start with triplet_count set to 0. This variable will store the number of valid arithmetic triplets. 2. Iterate
aishvgandhi
NORMAL
2025-01-23T06:14:17.230517+00:00
2025-01-23T06:14:17.230517+00:00
113
false
# Step-by-Step Explanation ## 1. **Initialize a Counter** ```python triplet_count = 0 ``` - Start with `triplet_count` set to 0. This variable will store the number of valid arithmetic triplets. ## 2. **Iterate Over `nums` Using a Set** ```python for i in set(nums): ``` - Convert `nums` into a ...
2
0
['Array', 'Two Pointers', 'Python3']
1
number-of-arithmetic-triplets
SIMPLE BRUTEFORCE C++ SOLUTION
simple-bruteforce-c-solution-by-jeffrin2-5ce3
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
Jeffrin2005
NORMAL
2024-07-18T12:39:56.990900+00:00
2024-07-18T12:39:56.990932+00:00
150
false
# 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:o(n^3)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:o(1)\n<!-- Add your space complexity here, e.g. $...
2
0
['C++']
0
number-of-arithmetic-triplets
⭐Creative solution using HashMap (w/ detailed explanation)
creative-solution-using-hashmap-w-detail-2lse
Observation:\n\nIn an arithmetic sequence of length n, there are n - 3 + 1 arithmetic triplets. For example, the sequence [2, 4, 6, 8] contains 2 arithmetic tri
Nature711
NORMAL
2024-05-25T07:56:33.402993+00:00
2024-05-25T07:56:33.403013+00:00
124
false
**Observation**:\n\nIn an arithmetic sequence of length `n`, there are `n - 3 + 1` arithmetic triplets. For example, the sequence `[2, 4, 6, 8]` contains 2 arithmetic triplets: `[2, 4, 6]` and `[4, 6, 8]`.\n\nGiven that the input array is sorted, we can scan from the start to find all possible "longest" arithmetic sequ...
2
0
['Java']
1
number-of-arithmetic-triplets
Efficient O(n) Solution for Counting Arithmetic Triplets in TypeScript | JavaScript
efficient-on-solution-for-counting-arith-e5jb
Intuition\n Describe your first thoughts on how to solve this problem. \n\n1. Create a Counter Hash:\nInitialize a hash to keep track of the number of arithmeti
sameeneeti
NORMAL
2023-09-22T18:54:43.905645+00:00
2023-09-26T14:40:47.650834+00:00
314
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n1. Create a Counter Hash:\nInitialize a hash to keep track of the number of arithmetic triplets found so far.\n\n2. Iterate Through the Array:\nIterate through the array nums. For each element nums[j], calculate the potential previous e...
2
0
['TypeScript', 'JavaScript']
0
number-of-arithmetic-triplets
[C++] check out this simple constant space solution.
c-check-out-this-simple-constant-space-s-3d8a
Code - 1 (Using Binary Search)\n- Time Complexity : O(nlogn)\n- Space Complexity : O(1)\n\n\nclass Solution {\npublic:\n int bs (vector<int> &nums, int &key,
vanshdhawan60
NORMAL
2023-09-07T22:43:47.953613+00:00
2023-09-07T22:43:47.953629+00:00
485
false
# Code - 1 (Using Binary Search)\n- **Time Complexity :** $$O(n*logn)$$\n- **Space Complexity :** $$O(1)$$\n\n```\nclass Solution {\npublic:\n int bs (vector<int> &nums, int &key, int start, int end) {\n int low = start, high = end;\n while (low <= high) {\n int mid = low+(high-low)/2;\n ...
2
0
['C++']
0
number-of-arithmetic-triplets
Binary Search approach in C++
binary-search-approach-in-c-by-hustling_-5yww
Intuition\n Describe your first thoughts on how to solve this problem. \nIntution is all about arithmatic mean concept . We should check wheather nums[i]+diff &
Hustling_lad
NORMAL
2023-07-23T07:53:56.210954+00:00
2023-07-23T07:53:56.210987+00:00
293
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntution is all about arithmatic mean concept . We should check wheather nums[i]+diff & nums[i]+2*diff is present in the vector or not.If true then increase count then return count\n\n# Approach\n<!-- Describe your approach to solving the...
