Formatted question description: https://leetcode.ca/all/925.html

925. Long Pressed Name (Easy)

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.

Constraints:

1 <= name.length <= 1000
1 <= typed.length <= 1000
name and typed contain only lowercase English letters.

Related Topics:
Two Pointers, String

Solution 1.

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        if (name == null || typed == null || name.length() == 0 || typed.length() == 0)
            return false;
        int nameLength = name.length(), typedLength = typed.length();
        int index2 = 0;
        for (int i = 0; i < nameLength; i++) {
            char c1 = name.charAt(i);
            if (c1 != typed.charAt(index2))
                return false;
            if (i == nameLength - 1 || c1 != name.charAt(i + 1)) {
                while (index2 < typedLength && typed.charAt(index2) == c1)
                    index2++;
            } else
                index2++;
            if (index2 == typedLength && i < nameLength - 1)
                return false;
        }
        return true;
    }
}

############

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        int m = name.length(), n = typed.length();
        int i = 0, j = 0;
        for (; i < m && j < n; ++i, ++j) {
            if (name.charAt(i) != typed.charAt(j)) {
                return false;
            }
            int cnt1 = 0, cnt2 = 0;
            char c = name.charAt(i);
            while (i + 1 < m && name.charAt(i + 1) == c) {
                ++i;
                ++cnt1;
            }
            while (j + 1 < n && typed.charAt(j + 1) == c) {
                ++j;
                ++cnt2;
            }
            if (cnt1 > cnt2) {
                return false;
            }
        }
        return i == m && j == n;
    }
}

// OJ: https://leetcode.com/problems/long-pressed-name/
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
    bool isLongPressedName(string name, string typed) {
        int i = 0, M = name.size(), j = 0, N = typed.size();
        while (i < M && j < N) {
            char c = name[i];
            int a = 0, b = 0;
            while (i < M && name[i] == c) ++i, ++a;
            while (j < N && typed[j] == c) ++j, ++b;
            if (a > b) return false;
        }
        return i == M && j == N;
    }
};

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        m, n = len(name), len(typed)
        i = j = 0
        while i < m and j < n:
            if name[i] != typed[j]:
                return False
            cnt1 = cnt2 = 0
            c = name[i]
            while i + 1 < m and name[i + 1] == c:
                i += 1
                cnt1 += 1
            while j + 1 < n and typed[j + 1] == c:
                j += 1
                cnt2 += 1
            if cnt1 > cnt2:
                return False
            i, j = i + 1, j + 1
        return i == m and j == n

############

class Solution(object):
    def isLongPressedName(self, name, typed):
        """
        :type name: str
        :type typed: str
        :rtype: bool
        """
        M = len(name)
        N = len(typed)
        i, j = 0, 0
        while i < M:
            c_i = name[i]
            count_i = 0
            count_j = 0
            while i < M and name[i] == c_i:
                i += 1
                count_i += 1
            while j < N and typed[j] == c_i:
                j += 1
                count_j += 1
            if count_j < count_i:
                return False
        return True

func isLongPressedName(name string, typed string) bool {
	m, n := len(name), len(typed)
	i, j := 0, 0
	for ; i < m && j < n; i, j = i+1, j+1 {
		if name[i] != typed[j] {
			return false
		}
		cnt1, cnt2 := 0, 0
		c := name[i]
		for i+1 < m && name[i+1] == c {
			i++
			cnt1++
		}
		for j+1 < n && typed[j+1] == c {
			j++
			cnt2++
		}
		if cnt1 > cnt2 {
			return false
		}
	}
	return i == m && j == n
}

925 - Long Pressed Name

925. Long Pressed Name (Easy)

Solution 1.

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925. Long Pressed Name (Easy)

Solution 1.

All Problems

All Solutions