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‎Day_13.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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"## 4.Define a function to convert a binary number to the corresponding decimal number\n",
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"### Test Cases:\n",
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"* binaryToDecimal(1101) -> 12\n",
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"* binaryToDecimal(1010) -> 10\n",
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"* binaryToDecimal(111000) -> 56"
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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": 235,
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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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"13\n",
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"10\n",
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"56\n"
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]
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}
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],
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"source": [
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"def BintoDec(B):\n",
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" s=0\n",
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" p=len(B)-1\n",
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" for i in range(len(B)):\n",
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" s=s+((int(B[i])) * (2**p))\n",
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" p=p-1\n",
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" return s\n",
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"print(BintoDec('1101'))\n",
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"print(BintoDec('1010'))\n",
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"print(BintoDec('111000'))"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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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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},
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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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"## 5. Define a function to convert a decimal number to the corresponding binary number\n",
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"\n",
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"\n",
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"### Test Cases:\n",
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"* decimalToBinary(15) -> 1111\n",
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"* decimalToBinary(1) -> 1"
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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": 314,
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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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"[1, 1, 1, 1]\n",
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"[1]\n",
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"[1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0]\n"
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]
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}
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],
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"source": [
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"def DectoBin(D,m):\n",
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" if D==0:\n",
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" breakpoint\n",
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" else:\n",
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" m.append(D%2)\n",
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" DectoBin(D//2,m)\n",
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" return m[::-1]\n",
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"print(DectoBin(15,[]))\n",
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"print(DectoBin(1,[]))\n",
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"print(DectoBin(6566564564,[]))"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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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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"## 7. Design a Python script to determine the difference in date for given two dates in YYYY:MM:DD format(0 <= YYYY <= 9999, 1 <= MM <= 12, 1 <= DD <= 31) following the leap year rules. Return the total number of days existing between the two dates.\n",
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"\n",
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"### Test Cases:\n",
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"* dateDifference('.&checktime(2019,05,10,':')', '.&checktime(2019,05,01,':')') -> 9\n",
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"* dateDifference('.&checktime(0003,03,03,':')', '.&checktime(0003,06,06,':')') -> 95"
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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": 339,
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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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"9\n",
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"96\n"
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]
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}
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],
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"source": [
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"def leapYear(L):\n",
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" return(L% 4==0 and L%100!=0 or L%400==0)\n",
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"def dateDifference(p,q):\n",
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" y= int(p[0:4])-int(q[0:4])\n",
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" m= int(p[5:7])-int(q[5:7])\n",
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" leapYear(y)\n",
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" if True:\n",
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" yd=y*366\n",
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" else:\n",
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" yd=y*365\n",
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" d= int(p[8:10])-int(q[8:10])\n",
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" md=m*31\n",
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" return abs(yd+md+d)\n",
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"\n",
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"print(dateDifference('.&checktime(2019,05,10,':')', '.&checktime(2019,05,01,':')'))\n",
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"print(dateDifference('.&checktime(0003,03,03,':')', '.&checktime(0003,06,06,':')'))\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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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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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"## 7. Define a function to print the sequence of spiral pattern elements for a given N x N matrix\n",
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"\n",
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"### Test Cases:\n",
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"spiralPattern([[1,2,3], [4,5,6], [7,8,9]]) -> 1 2 3 6 9 8 7 4 5"
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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": 350,
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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, 0], 9, 5, 8, 5, 6, 1, 7, 4)"
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]
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},
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"execution_count": 350,
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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 spiral(m):\n",
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" return m[0],m[1][3],m[2][3],m[3][3],m[3][2],m[3][1],m[3][0],m[2][0],m[1][0]\n",
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"spiral([[1,2,3,0],[4,5,6,9],[7,8,9,5],[1,6,5,8]]) "
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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": 18,
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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, 0, 9, 5, [1, 6, 5, 8], 7, 4]"
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]
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},
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"execution_count": 18,
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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 Spiral(m,s):\n",
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" for i in range(len(m[0])):\n",
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" s.append(m[0][i])\n",
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" #for i in range(len(m[1])):\n",
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" s.append(m[1][3])\n",
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" #for i in range(len(m[1])):\n",
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" s.append(m[2][3])\n",
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" #for i in range(len(m[3]),0):\n",
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" s.append(*m[3:2:-1])\n",
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" #for i in range(len(m[2])):\n",
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" s.append(m[2][0])\n",
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" #for i in range(len(m[1])):\n",
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" s.append(m[1][0])\n",
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" return(s) \n",
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"Spiral([[1,2,3,0],[4,5,6,9],[7,8,9,5],[1,6,5,8]],[]) "
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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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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},
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{
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"cell_type": "code",
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"execution_count": 445,
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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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"2\n",
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"2\n"
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]
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}
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],
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"source": [
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"def StringCount(s,r):\n",
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" print( len(max(s,r)))\n",
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" #for _ in range(len(max(s,r))):\n",
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" return s.count(r)\n",
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" \n",
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"print(StringCount(\"abccddccc\", \"cc\"))\n",
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"print(StringCount(\"1234567891122334455\", \"3\"))\n",
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"print(StringCount('str','substr'))\n",
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"print(StringCount('aaaaaaa', 'aaa'))"
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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": 459,
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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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"'[1,2,3,0] [4,5,6,9] [7,8,9,5] [1,6,5,8]'"
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]
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},
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"execution_count": 459,
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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 spiral(m):\n",
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" return ' '.join(m)\n",
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"spiral([\"[1,2,3,0]\",\"[4,5,6,9]\",\"[7,8,9,5]\",\"[1,6,5,8]\"]) "
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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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"'aabbcsdd'"
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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 merge(m,n,o):\n",
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" for i in range (len(max(m,n))):\n",
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" o.append(m[i]+n[i])\n",
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" \n",
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" return \"\".join(o)\n",
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"merge('abcd', 'absd',[])"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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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": "3.7.3"
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}
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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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