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### by Sumit Chourasia | May 06, 2020 | Category :coding | Tags : algorithmडेटा-संरचनाstringआसानleetcode #### Long Pressed Name - मिनी टीवी

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 friend's 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 wasn't in the typed output.``````

Example 3:

``````Input: name = "leelee", typed = "lleeelee"

Output: true``````

Example 4:

``````Input: name = "laiden", typed = "laiden"

Output: true

Explanation: It's not necessary to long press any character.``````

Constraints:

``````1 <= name.length <= 1000

1 <= typed.length <= 1000

The characters of name and typed are lowercase letters.``````

Solution:

``````using System;
using System.Collections.Generic;
using System.Text;

{
class IsLongPressedNameSoln
{
public void execute()
{
var name = "alex";
var typed = "aaleelx";

var res = IsLongPressedName(name, typed);
}

public bool IsLongPressedName(string name, string typed)
{
if (name != typed) return false;
for (int i = 0, j=0; i < name.Length;)
{
if (name[i] == typed[j])
{
j++;
i++;
}
else
{
if (typed[j] == typed[j - 1])
j++;
else
return false;
}
if (j == typed.Length && i == name.Length)
return true;
if (j >= typed.Length)
return false;
if (i == name.Length)
i--;

}
return false;
}
}
}
``````

Time Complexity: O(n)

Space Complexity: O(1)