Description
Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
You examine the typed characters of the keyboard. Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
Example 1:
Input: name = "alex", typed = "aaleex"
Output: true
Explanation: 'a' and 'e' in 'alex' were long pressed.
Example 2:
Input: name = "saeed", typed = "ssaaedd"
Output: false
Explanation: 'e' must have been pressed twice, but it was not in the typed output.
Constraints:
1 <= name.length, typed.length <= 1000name and typed consist of only lowercase English letters.
Solutions
Solution 1: Two Pointers
We use two pointers i and j to point to the first character of the strings typed and name respectively, and then start traversing. If typed[j] is not equal to name[i], it means the two strings do not match, and we directly return False. Otherwise, we find the next position of the continuous identical characters, denoted as x and y respectively. If x - i > y - j, it means the number of characters in typed is less than the number of characters in name, and we directly return False. Otherwise, we update i and j to x and y respectively, continue traversing, until i and j have traversed name and typed respectively, and return True.
The time complexity is O(m + n), where m and n are the lengths of the strings name and typed respectively. The space complexity is $O(1)`.
PythonJavaC++GoTypeScriptRust
class Solution:
def isLongPressedName(self, name: str, typed: str) -> bool:
m, n = len(name), len(typed)
i = j = 0
while i < m and j < n:
if name[i] != typed[j]:
return False
x = i + 1
while x < m and name[x] == name[i]:
x += 1
y = j + 1
while y < n and typed[y] == typed[j]:
y += 1
if x - i > y - j:
return False
i, j = x, y
return i == m and j == n(code-box)
class Solution {
public boolean isLongPressedName(String name, String typed) {
int m = name.length(), n = typed.length();
int i = 0, j = 0;
while (i < m && j < n) {
if (name.charAt(i) != typed.charAt(j)) {
return false;
}
int x = i + 1;
while (x < m && name.charAt(x) == name.charAt(i)) {
++x;
}
int y = j + 1;
while (y < n && typed.charAt(y) == typed.charAt(j)) {
++y;
}
if (x - i > y - j) {
return false;
}
i = x;
j = y;
}
return i == m && j == n;
}
}(code-box)
class Solution {
public:
bool isLongPressedName(string name, string typed) {
int m = name.length(), n = typed.length();
int i = 0, j = 0;
while (i < m && j < n) {
if (name[i] != typed[j]) {
return false;
}
int x = i + 1;
while (x < m && name[x] == name[i]) {
++x;
}
int y = j + 1;
while (y < n && typed[y] == typed[j]) {
++y;
}
if (x - i > y - j) {
return false;
}
i = x;
j = y;
}
return i == m && j == n;
}
};(code-box)
func isLongPressedName(name string, typed string) bool {
m, n := len(name), len(typed)
i, j := 0, 0
for i < m && j < n {
if name[i] != typed[j] {
return false
}
x, y := i+1, j+1
for x < m && name[x] == name[i] {
x++
}
for y < n && typed[y] == typed[j] {
y++
}
if x-i > y-j {
return false
}
i, j = x, y
}
return i == m && j == n
}(code-box)
function isLongPressedName(name: string, typed: string): boolean {
const [m, n] = [name.length, typed.length];
let i = 0;
let j = 0;
while (i < m && j < n) {
if (name[i] !== typed[j]) {
return false;
}
let x = i + 1;
while (x < m && name[x] === name[i]) {
x++;
}
let y = j + 1;
while (y < n && typed[y] === typed[j]) {
y++;
}
if (x - i > y - j) {
return false;
}
i = x;
j = y;
}
return i === m && j === n;
}(code-box)
impl Solution {
pub fn is_long_pressed_name(name: String, typed: String) -> bool {
let (m, n) = (name.len(), typed.len());
let (mut i, mut j) = (0, 0);
let s: Vec<char> = name.chars().collect();
let t: Vec<char> = typed.chars().collect();
while i < m && j < n {
if s[i] != t[j] {
return false;
}
let mut x = i + 1;
while x < m && s[x] == s[i] {
x += 1;
}
let mut y = j + 1;
while y < n && t[y] == t[j] {
y += 1;
}
if x - i > y - j {
return false;
}
i = x;
j = y;
}
i == m && j == n
}
}(code-box)
