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‎#hacktober2k19#cci‎

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Thanks to Mukul And Rajat
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Thanks to Mukul And Rajat. Your video is still helpful in 2021
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Thanks to Mukul And Rajat.
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Thanks to Mukul And Rajat. Your video is still helpful in 2021.

‎01-Regression.ipynb‎

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‎1's&2'sCompliment.cpp‎

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#include <iostream>
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#include<string>
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#include<cstdio>
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using namespace std;
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int main()
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{
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int len=8;
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char bin[len],one[len],two[len];
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cout<<"Enter the Binary Number:";
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gets(bin);
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int flag =0;
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for(int i=0;i<len;i++)//one's compliment
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{
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if(bin[i]=='1')
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{
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one[i]='0';
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}
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else if(bin[i]=='0')
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{
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one[i]='1';
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}
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else
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{
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cout<<"Invalid Binary Digits";
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flag=1;
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return 0;
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}
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}
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int carry=1;
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for(int i=len-1;i>=0;i--)//two's compliment
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{
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if(one[i]=='1' && carry==1)
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{
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two[i]='0';
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}
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else if(one[i]=='0' && carry==1)
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{
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two[i]='1';
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carry=0;
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}
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else
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{
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two[i]=one[i];
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}
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}
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if(flag==0)
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{
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cout<<"Binary number is:"<<bin<<endl;
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cout<<"1's Compliment of the number is:"<<one<<endl;
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cout<<"2's Compliment of the number is:"<<two<<endl;
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}
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return 0;
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}

‎2433. Find The Original Array of Prefix Xor‎

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You are given an integer array pref of size n. Find and return the array arr of size n that satisfies:
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pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i].
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Note that ^ denotes the bitwise-xor operation.
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It can be proven that the answer is unique.
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Example 1:
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Input: pref = [5,2,0,3,1]
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Output: [5,7,2,3,2]
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Explanation: From the array [5,7,2,3,2] we have the following:
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- pref[0] = 5.
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- pref[1] = 5 ^ 7 = 2.
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- pref[2] = 5 ^ 7 ^ 2 = 0.
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- pref[3] = 5 ^ 7 ^ 2 ^ 3 = 3.
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- pref[4] = 5 ^ 7 ^ 2 ^ 3 ^ 2 = 1.
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Example 2:
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Input: pref = [13]
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Output: [13]
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Explanation: We have pref[0] = arr[0] = 13.
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Constraints:
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1 <= pref.length <= 105
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0 <= pref[i] <= 106
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Solution
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class Solution {
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public:
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vector<int> findArray(vector<int>& pref) {
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for(int i=pref.size()-1;i>0;i--){
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pref[i]=pref[i]^pref[i-1];
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}
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return pref;
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}
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};

‎30_Substring_with_Concatenation.cpp‎

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30. Substring with Concatenation of All Words
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You are given a string s and an array of strings words. All the strings of words are of the same length.
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A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated.
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For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words.
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Return the starting indices of all the concatenated substrings in s. You can return the answer in any order.
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Example 1:
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Input: s = "barfoothefoobarman", words = ["foo","bar"]
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Output: [0,9]
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Explanation: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6.
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The substring starting at 0 is "barfoo". It is the concatenation of ["bar","foo"] which is a permutation of words.
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The substring starting at 9 is "foobar". It is the concatenation of ["foo","bar"] which is a permutation of words.
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The output order does not matter. Returning [9,0] is fine too.
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Example 2:
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Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
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Output: []
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Explanation: Since words.length == 4 and words[i].length == 4, the concatenated substring has to be of length 16.
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There is no substring of length 16 is s that is equal to the concatenation of any permutation of words.
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We return an empty array.
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Example 3:
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Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
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Output: [6,9,12]
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Explanation: Since words.length == 3 and words[i].length == 3, the concatenated substring has to be of length 9.
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The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"] which is a permutation of words.
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The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"] which is a permutation of words.
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The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"] which is a permutation of words.
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class Solution {
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public:
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vector<int> findSubstring(string s, vector<string>& words) {
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unordered_map<string, int> total;
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for(string s : words){
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total[s]+=1;
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}
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vector<int> ans;
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int wordSize = words[0].size(), noOfWords = words.size();
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for(int i = 0 ; i < s.size()-(wordSize*noOfWords)+1; i++){
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unordered_map <string,int>current;
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int j =0;
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for(; j < noOfWords; j++){
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string sub = s.substr(i+j*wordSize, wordSize);
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current[sub]+=1;
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if(total.find(sub) == total.end())
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break;
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else if(current[sub] > total[sub])
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break; // this substring occurs more than the original no of occ.
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}
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if(j == noOfWords) ans.push_back(i);
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}
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return ans;
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}
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};

