Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
Example 1:
Input: S = “ab#c”, T = “ad#c”
Output: true
Explanation: Both S and T become “ac”.Example 2:
Input: S = “ab##”, T = “c#d#”
Output: true
Explanation: Both S and T become “”.Example 3:
Input: S = “a##c”, T = “#a#c”
Output: true
Explanation: Both S and T become “c”.Example 4:
Input: S = “a#c”, T = “b”
Output: false
Explanation: S becomes “c” while T becomes “b”.Note:
1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and ‘#’ characters. Can you solve it in O(N) time and O(1) space?
Javascript algorithm to perform backspace string comparison
To simulate the backspace key, we can use a stack, and perform a pop operation when we want to delete previous character. In Javascript, we can use Array.prototype.pop() to remove the last element (which can be called on empty array and that returns undefined). When we iterate all characters, we need to join the stack/array as a string. Then, the last step is to perform a string comparisons.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | /** * @param {string} S * @param {string} T * @return {boolean} */ var backspaceCompare = function(S, T) { const build = (S) => { let st = []; for (let i = 0, len = S.length; i < len; ++ i) { if (S[i] == '#') { st.pop(); // if called on empty array, return undefined. } else { st.push(S[i]); } } return st.join(''); } return build(S) === build(T); }; |
/**
* @param {string} S
* @param {string} T
* @return {boolean}
*/
var backspaceCompare = function(S, T) {
const build = (S) => {
let st = [];
for (let i = 0, len = S.length; i < len; ++ i) {
if (S[i] == '#') {
st.pop(); // if called on empty array, return undefined.
} else {
st.push(S[i]);
}
}
return st.join('');
}
return build(S) === build(T);
};String backspace comparison in C++
Slightly differently, we do not construct/join the characters in the stack, instead, we can perform comparisons on both stacks generated from two input strings.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | class Solution { public: bool backspaceCompare(string S, string T) { stack<char> st1 = realString(S); stack<char> st2 = realString(T); if (st1.size() != st2.size()) return false; while (st1.size() > 0) { char p1 = st1.top(); char p2 = st2.top(); st1.pop(); st2.pop(); if (p1 != p2) return false; } return true; } private: stack<char> realString(string S) { stack<char> st; for (int i = 0; i < S.size(); ++ i) { if (S[i] == '#') { if (st.size() > 0) { st.pop(); } } else { st.push(S[i]); } } return st; } }; |
class Solution {
public:
bool backspaceCompare(string S, string T) {
stack<char> st1 = realString(S);
stack<char> st2 = realString(T);
if (st1.size() != st2.size()) return false;
while (st1.size() > 0) {
char p1 = st1.top();
char p2 = st2.top();
st1.pop();
st2.pop();
if (p1 != p2) return false;
}
return true;
}
private:
stack<char> realString(string S) {
stack<char> st;
for (int i = 0; i < S.size(); ++ i) {
if (S[i] == '#') {
if (st.size() > 0) {
st.pop();
}
} else {
st.push(S[i]);
}
}
return st;
}
};Instead of using stack, we can use a string. And the implementation is similar. We can push (string concatenation) and pop_back (to remove the last character).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | class Solution { public: bool backspaceCompare(string S, string T) { return w(S) == w(T); } private: string w(const string &s) { string ans = ""; for (const auto &n: s) { if (n == '#') { if (!ans.empty()) { ans.pop_back(); } } else { ans.push_back(n); } } return ans; } }; |
class Solution {
public:
bool backspaceCompare(string S, string T) {
return w(S) == w(T);
}
private:
string w(const string &s) {
string ans = "";
for (const auto &n: s) {
if (n == '#') {
if (!ans.empty()) {
ans.pop_back();
}
} else {
ans.push_back(n);
}
}
return ans;
}
};Java implementation of string backspace comparisons
We can use String.valueOf(stack) to convert the stack of characters into String.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | class Solution { public boolean backspaceCompare(String S, String T) { return build(S).equals(build(T)); } public String build(String S) { Stack<Character> ans = new Stack(); for (char c: S.toCharArray()) { if (c != '#') ans.push(c); else if (!ans.empty()) ans.pop(); } return String.valueOf(ans); } } |
class Solution {
public boolean backspaceCompare(String S, String T) {
return build(S).equals(build(T));
}
public String build(String S) {
Stack<Character> ans = new Stack();
for (char c: S.toCharArray()) {
if (c != '#')
ans.push(c);
else if (!ans.empty())
ans.pop();
}
return String.valueOf(ans);
}
}All above implementations run at complexity O(M + N) in both time and space where M and N are the lengths of input strings S and T respectively.
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