Difference between revisions of "Neet Code Problems"

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==Products of Array Discluding Self==
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==Product of Array Except Self==
  
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Video: https://www.youtube.com/watch?v=bNvIQI2wAjk [I ❤ APL and Haskell #2 ]
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Video: https://www.youtube.com/watch?v=bNvIQI2wAjk [Product of Array Except Self - Leetcode 238 - Python]
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===z^3 Solution===
 
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<pre>
 
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a=[-1,-2,-3,4,3,2,1];
 
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∏±a
 
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</pre>
 
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Gives answer -1
 
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∏ is Product, ± is SIGN function. APL has 3 characters it its solution (selected as the best solution in the video because of its brevity). Interesting that the solution with z^3 has even less characters, just 2.
 
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==Powerful Solution to Indexing Problem (1 character in z^3 vs 3 in APL)==
 
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Video: https://www.youtube.com/watch?v=ctbGMuakpHk [Product of Array Except Self - Leetcode 238 - Python]
 
  
 
Problem Description:
 
Problem Description:
Line 165: Line 151:
 
x=>n⒤x Function x, defined as n find indices that contain x  
 
x=>n⒤x Function x, defined as n find indices that contain x  
 
≡ can be used to loop to check greater than 1 etc
 
≡ can be used to loop to check greater than 1 etc
 +
 +
<pre>
 +
n=[1,2,3,1];
 +
((n∪).$(x=>n⒤x)≡(x=>x#>1))▣OR
 +
</pre>
 +
 +
This approach checks for lengths greater than 1 byrow, and cumuluate with OR over the entire array.
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 +
Another solution:
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n=[1,2,3,1];
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∏((n⍣)#1n)>1
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 +
Answer: true
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∏ Product
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n⍣ gives FREQUENCY of each element of n
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#1n extracts all elements from column 1 (note use of 1n for column n)
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>1 Checking if the product is greater than 1
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 +
 +
==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:
 +
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Input: coins = [1,5,10], amount = 12
 +
 +
Output: 3
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Explanation: 12 = 10 + 1 + 1. Note that we do not have to use every kind coin available.
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 +
Example 2:
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Input: coins = [2], amount = 3
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 +
Output: -1
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Explanation: The amount of 3 cannot be made up with coins of 2.
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 +
Example 3:
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Input: coins = [1], amount = 0
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Output: 0
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Explanation: Choosing 0 coins is a valid way to make up 0.
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 +
Constraints:
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1 <= coins.length <= 10
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1 <= coins[i] <= 2^31 - 1
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0 <= amount <= 10000
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 +
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===z^3 Solution===
 +
<pre>
 +
c= [2];o=t=3;
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m=0;
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s=[];
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(c⋱).$(i=>[n=i⟌t,m+=n*i;t-=n*i,s.push([n,i])]);
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m==o?∑(s#0n):-1;
 +
    // Gives Answer -1
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 +
Other inputs:
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c= [1,5,10];o=t=12;
 +
    // Gives Answer 3
 +
 +
c= [2];o=t=3;
 +
    // Gives Answer -1
 +
   
 +
   
 +
c= [1];o=t=0;
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    // Gives Answer 0
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</pre>
 +
 +
Notes:
 +
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⋱  Sorts Descending
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 +
.$  Apply function to each element
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i⟌t Integer Division of t by i
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m+=n*i Cumulate total used in m
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t-=n*i Reduce Outstanding
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s      Keeps track of coins and counts used
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m==o  If total used matches original amount exactly
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 +
∑(s#0n) Give total sum of the 1st column of s (Could have used o or m here)
 +
 +
-1    Or return -1
  
 
=Other Interesting Solutions=
 
=Other Interesting Solutions=

Latest revision as of 12:23, 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.

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

  1. 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

Other Interesting Solutions

See Related

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