String Expansion - Amazon Top Interview Questions
Problem Statement :
You are given a string s consisting of lowercase alphabet characters, digits, and brackets"(" and ")". s encodes a longer string and is represented as concatenation of n(t), where n is the number of times t is repeated, and t is either a regular string or it's another encoded string recursively.
Return the expanded version of s. Note that t can be the empty string.
Example 1
Input
s = "2(ye)0(z)2(2(po)w)"
Output
"yeyepopowpopow"
Solution :
Solution in C++ :
string solve(string s) {
stack<string> sk;
stack<int> cnt;
sk.push("");
for (int i = 0; i < s.size(); i++) {
if (isdigit(s[i])) {
int c = atoi(&s[i]);
while (i < s.size() && isdigit(s[i])) i++;
cnt.push(c);
sk.push("");
} else if (s[i] == ')') {
int c = cnt.top();
cnt.pop();
string ss = sk.top();
sk.pop();
while (c-- > 0) {
sk.top().append(ss);
}
} else {
sk.top().push_back(s[i]);
}
}
return sk.top();
}
Solution in Java :
import java.util.*;
class Solution {
public String solve(String s) {
return recurse(s);
}
public String recurse(String s) {
if (s.length() == 0)
return s;
String ret = "";
for (int j = 0; j < s.length(); j++) {
if (Character.isDigit(s.charAt(j))) {
int start = s.substring(j, s.length()).indexOf("(") + j;
int x = Integer.parseInt(s.substring(j, start));
int count = 1;
int end;
for (end = start + 1; end < s.length(); end++) {
if (s.charAt(end) == '(') {
count++;
} else if (s.charAt(end) == ')') {
count--;
}
if (count == 0) {
break;
}
}
String embed = s.substring(start + 1, end);
ret += duplicate(recurse(embed), x);
j += embed.length() + 2;
} else {
ret += String.valueOf(s.charAt(j));
}
}
return ret;
}
public String duplicate(String s, int x) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < x; i++) {
sb.append(s);
}
return sb.toString();
}
}
Solution in Python :
class Solution:
def solve(self, s):
if "(" not in s:
return s
# where it all starts
start = s.index("(")
# i is the position at which digits start,
# so we can simply take the string before that as-is
for i in range(len(s)):
if s[i].isdigit():
break
# number of times we need to multiply this string
count = int(s[i:start])
# we need to find the position in string that balances the current set of braces
# so, currently since we already have a '(', balance is 1
# when balance becomes 0, we have reached the end of the string
# that we will multiply
balance = 1
for end in range(start + 1, len(s)):
if s[end] == "(":
balance += 1
elif s[end] == ")":
balance -= 1
if not balance:
break
# now we simply return
# 1. s[:i] = string before integer part
# 2. count * self.solve(s[start+1: end]) = simply multiply the string between the braces
# but also solve this substring recursively
# 3. self.solve(s[end+1:]) = run the recursive function for the remaining string on the right
return s[:i] + count * self.solve(s[start + 1 : end]) + self.solve(s[end + 1 :])
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