Ifihan

from flask import Flask
from flask_restful import Resource, Api

app = Flask(__name__)
api = Api(app)

class Student(Resource):
    def get(self, name):
        return {"student": name}

Ifihan

import random

# prb_score = int(input("Enter the probability score: "))
def cross_over(parent1, parent2):
    child1 = []
    child2 = []
    for i in range(len(parent1)):
        if random.random() < 0.5:
            child1.append(parent1[i])
            child2.append(parent2[i])

Ifihan

import random

def mutation(individual, mutation_rate):
    for i in range(len(individual)):
        if random.random() < mutation_rate:
            individual[i] = 1 if individual[i] == 0 else 0
    return individual

Ifihan

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(len(nums) - 1):
            for j in range(i + 1, len(nums)):
                if nums[i] + nums[j] == target:
                    return i,j     
            
        

Ifihan

Permutations II

This is a follow up problem from Permutations. This goal of this problem is to return all possible unique results from the input of an array.

Solution

The solution I followed was Backtracking. Looks like a DFS (Depth-First Search) problem 🤔

Create a hashmap to take count of all number to leave have a non-reapeating digit or eliminate duplicate. Then the bactrack the decision tree.

Ifihan

using Plots
# define the Lorenz attractor
Base.@kwdef mutable struct Lorenz
    dt::Float64 = 0.02
    σ::Float64 = 10
    ρ::Float64 = 28
    β::Float64 = 8/3
    x::Float64 = 1
    y::Float64 = 1
    z::Float64 = 1

Ifihan

Two Sum

Question on Leetcode - Easy

Approach

There were three approaches taken to solve this question.

1. Use of two for-loops.
2. I call it the x = z - y (from the equation x + y = z).
3. Two pointer approach.

The two pointer approach was the approach used. In this approach, the array is first sortted from the lowest to the highest number.

Ifihan

Search Insert Position

Question on Leetcode - Easy

Approach

The approach taken was the binary search to find the target. I initialized two pointers, left and right at the beginning and end of the array respectfully. Then I used a while loop left < right. Then in the loop, I calculate the mid and compare it to the target. If the mid is the target, I return mid. If the mid is less than the target, I add a unit step to the left and if the mid is grater than one, I remove a unit step from the

Ifihan

21. Merge Two Sorted Lists

Question: Leetcode - Easy

Approach

The approach to use two pointers: The first one was a dummy ListNode() and the second is a tail. Then I traverse throught the list by appending the smaller values of either lists in the tail pointer.

Once the end of the list is gotten to, I then append the remaining nodes to the linked list.

Ifihan

Is Subsequence

Question on Leetcode - Easy

Approach

The approach taken was to use a pointer intialized at 0, for the sequence s string. This pointer servers as a counter which the increases for every positive match against the main string. Afterwards, I traversed through the main string t and returned the a boolean of if the pointer value was the same as the length of the sequence s.

Code

class Solution: