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| 1 | +// https://leetcode.com/problems/decode-string/ |
| 2 | +import 'package:test/test.dart'; |
| 3 | + |
| 4 | +void main() { |
| 5 | + final stopwatch = Stopwatch()..start(); |
| 6 | + runTests(); |
| 7 | + |
| 8 | + final Solution s = Solution(); |
| 9 | + s.decodeString("1[1[1[1[1[a]]]]]"); |
| 10 | + |
| 11 | + /// |
| 12 | + stopwatch.stop(); |
| 13 | + print('Function Execution Time : ${stopwatch.elapsedMicroseconds} micro sec'); |
| 14 | + // Function Execution Time : 365384 micro s |
| 15 | + // Function Execution Time : 394717 micro s |
| 16 | +} |
| 17 | + |
| 18 | +class Solution { |
| 19 | + ////! Accepted from the first time |
| 20 | + ////! Try to find another solution |
| 21 | + String decodeString(String s) { |
| 22 | + final StringBuffer sbResult = StringBuffer(); |
| 23 | + String subString = ""; |
| 24 | + final List<String> stack = []; |
| 25 | + for (var i = 0; i < s.length; i++) { |
| 26 | + final char = s[i]; |
| 27 | + if (char == ']') { |
| 28 | + String lastChar = stack.removeLast(); |
| 29 | + |
| 30 | + /// Get the substring needed to be repeated |
| 31 | + /// Add all the chars in the [] in the substring to repeat |
| 32 | + while (lastChar != '[') { |
| 33 | + subString = lastChar + subString; |
| 34 | + lastChar = stack.removeLast(); |
| 35 | + } |
| 36 | + lastChar = stack.last; |
| 37 | + int number = 0; |
| 38 | + int tens = 1; |
| 39 | + |
| 40 | + /// Get the repeat number |
| 41 | + while (lastChar.startsWith(RegExp(r'[0-9]'))) { |
| 42 | + stack.removeLast(); |
| 43 | + number += int.parse(lastChar) * tens; |
| 44 | + tens *= 10; |
| 45 | + if (stack.isNotEmpty) { |
| 46 | + lastChar = stack.last; |
| 47 | + } else { |
| 48 | + break; |
| 49 | + } |
| 50 | + } |
| 51 | + |
| 52 | + /// Check if the stack is empty: |
| 53 | + /// Then I will add the substring to the result |
| 54 | + /// Else Then I need to repeat this again and add this to the stack |
| 55 | + if (stack.isNotEmpty) { |
| 56 | + stack.add(subString * number); |
| 57 | + } else { |
| 58 | + sbResult.write(subString * number); |
| 59 | + } |
| 60 | + subString = ""; |
| 61 | + number = 0; |
| 62 | + tens = 1; |
| 63 | + } else if (char == '[') { |
| 64 | + stack.add(char); |
| 65 | + } else if (char.startsWith(RegExp(r'[0-9]'))) { |
| 66 | + stack.add(char); |
| 67 | + } else { |
| 68 | + /// If the string not in the [] then add it to the string result without repeat |
| 69 | + if (stack.isEmpty) { |
| 70 | + sbResult.write(char); |
| 71 | + } else { |
| 72 | + stack.add(char); |
| 73 | + } |
| 74 | + } |
| 75 | + } |
| 76 | + |
| 77 | + return sbResult.toString(); |
| 78 | + } |
| 79 | +} |
| 80 | + |
| 81 | +void runTests() { |
| 82 | + final Solution s = Solution(); |
| 83 | + |
| 84 | + group('Decode String', () { |
| 85 | + // Basic examples from problem statement |
| 86 | + test('Example 1: "3[a]2[bc]" → "aaabcbc"', () { |
| 87 | + expect(s.decodeString("3[a]2[bc]"), equals("aaabcbc")); |
| 88 | + }); |
| 89 | + |
| 90 | + test('Example 2: "3[a2[c]]" → "accaccacc"', () { |