2
0
['C++']
0
number-of-arithmetic-triplets
Easy-Simple Java Solution in O(n^2)
easy-simple-java-solution-in-on2-by-tauq-dju4
\n# Approach\n Describe your approach to solving the problem. \nWe have to find a triplet of (i, j, k) such that \ni<j=2 we add 1 to ans and at last return it.\
tauqueeralam42
NORMAL
2023-07-02T09:58:16.819238+00:00
2023-07-02T09:58:16.819264+00:00
492
false
\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe have to find a triplet of (i, j, k) such that \ni<j<k , \nnums[j] - nums[i] == diff, and \nnums[k] - nums[j] == diff\n\nin other word we can also say nums[k]-nums[i]== 2*diff. To reduce the time complexity I use only 2 loop remove the loop for k ...
2
0
['Array', 'Java']
0
number-of-arithmetic-triplets
Simple-Easy Brute Force Approach
simple-easy-brute-force-approach-by-tauq-f8hy
\n\n# Approach\n Describe your approach to solving the problem. \nHere I use simple brute force approach. Use 3 for loop iterarte i, j, k one by one and search
tauqueeralam42
NORMAL
2023-07-02T09:36:23.699190+00:00
2023-07-02T09:36:23.699218+00:00
694
false
\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nHere I use simple brute force approach. Use 3 for loop iterarte i, j, k one by one and search a combinantion of i, j, k which can satisfied these given conditions and count that combinations.\n\n# Complexity\n- Time complexity: O(n^3)\n<!-- Add yo...
2
0
['Array', 'Java']
0
number-of-arithmetic-triplets
Faster than 100%✅✅ || C++ || Very easy-logic with explanation
faster-than-100-c-very-easy-logic-with-e-oklg
Proof :please upvote if you liked my solution\n\n\nLogic: \nIf you have attempted 2sum leetcode question this code will be quite easy for you.The simple logic b
40metersdeep
NORMAL
2023-05-29T17:44:39.908014+00:00
2023-05-29T17:52:01.092773+00:00
437
false
**Proof :**_please upvote if you liked my solution_\n![image](https://assets.leetcode.com/users/images/5a09d233-c846-4b86-9c48-d5ec8f70b000_1685382160.8579662.png)\n\n**Logic:** \nIf you have attempted 2sum leetcode question this code will be quite easy for you.The simple logic behind the code is that there for every e...
2
0
['C']
0
number-of-arithmetic-triplets
Fastest at 1ms
fastest-at-1ms-by-mohammad__irfan-5rqu
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
mohammad__irfan
NORMAL
2023-05-02T11:33:19.669124+00:00
2023-05-02T11:33:19.669166+00:00
1,134
false
# 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
0
['Java']
3
number-of-arithmetic-triplets
JAVA | 1ms | O(N^2)
java-1ms-on2-by-firdavs06-ivdd
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
Firdavs06
NORMAL
2023-02-27T07:19:04.038382+00:00
2023-02-27T07:19:04.038426+00:00
826
false
# 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
0
['Java']
0
number-of-arithmetic-triplets
pointer manipulation in C
pointer-manipulation-in-c-by-fantsai8686-hhdp
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nex:\n[0,1,4,6,7,10]\ntr
fantsai8686
NORMAL
2023-02-21T11:27:39.885087+00:00
2023-03-29T08:21:09.288216+00:00
468
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nex:\n[0,1,4,6,7,10]\ntraversing nums in for loop from index1= 0~numsSize-3 (0,1,4,6).\n\nThere is a while loop in each for loop starting from index2= 1~numsSize-1 to c...
2
0
['C']
0
number-of-arithmetic-triplets
Solution in python easy to understand
solution-in-python-easy-to-understand-by-3prr
Input: nums = [0,1,4,6,7,10], diff = 3\nOutput: 2\nExplanation:\n(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.\n(2, 4, 5) is an ari
mittalaparna
NORMAL
2023-01-31T19:45:13.189850+00:00
2023-01-31T19:45:13.189887+00:00
83
false
Input: nums = [0,1,4,6,7,10], diff = 3\nOutput: 2\nExplanation:\n(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.\n(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.\n\nCode explaination:\n((nums[i] + diff) in arr) and ((nums[i] + 2 * diff) in arr)\n = 1 + 3 i...
2
0
['Hash Table', 'Python3']
0
number-of-arithmetic-triplets
Hashset n-diff and n-2*diff
hashset-n-diff-and-n-2diff-by-mohan_66-a6xg
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
Mohan_66
NORMAL
2023-01-28T05:26:46.572858+00:00
2023-01-28T05:26:46.572903+00:00
318
false
# 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
0
['Python3']
0
number-of-arithmetic-triplets
Binary Search || by searching || without any space🔥
binary-search-by-searching-without-any-s-8l31
INTUTION: \n\n1. search no+diff and no+2diff by the help of lower bound. \n1. if both numbers present in array then just count++\n2. same thing is done for each
suyashdewangan9
NORMAL
2022-10-18T16:40:41.814627+00:00
2022-10-18T16:51:54.188030+00:00
195
false
##### INTUTION: \n\n1. search no+diff and no+2diff by the help of lower bound. \n1. if both numbers present in array then just count++\n2. same thing is done for each element of nums\n\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n int count=0;\n for(int i=0;i...
