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Xiaopei
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leecode
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‎README.md

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# Interview-Coding
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Archive some common interview questions
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‎binary_search.ipynb

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{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"import os\n",
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"print(\"hello world\")"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"*************\n"
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]
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}
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],
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"source": [
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"print(\"*************\")"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"A message containing letters from A-Z can be encoded into numbers using the following mapping:\n",
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"\n",
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"'A' -> \"1\"\n",
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"'B' -> \"2\"\n",
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"...\n",
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"'Z' -> \"26\"\n",
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"To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, \"11106\" can be mapped into:\n",
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"\n",
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"\"AAJF\" with the grouping (1 1 10 6)\n",
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"\"KJF\" with the grouping (11 10 6)\n",
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"Note that the grouping (1 11 06) is invalid because \"06\" cannot be mapped into 'F' since \"6\" is different from \"06\".\n",
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"\n",
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"Given a string s containing only digits, return the number of ways to decode it.\n",
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"\n",
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"The test cases are generated so that the answer fits in a 32-bit integer.\n",
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"\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"3"
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]
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},
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"execution_count": 8,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"def NumDecodings(s:str)->int:\n",
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"\n",
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" if s.strip()==\"\":\n",
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" return 0\n",
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"\n",
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" ALL_DIGITS = [ i for i in range(10)]\n",
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"\n",
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" VALID_2ND_DIGITS = [ i for i in range(7)]\n",
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"\n",
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" num = 0\n",
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"\n",
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" pre_i = \"\"\n",
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"\n",
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"\n",
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" for i in s:\n",
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" if pre_i == \"1\" and int(i) in ALL_DIGITS:\n",
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" num +=2\n",
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" if pre_i == \"2\" and int(i) in VALID_2ND_DIGITS:\n",
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" num +=2\n",
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" if pre_i in [\"\",\"0\"] and int(i) in ALL_DIGITS and int(i) !=0:\n",
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" num +=1\n",
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" pre_i = i\n",
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" \n",
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" return num\n",
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"\n",
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"NumDecodings(\"12\")\n",
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"\n",
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"\n",
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"\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"[1, 2, 3, 4, 5, 6, 7, 8, 9]"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"NON_ZERO_DIGITS = [ 1 + i for i in range(9)]\n",
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"NON_ZERO_DIGITS"
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]
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}
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],
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"metadata": {
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"interpreter": {
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"hash": "e8d2d80473aa95d50a1c81516f1d5b4a59533016ed6d41e75f06aec8c08c83aa"
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},
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"kernelspec": {
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"display_name": "Python 3.10.1 64-bit",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.10.1"
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},
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"orig_nbformat": 4
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎dp.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Dynamic programing"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"A message containing letters from A-Z can be encoded into numbers using the following mapping:\n",
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"\n",
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"'A' -> \"1\"\n",
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"'B' -> \"2\"\n",
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"...\n",
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"'Z' -> \"26\"\n",
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"To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, \"11106\" can be mapped into:\n",
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"\n",
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"\"AAJF\" with the grouping (1 1 10 6)\n",
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"\"KJF\" with the grouping (11 10 6)\n",
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"Note that the grouping (1 11 06) is invalid because \"06\" cannot be mapped into 'F' since \"6\" is different from \"06\".\n",
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"\n",
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"Given a string s containing only digits, return the number of ways to decode it.\n",
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"\n",
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"The test cases are generated so that the answer fits in a 32-bit integer.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"3"
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]
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},
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"execution_count": 1,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"def NumDecodings(s:str)->int:\n",
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"\n",
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" if s.strip()==\"\":\n",
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" return 0\n",
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"\n",
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" ALL_DIGITS = [ i for i in range(10)]\n",
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"\n",
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" VALID_2ND_DIGITS = [ i for i in range(7)]\n",
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"\n",
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" num = 0\n",
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"\n",
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" pre_i = \"\"\n",
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"\n",
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"\n",
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" for i in s:\n",
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" if pre_i == \"1\" and int(i) in ALL_DIGITS:\n",
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" num +=2\n",
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" if pre_i == \"2\" and int(i) in VALID_2ND_DIGITS:\n",
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" num +=2\n",
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" if pre_i in [\"\",\"0\"] and int(i) in ALL_DIGITS and int(i) !=0:\n",
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" num +=1\n",
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" pre_i = i\n",
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" \n",
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" return num\n",
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"\n",
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"NumDecodings(\"12\")"
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]
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}
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],
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"metadata": {
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"interpreter": {
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"hash": "e8d2d80473aa95d50a1c81516f1d5b4a59533016ed6d41e75f06aec8c08c83aa"
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},
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"kernelspec": {
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"display_name": "Python 3.10.1 64-bit",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.10.1"
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},
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"orig_nbformat": 4
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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