| 91 | + expect(s.decodeString("3[a2[c]]"), equals("accaccacc")); |
| 92 | + }); |
| 93 | + |
| 94 | + test('Example 3: "2[abc]3[cd]ef" → "abcabccdcdcdef"', () { |
| 95 | + expect(s.decodeString("2[abc]3[cd]ef"), equals("abcabccdcdcdef")); |
| 96 | + }); |
| 97 | + |
| 98 | + // Edge cases |
| 99 | + test('Single character: "a" → "a"', () { |
| 100 | + expect(s.decodeString("a"), equals("a")); |
| 101 | + }); |
| 102 | + |
| 103 | + test('Single repetition: "1[a]" → "a"', () { |
| 104 | + expect(s.decodeString("1[a]"), equals("a")); |
| 105 | + }); |
| 106 | + |
| 107 | + test('Multiple digits for k: "10[a]" → "aaaaaaaaaa"', () { |
| 108 | + expect(s.decodeString("10[a]"), equals("aaaaaaaaaa")); |
| 109 | + }); |
| 110 | + |
| 111 | + // Nested cases |
| 112 | + test('Double nesting: "2[3[a]b]" → "aaabaaab"', () { |
| 113 | + expect(s.decodeString("2[3[a]b]"), equals("aaabaaab")); |
| 114 | + }); |
| 115 | + |
| 116 | + test('Triple nesting: "2[a3[b2[c]]]" → "abccbccbccabccbccbcc"', () { |
| 117 | + expect(s.decodeString("2[a3[b2[c]]]"), equals("abccbccbccabccbccbcc")); |
| 118 | + }); |
| 119 | + |
| 120 | + // Multiple segments |
| 121 | + test('Multiple segments: "2[a]3[b]4[c]" → "aabbbcccc"', () { |
| 122 | + expect(s.decodeString("2[a]3[b]4[c]"), equals("aabbbcccc")); |
| 123 | + }); |
| 124 | + |
| 125 | + test('Mixed segments: "a2[b3[c]d]e" → "abcccdbcccde"', () { |
| 126 | + expect(s.decodeString("a2[b3[c]d]e"), equals("abcccdbcccde")); |
| 127 | + }); |
| 128 | + |
| 129 | + // Maximum constraints |
| 130 | + test('Maximum k value: "300[a]" → 300 a\'s', () { |
| 131 | + expect(s.decodeString("300[a]"), equals("a" * 300)); |
| 132 | + }); |
| 133 | + |
| 134 | + test('Maximum nesting depth', () { |
| 135 | + expect(s.decodeString("1[1[1[1[1[a]]]]]"), equals("a")); |
| 136 | + }); |
| 137 | + |
| 138 | + // No encoding cases |
| 139 | + test('No encoding: "abc" → "abc"', () { |
| 140 | + expect(s.decodeString("abc"), equals("abc")); |
| 141 | + }); |
| 142 | + |
| 143 | + test('Empty brackets: "3[]" → ""', () { |
| 144 | + expect(s.decodeString("3[]"), equals("")); |
| 145 | + }); |
| 146 | + |
| 147 | + // Complex cases |
| 148 | + test('Complex case 1: "3[z]2[2[y]pq4[2[jk]e1[f]]]ef"', () { |
| 149 | + expect( |
| 150 | + s.decodeString("3[z]2[2[y]pq4[2[jk]e1[f]]]ef"), |
| 151 | + equals( |
| 152 | + "zzzyypqjkjkefjkjkefjkjkefjkjkefyypqjkjkefjkjkefjkjkefjkjkefef")); |
| 153 | + }); |
| 154 | + |
| 155 | + test('Complex case 2: "2[2[y]a3[b2[c]d]]"', () { |
| 156 | + expect(s.decodeString("2[2[y]a3[b2[c]d]]"), |
| 157 | + equals("yyabccdbccdbccdyyabccdbccdbccd")); |
| 158 | + }); |
| 159 | + |
| 160 | + // Random cases |
| 161 | + test('Random case 1: "2[a3[b]]" → "abbbabbb"', () { |
| 162 | + expect(s.decodeString("2[a3[b]]"), equals("abbbabbb")); |
| 163 | + }); |
| 164 | + |
| 165 | + test('Random case 2: "a2[bc3[d]]e" → "abcdddbcddde"', () { |
| 166 | + expect(s.decodeString("a2[bc3[d]]e"), equals("abcdddbcddde")); |
| 167 | + }); |
| 168 | + }); |
| 169 | +} |
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