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That is not to say you can’t solve them without reading the help/guidelines – in fact I you’ll notice \nthat some of the solutions deviate from the section’s topic. At the end of the day, \nthe solution needs to pass the tests, regardless of implementation.</p>\n<p>With that said, all of the code is in Python 3.</p>\n<h4 id="lesson-1-iterations">Lesson 1 - Iterations</h4>\n<ul>\n<li><em>binary gap:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_1/binary_gap.py\n\n# Task score: perfect score!\ndef solution(N):\n    n_binary = bin(N)\n    n_binary = n_binary.replace(\'0b\', \'\')\n    zeroes_between_ones = n_binary.split(\'1\')\n    # to handle case where an empty sequence is returned (when there\'s no binary gaps)\n    zeroes_between_ones = zeroes_between_ones[0:-1]\n    return len(max(zeroes_between_ones))\n```\n</code>\n</pre>\n\n<h4 id="lesson-2-arrays">Lesson 2 - Arrays</h4>\n<ul>\n<li><em>cyclic rotation:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_2/cyclic_rotation.py\n\n# Task score: perfect score!\ndef solution(A, K):\n    n = len(A)\n    if n == 0 or n == 1:\n        return A\n\n    for k_shift in range(K):\n        first_element = A[0]\n        last_element = A.pop()\n        A = [last_element] + [first_element] + A[1:]\n\n    return A\n```\n</code>\n</pre>\n\n<ul>\n<li><em>odd occurrences in array:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_2/odd_occurrences_in_array.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    if n == 1:\n        return A[0]\n    occurrences = {}\n\n    for i in range(0, n):\n        occurrences[A[i]] = occurrences.get(A[i], 0) + 1\n\n    for key, val in occurrences.items():\n        if val % 2 != 0:\n            return key\n```\n</code>\n</pre>\n\n<h4 id="lesson-3-time-complexity">Lesson 3 - Time Complexity</h4>\n<ul>\n<li><em>frog jump:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_3/frog_jump.py\n\nfrom math import ceil\n\n# Task score: perfect score!\ndef solution(X, Y, D):\n    distance = Y - X\n    return ceil(distance/D)\n```\n</code>\n</pre>\n\n<ul>\n<li><em>perm missing elem:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_3/perm_missing_elem.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    # + 1 because end of range is exclusive, and + 1 again because the exercise says range is [1, ...(N + 1)]\n    s = {x for x in range(1, n + 1 + 1)}\n    a = set(A)\n    return s.difference(a).pop()\n```\n</code>\n</pre>\n\n<ul>\n<li><em>tape equilibrium:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_3/tape_equilibrium.py\n\ndef prefix_sums(A):\n    n = len(A)\n    P = [0] * (n + 1)\n    for k in range(1, n + 1):\n        P[k] = P[k - 1] + A[k - 1]\n    return P\n\ndef count_total(P, x, y):\n    return P[y + 1] - P[x]\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)                                                        # O(1)\n    prefix_sums_array = prefix_sums(A)                                # O(n)\n\n    diffs = []\n    for p in range(1, n):\n        first_part_total = count_total(prefix_sums_array, 0, p - 1)   # O(1)\n        second_part_total = count_total(prefix_sums_array, p, n - 1)  # O(1)\n        diff = abs(first_part_total - second_part_total)              # O(1)\n        diffs.append(diff)                                            # O(1)\n\n    return min(diffs)                                                 # O(n)\n```\n</code>\n</pre>\n\n<h4 id="lesson-4-counting-elements">Lesson 4 - Counting Elements</h4>\n<ul>\n<li><em>perm check:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_4/perm_check.