Difference between revisions of "Neet Code Problems"
| Line 158: | Line 158: | ||
This approach checks for lengths greater than 1 byrow, and cumuluate with OR over the entire array. | This approach checks for lengths greater than 1 byrow, and cumuluate with OR over the entire array. | ||
| + | |||
| + | Another solution: | ||
| + | n=[1,2,3,1]; | ||
| + | ∏((n⍣)#1n)>1 | ||
| + | |||
| + | Answer: true | ||
| + | |||
| + | ∏ Product | ||
| + | n⍣ gives FREQUENCY of each element of n | ||
| + | #1n extracts all elements from column 1 (note use of 1n for column n) | ||
| + | >1 Checking if the product is greater than 1 | ||
=Other Interesting Solutions= | =Other Interesting Solutions= | ||
Revision as of 11:54, 12 September 2024
Neet Code Solutions
Thinking in z^3
This is a quick introduction to the z^3 language (pronounced as "zcubes language").
ZCubes Language (z^3) is an easy to write, natural to read, omni-functional language.
Learn z^3 Language in detail by clicking this link.
- ZCubes Language Documentation
- z^3 Language Detail
- z^3 Commands and Operators
- Z Functions, Member Functions, etc.
Introduction
Following are z^3 Solutions to YouTube Code Report Channel Videos. To learn how to code in ZCubes language, please visit [https://coding.zcubes.com]. ZCubes Web Platform can be started in Code Mode at [https://code.zcubes.com].
Problems and Solutions
Product of Array Except Self
Video: https://www.youtube.com/watch?v=bNvIQI2wAjk [Product of Array Except Self - Leetcode 238 - Python]
Problem Description:
Given an integer array nums, return an array output where output[i] is the product of all the elements of nums except nums[i].
Each product is guaranteed to fit in a 32-bit integer.
Follow-up: Could you solve it in O ( n ) O(n) time without using the division operation?
Example 1:
Input: nums = [1,2,4,6]
Output: [48,24,12,8] Example 2:
Input: nums = [-1,0,1,2,3]
Output: [0,-6,0,0,0]
z^3 Solution
n=[1,2,4,6]; n⪥n≡∏
Gives answer
| 24 |
| 12 |
| 48 |
| 24
|
n=[-1,1,0,-3,3]; n⪥n≡∏
Gives answer:
| 0 |
| 0 |
| 0 |
| 0 |
| 0
|
Notes: ⪥ is the .around operator on an array, which takes one or more indices from the second argument to give you an array except that index. ≡∏ Product (∏) conducted by row (≡)
n⪥n can be written as n.around(n) also.
Duplicate Integer
Video: https://www.youtube.com/watch?v=3OamzN90kPg [Contains Duplicate - Leetcode 217 - Python]
Problem Description:
Given an integer array nums, return true if any value appears more than once in the array, otherwise return false.
Example 1:
Input: nums = [1, 2, 3, 3]
Output: true Example 2:
Input: nums = [1, 2, 3, 4]
Output: false
z^3 Solution
n=[1,2,3,1]; (n∪).$(x=>n⒤x)
Gives answer
| 0 | 3 |
| 1 | |
| 2
|
Notes: n∪ Uniques of n array .$ Apply function to each element x=>n⒤x Function x, defined as n find indices that contain x ≡ can be used to loop to check greater than 1 etc
n=[1,2,3,1]; ((n∪).$(x=>n⒤x)≡(x=>x#>1))▣OR
This approach checks for lengths greater than 1 byrow, and cumuluate with OR over the entire array.
Another solution: n=[1,2,3,1]; ∏((n⍣)#1n)>1
Answer: true
∏ Product n⍣ gives FREQUENCY of each element of n
- 1n extracts all elements from column 1 (note use of 1n for column n)
>1 Checking if the product is greater than 1