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| 1 | +想法:用 Recursiving Backtracking 的方法,找到所有可能的字串,並在找到第 k 個時停止並回傳 |
| 2 | + |
| 3 | +Time Complexity : O(n * 2^n) for the fist letter has 3 options , after that , each letter have two options |
| 4 | +So there are n * 2^(n - 1) possible strings = O(n * 2^n) |
| 5 | +Space Complexity : O(2^n) for the recursion (recursion depth = O(n)) |
| 6 | + |
| 7 | +class Solution { |
| 8 | +public: |
| 9 | + string ans ; |
| 10 | + int count = 0 ; |
| 11 | + void constructHappyString(int n , int k , string& result , int index) { |
| 12 | + if (count >= k) |
| 13 | + return ; |
| 14 | + if (index == n) { |
| 15 | + if (++count == k) |
| 16 | + ans = result ; |
| 17 | + return ; |
| 18 | + } |
| 19 | + |
| 20 | + for(char c = 'a' ; c <= 'c' ; c++) { |
| 21 | + if (index == 0 || result[index - 1] != c) { |
| 22 | + result[index] = c ; |
| 23 | + constructHappyString(n , k , result , index + 1) ; |
| 24 | + } |
| 25 | + } |
| 26 | + return ; |
| 27 | + } |
| 28 | + string getHappyString(int n, int k) { |
| 29 | + string result ; |
| 30 | + result.resize(n) ; |
| 31 | + constructHappyString(n , k , result , 0 ) ; |
| 32 | + return ans ; |
| 33 | + } |
| 34 | +}; |
| 35 | + |
| 36 | +// 法二:我們可以知道前 2^(n - 1) 個的第一個字母一定是 a , 2^(n - 1) + 1 ~ 2 * 2^(n - 1)的第一個字母是 b , 接著是 c |
| 37 | +因此,我們可以從 k 得知各個位置是什麼字母並輸出 |
| 38 | + |
| 39 | +Time Complexity : O(n) for constructing the string |
| 40 | +Space Complexity : O(n) for the answer string |
| 41 | + |
| 42 | +class Solution { |
| 43 | +public: |
| 44 | + string getHappyString(int n, int k) { |
| 45 | + // Calculate the total number of happy strings of length n |
| 46 | + int total = 3 * (1 << (n - 1)); |
| 47 | + |
| 48 | + // If k is greater than the total number of happy strings, return an |
| 49 | + // empty string |
| 50 | + if (k > total) return ""; |
| 51 | + |
| 52 | + string result(n, 'a'); |
| 53 | + |
| 54 | + // Define mappings for the next smallest and greatest valid characters |
| 55 | + unordered_map<char, char> nextSmallest = { |
| 56 | + {'a', 'b'}, {'b', 'a'}, {'c', 'a'}}; |
| 57 | + unordered_map<char, char> nextGreatest = { |
| 58 | + {'a', 'c'}, {'b', 'c'}, {'c', 'b'}}; |
| 59 | + |
| 60 | + // Calculate the starting indices for strings beginning with 'a', 'b', |
| 61 | + // and 'c' |
| 62 | + int startA = 1; |
| 63 | + int startB = startA + (1 << (n - 1)); |
| 64 | + int startC = startB + (1 << (n - 1)); |
| 65 | + |
| 66 | + // Determine the first character based on the value of k |
| 67 | + if (k < startB) { |
| 68 | + result[0] = 'a'; |
| 69 | + k -= startA; |
| 70 | + } else if (k < startC) { |
| 71 | + result[0] = 'b'; |
| 72 | + k -= startB; |
| 73 | + } else { |
| 74 | + result[0] = 'c'; |
| 75 | + k -= startC; |
| 76 | + } |
| 77 | + |
| 78 | + // Iterate through the remaining positions in the result string |
| 79 | + for (int charIndex = 1; charIndex < n; charIndex++) { |
| 80 | + // Calculate the midpoint of the group for the current character |
| 81 | + // position |
| 82 | + int midpoint = (1 << (n - charIndex - 1)); |
| 83 | + |
| 84 | + // Determine the next character based on the value of k |
| 85 | + if (k < midpoint) { |
| 86 | + result[charIndex] = nextSmallest[result[charIndex - 1]]; |
| 87 | + } else { |
| 88 | + result[charIndex] = nextGreatest[result[charIndex - 1]]; |
| 89 | + k -= midpoint; |
| 90 | + } |
| 91 | + } |
| 92 | + |
| 93 | + return result; |
| 94 | + } |
| 95 | +}; |
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