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    # n + 1 because range is exclusive but need to include the last number in the sequence\n    s = {x for x in range(1, n + 1)}\n    a = set(A)\n    if len(s.difference(a)) > 0:\n        return 0\n    return 1\n```\n</code>\n</pre>\n\n<ul>\n<li><em>frog river one:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_4/frog_river_one.py\n\n# Task score: perfect score!\ndef solution(X, A):\n    complete_path = {position for position in range(1, X + 1)}\n    current_path = set()\n    for time in range(0, len(A)):\n        current_path.add(A[time])\n        # as soon as the two sets are equal, return the current index/time, which will be the solution\n        if current_path == complete_path:\n            return time\n\n    return -1\n```\n</code>\n</pre>\n\n<ul>\n<li><em>max counters:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_4/max_counters.py\n\n# Task score 66%: 100% correctness, 40% performance\ndef solution(N, A):\n    counters = [0] * N\n    for instruction in A:\n        if instruction == N + 1:\n            counters = [max(counters)] * N\n        else:\n            counters[instruction - 1] += 1\n\n    return counters\n```\n</code>\n</pre>\n\n<ul>\n<li><em>missing integer:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_4/missing_integer.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    # n + 1 because end of range is exclusive, and + 1 again because the exercise wants range [1, ...(N + 1)]\n    s = {x for x in range(1, n + 1 + 1)}\n    a = set(A)\n    return min(s.difference(a))\n```\n</code>\n</pre>\n\n<h4 id="lesson-5-prefix-sums">Lesson 5 - Prefix Sums</h4>\n<ul>\n<li><em>passing cars:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_5/passing_cars.py\n\n# Task score: perfect score! (prefix sums -- cumulative sum from the start of sequence)\n#\n# The crucial idea behind the algorithm is that each West car(1) creates a pair with each preceding East car(0).\n# Hence, anytime we encounter a 1, we add up all preceding 0s.\ndef solution(A):\n    n = len(A)\n\n    zeros_counter = 0\n    pairs_count = 0\n    for i in range(n):\n        if A[i] == 0:\n            zeros_counter += 1\n        if A[i] == 1:\n            pairs_count += zeros_counter\n        if pairs_count > 1000000000:\n            return -1\n\n    return pairs_count\n```\n</code>\n</pre>\n\n<ul>\n<li><em>genomic range query:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_5/genomic_range_query.py\n\n# Task score: perfect score!\ndef solution(S, P, Q):\n    m = len(P)\n\n    result = []\n    for k in range(0, m):\n        sub_sequence = S[P[k] : Q[k] + 1]\n        if \'A\' in sub_sequence:\n            result.append(1)\n        elif \'C\' in sub_sequence:\n            result.append(2)\n        elif \'G\' in sub_sequence:\n            result.append(3)\n        else:\n            result.append(4)\n\n    return result\n```\n</code>\n</pre>\n\n<ul>\n<li><em>min avg two slice:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_5/min_avg_two_slice.py\n\ndef prefix_sums(A):\n    n = len(A)\n    P = [0] * (n + 1)\n    for k in range(1, n + 1):\n        P[k] = P[k - 1] + A[k - 1]\n    return P\n\ndef count_total(P, x, y):\n    return P[y + 1] - P[x]\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    P = prefix_sums(A)\n    min_avg_idx = 0\n    min_avg = (count_total(P, 0, 1)) / 2\n\n    for i in range(1, n - 2):\n        local_avg_of_two = (count_total(P, i, i + 1)) / 2\n        local_avg_of_three = (count_total(P, i, i + 2)) / 3\n        # the core idea is that on every iteration, we assign to min_avg and min_avg_idx\n        # whichever is smaller between avg of the local duplet or triplet\n        local_min_avg = min(local_avg_of_two, local_avg_of_three)\n        if local_min_avg < min_avg:\n            min_avg = local_min_avg\n            min_avg_idx = i\n\n    last_pair_avg = (count_total(P, n - 2, n - 1)) / 2\n    if last_pair_avg < min_avg:\n        min_avg = last_pair_avg  # no need to set this but doing it for the sake of completeness\n        min_avg_idx = n - 2\n\n    return min_avg_idx\n```\n</code>\n</pre>\n\n<ul>\n<li><em>count div:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_5/count_div.py\n\n# Prefix sums approach. The crucial bit here is understanding that the number of divisors in slice\n# [0... A] and [0... B] is A/K and B/K respectively, with the added benefit of also finding the upper and lower\n# bounds of our "prefix sum array" (except we don\'t have to actually build the array as we\'re only concerned with the\n# slices here).\n#\n# Task score: perfect score!\ndef solution(A, B, K):\n    # Note: A/K = number of divisors from 1 to A\n    #       B/K = number of divisors from 1 to B\n    upper_bound = (B // K)\n    lower_bound = (A // K)\n    # because "A" is inclusive, and since the number of divisors has\n    # already been counted in the lower_bound, take away the divisor that was double counted\n    if A % K == 0:\n        lower_bound -= 1\n    return upper_bound - lower_bound  # all operations in the solution run in O(1) time\n```\n</code>\n</pre>\n\n<ul>\n<li><em>distinct:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_6/distinct.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    if n == 0:\n        return 0\n    A = sorted(A)            # time complexity O(n * log n), space complexity O(n)\n    distinct = 1\n    for i in range(1, n):\n        if A[i] != A[i - 1]:\n            distinct += 1    # time complex. O(n)\n\n    return distinct\n```\n</code>\n</pre>\n\n<ul>\n<li><em>max product of three:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_6/max_product_of_three.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    A = sorted(A)                                  # space complexity O(n), time complexity O(n * log n)\n    initial_triplet = A[0] * A[1] * A[n - 1]       # biggest triplet is made of negative and positive numbers\n    last_triplet = A[n - 1] * A[n - 2] * A[n - 3]  # biggest triplet is made of only positive numbers\n    return max(initial_triplet, last_triplet)\n```\n</code>\n</pre>\n\n<ul>\n<li><em>triangle:</em></li>\n</ul>\n<pre>\n<code>\n```\n# lesson_6/triangle.py\n\n# Task score: perfect score!