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‎Simple Questions to Hone Python Skills.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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"# Question 1: Number of Good Pairs, by Microsoft and Amazon\n",
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"- Given an array of integers nums.\n",
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"- A pair (i,j) is called good if nums[i] == nums[j] and i < j.\n",
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"- Return the number of good pairs.\n",
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"- https://leetcode.com/problems/number-of-good-pairs/"
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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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"4"
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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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"# solution 1: brutal force & a nested for loop\n",
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"def good_pair(nums):\n",
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" count = 0 \n",
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" for i in range(len(nums)):\n",
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" for j in range(i+1,len(nums)):\n",
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" if nums[i] ==nums[j]:\n",
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" count+=1\n",
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" return count\n",
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"\n",
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"# test case \n",
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"nums = [1,2,3,1,1,3]\n",
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"\n",
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"good_pair(nums)"
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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": 17,
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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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"4"
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]
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},
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"execution_count": 17,
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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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"# solution 2: dictionary & hashmap\n",
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"def good_pair(nums):\n",
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" \n",
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" hashmap = {}\n",
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" count = 0 \n",
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" \n",
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" for num in nums: # read elements from nums sequentially \n",
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" if num in hashmap:\n",
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" count += hashmap[num]\n",
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" hashmap[num]+=1\n",
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"\n",
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" else: \n",
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" hashmap[num]=1\n",
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" return count \n",
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"\n",
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"# test case \n",
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"nums = [1,2,3,1,1,3]\n",
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"\n",
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"good_pair(nums)"
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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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"12"
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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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"number = 123\n",
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"#num = number%10\n",
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"\n",
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"number/=10\n",
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"int(number)"
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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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"--- "
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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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"# Question 2: Armstrong Number, by Amazon\n",
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"- The k-digit number N is an Armstrong number if and only if the k-th power of each digit sums to N.\n",
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"- Given a positive integer N, return true if and only if it is an Armstrong number.\n",
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"- https://leetcode.com/problems/armstrong-number/"
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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": 19,
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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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"True"
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]
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},
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"execution_count": 19,
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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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"# solution 1: change data type using int()\n",
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"\n",
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"def Armstrong(num):\n",
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" \n",
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" k = len(str(num))\n",
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" \n",
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" digit_sum =0\n",
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" \n",
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" for i in str(num):\n",
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" digit_sum += int(i)**k\n",
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"\n",
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" return digit_sum == num\n",
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"\n",
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"# test case \n",
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"Armstrong(num = 153)"
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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": 2,
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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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"True"
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]
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},
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"execution_count": 2,
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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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"# solution 2: pop and push \n",
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"\n",
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"def Armstrong_1(num):\n",
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" num1 = num # prepare num1 for comparison\n",
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" k = len(str(num))\n",
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"\n",
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" digit_sum= 0 \n",
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" \n",
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" while num != 0:\n",
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" digit = num%10\n",
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" digit_sum += digit**k\n",
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" num = int(num/10)\n",
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" return digit_sum == num1\n",
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"\n",
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"# test case\n",
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"Armstrong_1(num = 153)"
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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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"---"
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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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"# Question 3: Count Primes, by FAANG\n",
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"- Count the number of prime numbers less than a non-negative number, n.\n",
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"- https://leetcode.com/problems/count-primes/"
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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": 32,
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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": 32,
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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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"# solution 1: brutal force not efficient\n",
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"def countPrimes(n):\n",
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" if n <=2:\n",
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" return 0\n",
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" count = 0 \n",
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" for i in range(2,n):\n",
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" for j in range(2,i):\n",
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" if i%j ==0:\n",
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" break\n",
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" else:\n",
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" count+=1\n",
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" return count\n",
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"\n",
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"# test case\n",
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"countPrimes(7)# primes: 2,3,5"
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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": 31,
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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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"9592"
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]
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},
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"execution_count": 31,
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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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"# solution 2: Sieve of Eratosthene\n",
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"# super fast but have to remember the math\n",
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"def countPrimes(n):\n",
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" if n < 3:\n",
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" return 0\n",
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" \n",
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" primes = [True] * n\n",
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" \n",
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" primes[0] = primes[1] = False\n",
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" \n",
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" for i in range(2, int(n ** 0.5) + 1):\n",
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" \n",
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" if primes[i]:\n",
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" \n",
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" primes[i * i: n: i] = [False] * len(primes[i * i: n: i])\n",
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" \n",
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" return sum(primes)\n",
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"\n",
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"# test case\n",
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"countPrimes(100000)"
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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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"---"
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]
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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": "3.7.4"
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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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