Recursively decode a string encoded as a count followed by a substring
In this problem, we need to decode the given string by repeatedly adding the total count number of times.
We can approach the problem in three different ways, and we can use two stacks or one stack to solve the problem. Also, we can solve the problem without using two stacks.
Problem Statement - We are given a string str containing left and right brackets, alpha and numeric characters. We need to decode the string recursively.
The following are patterns or rules for decoding strings.
[chars] - "chars" should appear count times in the result string.
Example
enter
str = "2[d]3[c2[n]]";
Output
ddcnncnncnn
illustrate
First, we decode 2[n] and get "2[d]3[cnn]".
Next, we decode 3[cnn]. So, we get "2[d]cnnncnncnn".
Next, we decode 2[d]. So, we get "ddcnnncnncnn".
enter
5[i]
Output
iiiii
Explanation- When we decode the given string, we get 5 "I"s.
enter
3[fg]
Output
fgfgfg
Explanation- When we decode the input string, we get "fg" 3 times.
method 1
We will use two stacks to solve the problem in this method. When we get an opening parenthesis, we push it onto the stack. Additionally, when we get numeric characters, we append all numeric characters to a valid positive integer and add them to the integer stack. Afterwards, when we get the closing bracket, we pop the integer and character from the stack.
algorithm
Step 1- Define the "instSt" stack to store numbers and "strSt" to store string characters and left brackets. Additionally, "answer" is initialized to store the result string and "tempStr" is initialized to store the temporary string.
Step 2- Start traversing the string.
Step 3 - If the current character is a number, initialize "num" with 0 to store the number.
Step 3.1 − If the character at the pthth index is a number, traverse the string until you get an alphabetic character or a bracket. Within the loop, multiply the previous value of "num" by 10 and add the current number to it.
Step 3.2− Increase the value of "p" by 1.
Step 3.3 - Push the numeric value to the "instSt" stack.
Step 4 - If the character at the pth index is a right bracket, follow these steps.
Step 4.1- Initialize "temp_str" with an empty string. Afterwards, if 'instSt' is not empty, pop the top integer from the stack.
Step 4.2 - Now, use a loop until we get the left bracket or the stack becomes empty from the "strSt" stack. Also, append characters to "temp_str".
Step 4.3 - If we stopped pooping the character due to "[", remove it.
Step 4.4 - Next, we need to append "temp_Str" "num" times to the "answer" string.
Step 4.5 - Insert each character of the "answer" string into the "strSt" stack and reinitialize it with an empty string.
Step 5 − If the current character is a left bracket, please follow the steps below.
Step 5.1 − If the previous character is a number, push "[" onto the stack "StrSt". Otherwise, push '[' onto the StrSt stack and push 1 onto the 'instSt' stack.
Step 6− If we get an alphabetic character, push it onto the "strSt" stack.
Step 7 - Finally, use a loop to remove all characters from the "strSt" stack, append to the "answer" string, and return it.
Example
#includeusing namespace std; string decodeTheString(string alpha) { stack instSt; stack StrSt; string tempStr = "", answer = ""; // Iterate the string for (int p = 0; p < alpha.length(); p++) { int num = 0; // If we find the number, extract the number and push it to the stack if (alpha[p] >= '0' && alpha[p] <= '9') { // Making iterations until we get an alphabetic character while (alpha[p] >= '0' && alpha[p] <= '9') { num = num * 10 + alpha[p] - '0'; p++; } p--; instSt.push(num); } // If the character at the pth index is closing bracket else if (alpha[p] == ']') { tempStr = ""; num = 0; // Pop the number from the stack if (!instSt.empty()) { num = instSt.top(); instSt.pop(); } // Pop the character until we get the opening bracket while (!StrSt.empty() && StrSt.top() != '[') { tempStr = StrSt.top() + tempStr; StrSt.pop(); } // remove the opening bracket if (!StrSt.empty() && StrSt.top() == '[') StrSt.pop(); // Append string to answer for num times for (int j = 0; j < num; j++) answer = answer + tempStr; // Insert the updated string again into the stack for (int j = 0; j < answer.length(); j++) StrSt.push(answer[j]); answer = ""; } // If the character at the pth index is an opening bracket else if (alpha[p] == '[') { if (alpha[p - 1] >= '0' && alpha[p - 1] <= '9') { StrSt.push(alpha[p]); } else { StrSt.push(alpha[p]); instSt.push(1); } } else { // Push alphabetic character in the string stack. StrSt.push(alpha[p]); } } // Pop all the elements, make a string, and return. while (!StrSt.empty()) { answer = StrSt.top() + answer; StrSt.pop(); } return answer; } // starting code int main() { string str = "2[d]3[c2[n]]"; cout << "The resultant string after decoding it is - " << decodeTheString(str) << endl; return 0; }
Output
The resultant string after decoding it is - ddcnncnncnn
Time complexity− O(n^2), because we iterate over the string and keep pushing and popping elements to the stack.
Space Complexity− O(n) to store elements in the stack.
Method 2
We will solve the problem without using stack in this method. Additionally, we will use the reverse() method to reverse the string.
algorithm
Step 1- Start iterating the string.
Step 2− If the i-th character is "]", push it into the "answer" string. Otherwise, follow the steps below.
Step 3 - Initialize "temp_Str" with an empty string.
Step 4 - Continue iterating through the "answer" string until the string is empty or the "[" character is found. Also, continue popping the last character from the "answer" string and appending it to the "temp_Str" string.
Step 5 - Reverse the "temp_Str" string as we traverse backward from where we found the "]" bracket.
Step 6 - Remove the last character from the "answer" string to remove the "[" character.
第 7 步- 如果“答案”字符串顶部包含数字,则使用数字生成一个整数,并将其存储在 number 变量中。
第 8 步- 反转数字字符串。
第 9 步- 使用 stoi() 方法将字符串转换为数字。
步骤 10 - 将 temp_Str 字符串附加到答案字符串“number”次。
第 11 步- 返回“答案”字符串。
示例
#includeusing namespace std; string decodeTheString(string alpha) { string answer = ""; // iterate the string characters for (int i = 0; i < alpha.length(); i++) { // for all other characters except the closing bracket if (alpha[i] != ']') { answer.push_back(alpha[i]); } else { // Extract the substring lying within the pair string temp_str = ""; // Keep on popping characters until '[' is found. while (!answer.empty() && answer.back() != '[') { temp_str.push_back(answer.back()); answer.pop_back(); } // get original string by reversing the string reverse(temp_str.begin(), temp_str.end()); // open bracket removal answer.pop_back(); // get integer value before the '[' character string number = ""; // get the number before opening bracket while (!answer.empty() && answer.back() >= '0' && answer.back() <= '9') { number.push_back(answer.back()); answer.pop_back(); } // reverse number string reverse(number.begin(), number.end()); // convert string to integer int numInt = stoi(number); for (int p = 0; p < numInt; p++) { answer += temp_str; } } } return answer; } int main() { string str = "2[d]3[c2[n]]"; cout << "The resultant string after decoding it is - " << decodeTheString(str) << endl; return 0; }
输出
The resultant string after decoding it is − ddcnncnncnn
时间复杂度− O(N^2),因为我们遍历字符串并在循环内使用reverse()方法。
空间复杂度− O(N) 来存储数字和临时字符串。
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