\ndef solution(A):\n    n = len(A)\n    if n < 3:\n        return 0\n    A = sorted(A)\n\n    for i in range(0, n - 2):\n        if A[i] + A[i + 1] > A[i + 2]:\n            if A[i + 1] + A[i + 2] > A[i]:\n                if A[i + 2] + A[i] > A[i + 1]:\n                    return 1\n\n    return 0\n```\n</code>\n</pre>\n\n<br>\n<br>'},function(e,t,n){"use strict";n.r(t),n.d(t,"compiler",function(){return Ie}),n.d(t,"default",function(){return Le});var r=n(1),o=n.n(r),i=n(12),a=n.n(i),l=Object.assign||function(e){for(var t=1;t<arguments.length;t++){var n=arguments[t];for(var r in n)Object.prototype.hasOwnProperty.call(n,r)&&(e[r]=n[r])}return e};var u={accesskey:"accessKey",allowfullscreen:"allowFullScreen",allowtransparency:"allowTransparency",autocomplete:"autoComplete",autofocus:"autoFocus",autoplay:"autoPlay",cellpadding:"cellPadding",cellspacing:"cellSpacing",charset:"charSet",class:"className",classid:"classId",colspan:"colSpan",contenteditable:"contentEditable",contextmenu:"contextMenu",crossorigin:"crossOrigin",enctype:"encType",for:"htmlFor",formaction:"formAction",formenctype:"formEncType",formmethod:"formMethod",formnovalidate:"formNoValidate",formtarget:"formTarget",frameborder:"frameBorder",hreflang:"hrefLang",inputmode:"inputMode",keyparams:"keyParams",keytype:"keyType",marginheight:"marginHeight",marginwidth:"marginWidth",maxlength:"maxLength",mediagroup:"mediaGroup",minlength:"minLength",novalidate:"noValidate",radiogroup:"radioGroup",readonly:"readOnly",rowspan:"rowSpan",spellcheck:"spellCheck",srcdoc:"srcDoc",srclang:"srcLang",srcset:"srcSet",tabindex:"tabIndex",usemap:"useMap"},c=["style","script"],s=/([-A-Z0-9_:]+)(?:\s*=\s*(?:(?:"((?:\\.|[^"])*)")|(?:'((?:\\.|[^'])*)')|(?:\{((?:\\.|{[^}]*?}|[^}])*)\})))?/gi,f=/mailto:/i,p=/\n{2,}$/,d=/^( *>[^\n]+(\n[^\n]+)*\n*)+\n{2,}/,h=/^ *> ?/gm,m=/^ {2,}\n/,y=/^(?:( *[-*_]) *){3,}(?:\n *)+\n/,v=/^\s*(`{3,}|~{3,}) *(\S+)? *\n([\s\S]+?)\s*\1 *(?:\n *)+\n?/,g=/^(?: {4}[^\n]+\n*)+(?:\n *)+\n?/,b=/^(`+)\s*([\s\S]*?[^`])\s*\1(?!`)/,w=/^(?:\n *)*\n/,k=/\r\n?/g,x=/^\[\^(.*)\](:.*)\n/,_=/^\[\^(.*)\]/,E=/\f/g,T=/^\s*?\[(x|\s)\]/,C=/^ *(#{1,6}) *([^\n]+)\n{0,2}/,P=/^([^\n]+)\n *(=|-){3,} *(?:\n *)+\n/,S=/^ *<([A-Za-z][^ >/]*) ?([^>]*)\/{0}>\n?(\s*(?:<\1[^>]*?>[\s\S]*?<\/\1>|(?!<\1)[\s\S])*?)<\/\1>\n*/,O=/^<!--.*?-->/,R=/^(data|aria|x)-[a-z_][a-z\d_.-]*$/,N=/^ *<([A-Za-z][\w:]*)\s*((?:<.*?>|[^>])*)>(?!<\/\1>)\s*/,A=/^\{.*\}$/,j=/^(https?:\/\/[^\s<]+[^<.,:;"')\]\s])/,M=/^<([^ >]+@[^ >]+)>/,U=/^<([^ >]+:\/[^ >]+)>/,I=/ *\n+$/,L=/(?:^|\n)( *)$/,F=/-([a-z])?/gi,D=/^(.*\|?.*)\n *(\|? *[-:]+ *\|[-| :]*)\n((?:.*\|.*\n)*)\n?/,z=/^((?:[^\n]|\n(?! *\n))+)(?:\n *)+\n/,B=/^\[([^\]]*)\]:\s*(\S+)\s*("([^"]*)")?/,W=/^!\[([^\]]*)\] ?\[([^\]]*)\]/,H=/^\[([^\]]*)\] ?\[([^\]]*)\]/,V=/(\n|^[-*]\s|^#|^ {2,}|^-{2,}|^>\s)/,$=/\t/g,q=/(^ *\||\| *$)/g,K=/^ *:-+: *$/,Y=/^ *:-+ *$/,Q=/^ *-+: *$/,G=/ *\| */,X=/^([*_])\1((?:[^`~()\[\]<>]*?|(?:.*?([`~]).*?\3.*?)*|(?:.*?\([^)]*?\).*?)*|(?:.*?\[[^\]]*?\].*?)*|(?:.*?<.*?>.*?)*|[^\1]*?)\1?)\1{2}/,Z=/^([*_])((?:[^`~()\[\]<>]*?|(?:.*?([`~]).*?\3.*?)*|(?:.*?\([^)]*?\).*?)*|(?:.*?\[[^\]]*?\].*?)*|(?:.*?<.*?>.*?)*|[^\1]*?))\1/,J=/^~~((?:.*?([`~]).*?\2.*?)*|(?:.*?<.*?>.*?)*|.+?)~~/,ee=/^\\([^0-9A-Za-z\s])/,te=/^[\s\S]+?(?=[^0-9A-Z\s\u00c0-\uffff]|\d+\.|\n\n| {2,}\n|\w+:\S|$)/i,ne=/(^\n+|(\n|\s)+$)/g,re=/^(\s*)/,oe=/\\([^0-9A-Z\s])/gi,ie=new RegExp("^( *)((?:[*+-]|\\d+\\.)) +"),ae=new RegExp("( *)((?:[*+-]|\\d+\\.)) +[^\\n]*(?:\\n(?!\\1(?:[*+-]|\\d+\\.) )[^\\n]*)*(\\n|$)","gm"),le=new RegExp("^( *)((?:[*+-]|\\d+\\.)) 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