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📅  最后修改于: 2021-05-24 20:57:42             🧑  作者: Mango
给定两个字符串A和B由两种交换A [I]与B [I]或保持它不变最小化独特字符字符串A的数量。交换的数量可以大于或等于0。请注意,A [i]只能与B中相同的索引元素交换。打印最小数量的唯一字符。约束:0 例子: Input : A = ababa B = babab Output : 1 Swapping all b's in string A, with a's in string B results in string A having all characters as a. Input : A = abaaa B = bbabb Output : 2 Initially string A has 2 unique characters. Swapping at any index does not change this count. 推荐:请首先在IDE上尝试您的方法,然后查看解决方案。 方法:可以使用回溯解决问题。创建一个键为A [i]且值为对应字符的计数的映射。映射的大小表明了不同字符的数量,因为只有字符串A中存在的那些元素才作为映射中的键出现。在每个索引位置,都有两种选择:将A [i]与B [i]交换或保持A [i]不变。从索引0开始,并对每个索引执行以下操作: 保持A [i]不变,在映射中将A [i]的计数加1,然后递归调用下一个索引。 通过将A [i]的计数减1,将A [i]与B [i]交换,在映射中将A [i]的计数增1进行回溯,并再次递归调用下一个索引。 保留变量ans以存储不同字符的整体最小值。在上述两种情况下,遍历整个字符串时,将当前的不同字符与ans的最小值进行比较,并相应地更新ans。 执行: // CPP program to minimize number of // unique characters in a string. #include using namespace std; // Utility function to find minimum // number of unique characters in string. void minCountUtil(string A, string B, unordered_map& ele, int& ans, int ind) { // If entire string is traversed, then // compare current number of distinct // characters in A with overall minimum. if (ind == A.length()) { ans = min(ans, (int)ele.size()); return; } // swap A[i] with B[i], increase count of // corresponding character in map and call // recursively for next index. swap(A[ind], B[ind]); ele[A[ind]]++; minCountUtil(A, B, ele, ans, ind + 1); // Backtrack (Undo the changes done) ele[A[ind]]--; // If count of character is reduced to zero, // then that character is not present in A. // So remove that character from map. if (ele[A[ind]] == 0) ele.erase(A[ind]); // Restore A to original form. // (Backtracking step) swap(A[ind], B[ind]); // Increase count of A[i] in map and // call recursively for next index. ele[A[ind]]++; minCountUtil(A, B, ele, ans, ind + 1); // Restore the changes done // (Backtracking step) ele[A[ind]]--; if (ele[A[ind]] == 0) ele.erase(A[ind]); } // Function to find minimum number of // distinct characters in string. int minCount(string A, string B) { // Variable to store minimum number // of distinct character. // Initialize it with length of A // as maximum possible value is // length of A. int ans = A.length(); // Map to store count of distinct // characters in A. To keep // complexity of insert operation // constant unordered_map is used. unordered_map ele; // Call utility function to find // minimum number of unique // characters. minCountUtil(A, B, ele, ans, 0); return ans; } int main() { string A = "abaaa"; string B = "bbabb"; cout << minCount(A, B); return 0; } 输出: 2 时间复杂度: O(2 n )辅助空间: O(n)
例子:
Input : A = ababa B = babab Output : 1 Swapping all b's in string A, with a's in string B results in string A having all characters as a. Input : A = abaaa B = bbabb Output : 2 Initially string A has 2 unique characters. Swapping at any index does not change this count.
方法:可以使用回溯解决问题。创建一个键为A [i]且值为对应字符的计数的映射。映射的大小表明了不同字符的数量,因为只有字符串A中存在的那些元素才作为映射中的键出现。在每个索引位置,都有两种选择:将A [i]与B [i]交换或保持A [i]不变。从索引0开始,并对每个索引执行以下操作:
保留变量ans以存储不同字符的整体最小值。在上述两种情况下,遍历整个字符串时,将当前的不同字符与ans的最小值进行比较,并相应地更新ans。
执行:
// CPP program to minimize number of // unique characters in a string. #include using namespace std; // Utility function to find minimum // number of unique characters in string. void minCountUtil(string A, string B, unordered_map& ele, int& ans, int ind) { // If entire string is traversed, then // compare current number of distinct // characters in A with overall minimum. if (ind == A.length()) { ans = min(ans, (int)ele.size()); return; } // swap A[i] with B[i], increase count of // corresponding character in map and call // recursively for next index. swap(A[ind], B[ind]); ele[A[ind]]++; minCountUtil(A, B, ele, ans, ind + 1); // Backtrack (Undo the changes done) ele[A[ind]]--; // If count of character is reduced to zero, // then that character is not present in A. // So remove that character from map. if (ele[A[ind]] == 0) ele.erase(A[ind]); // Restore A to original form. // (Backtracking step) swap(A[ind], B[ind]); // Increase count of A[i] in map and // call recursively for next index. ele[A[ind]]++; minCountUtil(A, B, ele, ans, ind + 1); // Restore the changes done // (Backtracking step) ele[A[ind]]--; if (ele[A[ind]] == 0) ele.erase(A[ind]); } // Function to find minimum number of // distinct characters in string. int minCount(string A, string B) { // Variable to store minimum number // of distinct character. // Initialize it with length of A // as maximum possible value is // length of A. int ans = A.length(); // Map to store count of distinct // characters in A. To keep // complexity of insert operation // constant unordered_map is used. unordered_map ele; // Call utility function to find // minimum number of unique // characters. minCountUtil(A, B, ele, ans, 0); return ans; } int main() { string A = "abaaa"; string B = "bbabb"; cout << minCount(A, B); return 0; }
2
时间复杂度: O(2 n )辅助空间: O(n)