Algorithm
Problem Name: 925. Long Pressed Name
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 <= 1000nameandtypedconsist of only lowercase English letters.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
bool isLongPressedName(string name, string typed) {
int a = 0, b = 0, n = name.size(), m = typed.size();
while (a < n && b < m) {
if (name[a++] != typed[b++]) return false;
while (b > 0 && name[a] != typed[b] && typed[b] == typed[b - 1]) ++b;
}
return a == n && b == m;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean isLongPressedName(String name, String typed) {
int idxName = 0;
int idxTyped = 0;
while (idxName < name.length() && idxTyped < typed.length()) {
if (name.charAt(idxName) != typed.charAt(idxTyped)) {
return false;
}
char c = name.charAt(idxName);
int nameFreq = 0;
while (idxName < name.length() && name.charAt(idxName) == c) {
idxName++;
nameFreq++;
}
c = typed.charAt(idxTyped);
int typedFreq = 0;
while (idxTyped < typed.length() && typed.charAt(idxTyped) == c) {
idxTyped++;
typedFreq++;
}
if (nameFreq > typedFreq) {
return false;
}
}
return idxName == name.length() && idxTyped == typed.length();
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const isLongPressedName = function(name, typed) {
let i = 0, m = name.length, n = typed.length
for(let j = 0; j < n; j++) {
if(i < m && name[i] === typed[j]) i++
else if(j === 0 || typed[j] !== typed[j - 1]> return false
}
return i === m
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def isLongPressedName(self, name, typed):
pre, i = None, 0
for c in typed:
if i < len(name) and c == name[i]:
pre, i = name[i], i + 1
elif c != pre:
return False
return i == len(name)
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0925_LongPressedName
{
public bool IsLongPressedName(string name, string typed)
{
int i = 0, j = 0;
while (i < name.Length && j < typed.Length)
{
if (name[i] == typed[j])
{
i++;
j++;
}
else
{
if (j > 0 && typed[j] == typed[j - 1])
j++;
else
return false;
}
}
while (j < typed.Length && j > 0 && typed[j] == typed[j - 1])
j++;
return i == name.Length && j == typed.Length;
}
}
}
Input
Output