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3330. Find the Original Typed String I

Description

Alice is attempting to type a specific string on her computer. However, she tends to be clumsy and may press a key for too long, resulting in a character being typed multiple times.

Although Alice tried to focus on her typing, she is aware that she may still have done this at most once.

You are given a string word, which represents the final output displayed on Alice's screen.

Return the total number of possible original strings that Alice might have intended to type.

 

Example 1:

Input: word = "abbcccc"

Output: 5

Explanation:

The possible strings are: "abbcccc", "abbccc", "abbcc", "abbc", and "abcccc".

Example 2:

Input: word = "abcd"

Output: 1

Explanation:

The only possible string is "abcd".

Example 3:

Input: word = "aaaa"

Output: 4

 

Constraints:

  • 1 <= word.length <= 100
  • word consists only of lowercase English letters.

Solutions

Solution 1

  • class Solution {
    public int possibleStringCount(String word) {
        int f = 1;
        for (int i = 1; i < word.length(); ++i) {
            if (word.charAt(i) == word.charAt(i - 1)) {
                ++f;
            }
        }
        return f;
    }
}


  • class Solution {
public:
    int possibleStringCount(string word) {
        int f = 1;
        for (int i = 1; i < word.size(); ++i) {
            f += word[i] == word[i - 1];
        }
        return f;
    }
};


  • class Solution:
    def possibleStringCount(self, word: str) -> int:
        return 1 + sum(x == y for x, y in pairwise(word))


  • func possibleStringCount(word string) int {
	f := 1
	for i := 1; i < len(word); i++ {
		if word[i] == word[i-1] {
			f++
		}
	}
	return f
}


  • function possibleStringCount(word: string): number {
    let f = 1;
    for (let i = 1; i < word.length; ++i) {
        f += word[i] === word[i - 1] ? 1 : 0;
    }
    return f;
}

All Problems

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