2
0
['Binary Search', 'C']
0
number-of-arithmetic-triplets
100% faster || c++ || simple
100-faster-c-simple-by-recursive_coder-xvhd
\n\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n \n vector<int> hash(305,0); // for storing cnt of each
recursive_coder
NORMAL
2022-10-08T07:03:09.372355+00:00
2022-10-08T07:03:48.649782+00:00
908
false
![image](https://assets.leetcode.com/users/images/1408c3f5-e53c-4766-adc5-f66963609103_1665212620.8311636.png)\n```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n \n vector<int> hash(305,0); // for storing cnt of each element in the array\n for(int i = 0 ; i...
2
0
['C']
0
number-of-arithmetic-triplets
Python O(n log n) time, O(1) space solution with comments
python-on-log-n-time-o1-space-solution-w-36fh
Here is my optimized code that has:\nTime: O(n log n)\nSpace: O(1)\n\npython\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n n =
trpaslik
NORMAL
2022-09-16T10:53:06.383950+00:00
2022-09-16T10:53:40.646376+00:00
279
false
Here is my optimized code that has:\nTime: O(n log n)\nSpace: O(1)\n\n```python\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n n = len(nums)\n ans = 0\n for i in range(n-2):\n num0 = nums[i] # current num\n num1 = num0 + diff # expected nums[j]\n ...
2
0
['Python']
1
number-of-arithmetic-triplets
SIMPLE JAVA CODE
simple-java-code-by-priyankan_23-8gp2
\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n int count=0;\n for(int i=0;i<nums.length-2;i++){\n int
priyankan_23
NORMAL
2022-09-11T14:05:06.728079+00:00
2022-09-11T14:05:06.728114+00:00
188
false
```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n int count=0;\n for(int i=0;i<nums.length-2;i++){\n int j=i+1,check=0;\n while(j<nums.length){\n if(nums[j]-nums[i]==diff){\n check=1;\n break;\n ...
2
0
[]
0
number-of-arithmetic-triplets
Python | Linear | Fast | Explained with Comments
python-linear-fast-explained-with-commen-p3t9
\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n \n # a variable to store number of triplets\n # i
rohitsh26
NORMAL
2022-09-08T04:06:30.264182+00:00
2022-09-08T04:23:09.232989+00:00
160
false
```\nclass Solution:\n def arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n \n # a variable to store number of triplets\n # initializing with 0\n counterOfUniqueArithmeticTriplet = 0\n \n # create a set of number in the nums list\n _set = set(nums)\n ...
2
0
['Python']
1
number-of-arithmetic-triplets
Brute force solution of java
brute-force-solution-of-java-by-annshu_8-vq6z
\tclass Solution {\n\t\tpublic int arithmeticTriplets(int[] nums, int diff) {\n\t\t int count = 0;\n\t\t\tfor(int i= 0; i< nums.length ; i++){\n\t\t\t\tfor( i
annshu_8410
NORMAL
2022-09-07T19:25:58.885742+00:00
2022-09-07T19:25:58.885780+00:00
53
false
\tclass Solution {\n\t\tpublic int arithmeticTriplets(int[] nums, int diff) {\n\t\t int count = 0;\n\t\t\tfor(int i= 0; i< nums.length ; i++){\n\t\t\t\tfor( int j = i+1; j < nums.length ; j++){\n\t\t\t\t\tfor( int k = j+1; k < nums.length ; k++){\n\t\t\t\t\t\tif(i < j && j < k &&\n\t\t\t\t\t\t nums[j] - nums[i] == d...
2
0
['Java']
0
number-of-arithmetic-triplets
FULLY EXPLAINED - O(n^3), O(n^2) and O(n) Solutions
fully-explained-on3-on2-and-on-solutions-3ccu
1. O(n^3)\n\nThe straightforward way is to have three nested loops and then check the conditions. But ofcourse this won\'t be an efficient approach and will giv
itsarvindhere
NORMAL
2022-09-07T13:13:49.275515+00:00
2022-09-07T13:54:05.495455+00:00
223
false
## **1. O(n^3)**\n\nThe straightforward way is to have three nested loops and then check the conditions. But ofcourse this won\'t be an efficient approach and will give TLE in case of large input. \n\n```\ndef arithmeticTriplets(self, nums: List[int], diff: int) -> int:\n count = 0;\n for i,num1 in enumer...
