925.long-pressed-name.cpp

/*
 * @lc app=leetcode id=925 lang=cpp
 *
 * [925] Long Pressed Name
 *
 * https://leetcode.com/problems/long-pressed-name/description/
 *
 * algorithms
 * Easy (44.35%)
 * Total Accepted:    14.6K
 * Total Submissions: 32.9K
 * Testcase Example:  '"alex"\n"aaleex"'
 *
 * 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 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.
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 * Note:
 * 
 * 
 * name.length <= 1000
 * typed.length <= 1000
 * The characters of name and typed are lowercase letters.
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 */
class Solution {
public:
    bool isLongPressedName(string name, string typed) {
        int i = 0, j = 0;
        while(i-1 < (int)name.size() && j < (int)typed.size()){
            cout<<i<<":"<<j<<" ";
            if(i < (int)name.size() && name[i] == typed[j])
                i++, j++;
            else if(name[i-1] == typed[j])
                j++;
            else
                return false;
        }
        
        if(i == (int)name.size() && j == (int)typed.size())
            return true;
        return false;
    }
};