Neet Code Problems
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∪).$(x=>n⒤x)≡(x=>x#>1))▣OR
This approach checks for lengths greater than 1 byrow, and cumuluate with OR over the entire array.
Another solution: n=[1,2,3,1]; ∏((n⍣)#1n)>1
Answer: true
∏ Product n⍣ gives FREQUENCY of each element of n
- 1n extracts all elements from column 1 (note use of 1n for column n)
>1 Checking if the product is greater than 1
Coin Change Problem
Video: https://www.youtube.com/watch?v=H9bfqozjoqs [Coin Change - Dynamic Programming Bottom Up - Leetcode 322]
Problem Description:
You are given an integer array coins representing coins of different denominations (e.g. 1 dollar, 5 dollars, etc) and an integer amount representing a target amount of money.
Return the fewest number of coins that you need to make up the exact target amount. If it is impossible to make up the amount, return -1.
You may assume that you have an unlimited number of each coin.
Example 1:
Input: coins = [1,5,10], amount = 12
Output: 3 Explanation: 12 = 10 + 1 + 1. Note that we do not have to use every kind coin available.
Example 2:
Input: coins = [2], amount = 3
Output: -1 Explanation: The amount of 3 cannot be made up with coins of 2.
Example 3:
Input: coins = [1], amount = 0
Output: 0 Explanation: Choosing 0 coins is a valid way to make up 0.
Constraints:
1 <= coins.length <= 10 1 <= coins[i] <= 2^31 - 1 0 <= amount <= 10000
z^3 Solution
c= [2];o=t=3;
m=0;
s=[];
(c⋱).$(i=>[n=i⟌t,m+=n*i;t-=n*i,s.push([n,i])]);
m==o?∑(s#0n):-1;
// Gives Answer -1
Other inputs:
c= [1,5,10];o=t=12;
// Gives Answer 3
c= [2];o=t=3;
// Gives Answer -1
c= [1];o=t=0;
// Gives Answer 0
Notes:
⋱ Sorts Descending
.$ Apply function to each element
i⟌t Integer Division of t by i
m+=n*i Cumulate total used in m
t-=n*i Reduce Outstanding
s Keeps track of coins and counts used
m==o If total used matches original amount exactly
∑(s#0n) Give total sum of the 1st column of s (Could have used o or m here)
-1 Or return -1