2
0
['Python']
0
number-of-arithmetic-triplets
C+++ || 0ms Solution || Using Constant Space
c-0ms-solution-using-constant-space-by-k-hnzj
\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n \n int count = 0;\n \n for(int i = 0 ; i<num
Khwaja_Abdul_Samad
NORMAL
2022-09-02T13:04:59.492129+00:00
2022-09-02T13:04:59.492168+00:00
245
false
```\nclass Solution {\npublic:\n int arithmeticTriplets(vector<int>& nums, int diff) {\n \n int count = 0;\n \n for(int i = 0 ; i<nums.size()-2 ; i++)\n {\n int j = i+1 ;\n while(j<nums.size()-1 && nums[j]-nums[i] < diff)\n j++; \n ...
2
0
['C']
1
number-of-arithmetic-triplets
Java | beats 69% lol | easy and simple solution |
java-beats-69-lol-easy-and-simple-soluti-vz3y
```\nint ans = 0;\n //we take an answer that we will be returning ;\n \n for(int i = nums.length-1 ; i >=0 ; i--){//for the first element
k_io
NORMAL
2022-08-21T09:22:54.137440+00:00
2022-08-21T09:22:54.137467+00:00
214
false
```\nint ans = 0;\n //we take an answer that we will be returning ;\n \n for(int i = nums.length-1 ; i >=0 ; i--){//for the first element of the triplet\n \n meow :\n for(int j = nums.length-1 ; j >=0 ; j--){//for the second element of the triplet ;\n if...
2
0
['Array', 'Java']
0
number-of-arithmetic-triplets
Java solution | O(n)
java-solution-on-by-sourin_bruh-gt5i
\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n \n Set<Integer> set = new HashSet<>();\n \n for (int
sourin_bruh
NORMAL
2022-08-21T08:03:11.503787+00:00
2022-08-21T09:36:28.388880+00:00
115
false
```\nclass Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n \n Set<Integer> set = new HashSet<>();\n \n for (int x : nums) set.add(x);\n \n int ans = 0;\n \n for (int x : nums) {\n if (set.contains(x - diff) && set.contains(x + di...
2
0
['Java']
2
number-of-arithmetic-triplets
Java Solution | Simple solution
java-solution-simple-solution-by-shiladi-149f
class Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set set = new HashSet<>();\n for (int x : nums) {\n set.a
shiladityaroy
NORMAL
2022-08-14T20:01:26.366631+00:00
2022-08-14T20:01:26.366660+00:00
293
false
class Solution {\n public int arithmeticTriplets(int[] nums, int diff) {\n Set<Integer> set = new HashSet<>();\n for (int x : nums) {\n set.add(x);\n }\n \n int ans = 0;\n for (int x : nums) {\n if (set.contains(x - diff) && set.contains(x + diff)) {\n ...
2
0
['Java']
1
reordered-power-of-2
[C++/Java/Python] Straight Forward
cjavapython-straight-forward-by-lee215-5nse
counter will counter the number of digits 9876543210 in the given number.\nThen I just compare counter(N) with all counter(power of 2).\n1 <= N <= 10^9, so up t
lee215
NORMAL
2018-07-15T03:07:40.854887+00:00
2018-10-25T02:53:21.915764+00:00
23,923
false
`counter` will counter the number of digits 9876543210 in the given number.\nThen I just compare `counter(N)` with all `counter(power of 2)`.\n`1 <= N <= 10^9`, so up to 8 same digits.\nIf `N > 10^9`, we can use a hash map.\n\n**C++:**\n```\n bool reorderedPowerOf2(int N) {\n long c = counter(N);\n for...
325
4
[]
47
reordered-power-of-2
Simple Java Solution Based on String Sorting
simple-java-solution-based-on-string-sor-cg66
The idea here is similar to that of group Anagrams problem (Leetcode #49). \n\nFirst, we convert the input number (N) into a string and sort the string. Next, w
trotro
NORMAL
2018-07-20T23:27:49.802904+00:00
2018-07-20T23:27:49.802904+00:00
5,807
false
The idea here is similar to that of group Anagrams problem (Leetcode #49). \n\nFirst, we convert the input number (N) into a string and sort the string. Next, we get the digits that form the power of 2 (by using 1 << i and vary i), convert them into a string, and then sort them. As we convert the powers of 2 (and there...
