<p>Your friend is typing his <code>name</code> into a keyboard. Sometimes, when typing a character <code>c</code>, the key might get <em>long pressed</em>, and the character will be typed 1 or more times.</p>

<p>You examine the <code>typed</code> characters of the keyboard. Return <code>True</code> if it is possible that it was your friends name, with some characters (possibly none) being long pressed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> name = &quot;alex&quot;, typed = &quot;aaleex&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&#39;a&#39; and &#39;e&#39; in &#39;alex&#39; were long pressed.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> name = &quot;saeed&quot;, typed = &quot;ssaaedd&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>&#39;e&#39; must have been pressed twice, but it was not in the typed output.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= name.length, typed.length &lt;= 1000</code></li>
	<li><code>name</code> and <code>typed</code> consist of only lowercase English letters.</li>
</ul>

class Solution {
public:
    bool isLongPressedName(string name, string typed) {
        int i1=1;
        int i2=1;
        int n1=name.size();
        int n2=typed.size();
        if(n1>n2){
            return false;
        }
        if(name[0]!=typed[0]) return false;
        while(i1<n1){
                if(name[i1]==typed[i2]){
                    i1++; i2++;
                }
                else if(typed[i2]==typed[i2-1]){
                    i2++;
                }
                else{
                    return false;
                }
        }
        cout<<i2;
        //checking for rest characters of typed
        while(i2<n2){
            if(typed[i2]!=typed[i2-1]){
                return false;
            }
            i2++;
        }

        return true;
    }
};
7900