leetcode 925. Long Pressed Name

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        int i = 0, j = 0;
        while (i < name.length()) {
            if (j == typed.length() || name.charAt(i) != typed.charAt(j)) return false;
            int i1 = i;
            for (;i1 < name.length() && name.charAt(i1) == name.charAt(i); i1++);
            int i2 = j;
            for (;i2 < typed.length() && typed.charAt(i2) == typed.charAt(j);i2++);
            if (i2 - j < i1 - i) return false;
            i = i1;
            j = i2;
        }
        if (j < typed.length()) return false;
        return true;
    }
}

A shorter solution

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        int M = name.length(), N = typed.length();
        int j = 0;
        for (int i = 0; i < N; ++i) {
            if (j < M && typed.charAt(i) == name.charAt(j)) {
                ++j;
            }
            else if (i == 0 || typed.charAt(i - 1) != typed.charAt(i)) {
                return false;
            }
        }
        return j == M;
    }
}