143
0
[]
12
reordered-power-of-2
C++ Super Simple and Short Solution, 0 ms, Faster than 100%
c-super-simple-and-short-solution-0-ms-f-h8kl
\nclass Solution {\npublic:\n bool reorderedPowerOf2(int N) {\n string N_str = sorted_num(N);\n for (int i = 0; i < 32; i++)\n if (N
yehudisk
NORMAL
2021-03-21T08:35:30.265019+00:00
2021-03-21T08:35:36.558397+00:00
7,636
false
```\nclass Solution {\npublic:\n bool reorderedPowerOf2(int N) {\n string N_str = sorted_num(N);\n for (int i = 0; i < 32; i++)\n if (N_str == sorted_num(1 << i)) return true;\n return false;\n }\n \n string sorted_num(int n) {\n string res = to_string(n);\n sor...
108
13
['C']
12
reordered-power-of-2
JS, Python, Java, C++ | Easy, Short Solution w/ Explanation
js-python-java-c-easy-short-solution-w-e-hcjo
(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, please upvote this post.)\n\n---\n\n#### Idea:\n
sgallivan
NORMAL
2021-03-21T08:05:47.270873+00:00
2021-03-21T09:04:21.120695+00:00
3,967
false
*(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful,* ***please upvote*** *this post.)*\n\n---\n\n#### ***Idea:***\n\nThe easiest way to check if two things are shuffled versions of each other, which is what this problem is asking us to do, is to sort them bot...
71
12
['C', 'Python', 'Java', 'JavaScript']
2
reordered-power-of-2
C++ || 100% fast || 0ms
c-100-fast-0ms-by-manuraj_tomar-q8im
\nbool reorderedPowerOf2(int n) {\n string s = to_string(n);\n sort(s.begin(),s.end());\n\t\t\n vector<string> power;\n for(int i=0;
Manuraj_Tomar
NORMAL
2022-08-26T00:25:33.346170+00:00
2022-08-26T00:25:33.346214+00:00
11,209
false
```\nbool reorderedPowerOf2(int n) {\n string s = to_string(n);\n sort(s.begin(),s.end());\n\t\t\n vector<string> power;\n for(int i=0;i<=30;i++){\n int p = pow(2,i);\n power.push_back(to_string(p));\n }\n \n for(int i=0;i<=30;i++){\n sor...
70
0
['C']
17
reordered-power-of-2
Quick Math
quick-math-by-fllght-8pw1
The main idea is what powers of 2 look like in binary form:\n\n\n\nthis will help easily iterate over all powers of two\n\nSo all we need is to convert n to sor
FLlGHT
NORMAL
2022-08-26T03:07:13.630261+00:00
2023-05-12T05:19:51.907428+00:00
6,415
false
The main idea is what powers of 2 look like in binary form:\n\n<img width="200" src="https://assets.leetcode.com/users/images/ae9964b4-ffbd-40a5-a934-e3dc1de49d28_1661482426.3498342.png">\n\nthis will help easily iterate over all powers of two\n\nSo all we need is to convert n to sorted digits and then compare them wit...
59
0
['C', 'Python', 'Java']
14
reordered-power-of-2
✅Reordered Power of 2 | Short & Easy w/ Explanation | Beats 100%
reordered-power-of-2-short-easy-w-explan-sn76
Solution - I (Counting Digits Frequency)\n\nA simple solution is to check if frequency of digits in N and all powers of 2 less than 10^9 are equal. In our case,
archit91
NORMAL
2021-03-21T09:16:40.942964+00:00
2021-03-21T14:52:30.030476+00:00
3,360
false
***Solution - I (Counting Digits Frequency)***\n\nA simple solution is to check if frequency of digits in N and all powers of 2 less than `10^9` are equal. In our case, we need to check for all powers of 2 from `2^0` to `2^29` and if any of them matches with digits in `N`, return true.\n\n```\n// counts frequency of ea...
42
1
['C']
5
reordered-power-of-2
✅Easy Java Solution || 100% Faster 1ms 🔥🔥 || With Basic Explanation
easy-java-solution-100-faster-1ms-with-b-appb
If you really found my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community, if you have some queries
ganajayant
NORMAL
2022-08-26T05:22:19.464149+00:00
2022-08-26T06:04:02.776417+00:00
2,943
false
**If you really found my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community, if you have some queries or some improvements please feel free to comment and share your views.**\n\nbasically we are finding occurrence of each digit in n after that we are finding s...
36
0
['Math', 'Counting', 'Java']
12