‎Auto-File-Manager/Description.txt‎

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-> This program makes the any directory of the file manager looks cool by managing all the content of the specific directory. It store all content by creating seperate
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folder for images,media,document and also create Others folder to store the content which is not relevant to remaining 3 folders.
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-> It seperate the content by checking its extension.
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eg. if the extension of the file is .mp3 then that file will get store in media folder.
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-> The program takes any directory path as an input and then seperate the content according to their extension and store it in the given folder which is mentioned above.

‎Auto-File-Manager/README_Amaan.md‎

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# Auto_FileManager
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- The program makes the any directory of the file manager looks cool by managing all the content of the specific directory. It store all content by creating seperate
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folder for images,media,document and also create Others folder to store the content which is not relevant to remaining 3 folders.
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- It seperate the content by checking its extension.
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eg. if the extension of the file is .mp3 then that file will get store in media folder.
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- The program takes any directory path as an input and then seperate the content according to their extension and store it in the given folder which is mentioned above.

‎Auto-File-Manager/auto_file_manager.py‎

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import os
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directory = (input(r"Enter the path of the directory :"))
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def makedirectory(folder):
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os.chdir(path=directory)
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if not os.path.exists(folder):
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os.makedirs(folder)
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files = os.listdir(path=directory)
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makedirectory("Images")
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makedirectory("Docs")
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makedirectory("Media")
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makedirectory("Others")
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imgext = ['.jpg','.png','.jpeg']
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imgs = [file for file in files if os.path.splitext(file)[1].lower() in imgext]
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# print(imgs)
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docsext = ['.txt','.xml','.csv','.pdf','.docx','.doc','.xsl','.ppt']
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docs = [file for file in files if os.path.splitext(file)[1].lower() in docsext]
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# print(docs)
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mediaext = ['.mp4','.mkv','.mp3','.flv','.avi','.wav']
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medias = [file for file in files if os.path.splitext(file)[1].lower() in mediaext]
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# print(medias)
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others = []
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for file in files:
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ext = os.path.splitext(file)[1].lower()
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if (ext not in imgext) and (ext not in docsext) and (ext not in mediaext) and os.path.isfile(file):
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others.append(file)
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# print(others)
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def move_to_folder(foldername,files):
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for file in files:
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os.replace(file,f"{foldername}/{file}")
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move_to_folder("Media",medias)
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move_to_folder("Images",imgs)
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move_to_folder("Docs",docs)
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move_to_folder("Others",others)
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‎Book_Allocation_Problem.cpp‎

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/*
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Given an array ‘pages’ of integer numbers, where ‘pages[i]’ represents the number of pages in the ‘i-th’ book. There are ‘m’ number of students, and the task is to allocate all the books to their students.
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Allocate books in a way such that:
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1. Each student gets at least one book.
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2. Each book should be allocated to a student.
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3. Book allocation should be in a contiguous manner.
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You have to allocate the books to ‘m’ students such that the maximum number of pages assigned to a student is minimum.
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*/
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#include <bits/stdc++.h>
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using namespace std;
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bool isPossible(int arr[], int n, int m, int mid)
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{
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int studentCount = 1;
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int pageSum = 0;
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for (int i = 0; i < n; i++)
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{
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if (pageSum + arr[i] <= mid)
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pageSum += arr[i];
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else
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{
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studentCount++;
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if (studentCount > m || arr[i] > mid)
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{
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return false;
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}
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pageSum = arr[i];
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}
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}
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return true;
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}
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int findPages(int arr[], int n, int m)
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{
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int sum = 0;
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for (size_t i = 0; i < n; i++)
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{
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sum += arr[i];
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}
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int s = 0, e = sum;
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int ans = -1;
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while (s <= e)
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{
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int mid = s + (e - s) / 2;
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if (isPossible(arr, n, m, mid))
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{
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ans = mid;
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e = mid - 1;
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}
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else
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{
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s = mid + 1;
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}
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}
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return ans;
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}
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int main()
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{
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int arr[] = {10, 20, 30, 40};
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int n = sizeof(arr) / sizeof(arr[0]);
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int m = 2;
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cout << findPages(arr, n, m);
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return 0;
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}

‎CodingClubIndia-master.zip‎

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