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New Problem Solution - "1935. Maximum Number of Words You Can Type"
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algorithms/cpp/maximumNumberOfWordsYouCanType/MaximumNumberOfWordsYouCanType.cpp
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// Source : https://leetcode.com/problems/maximum-number-of-words-you-can-type/ | ||
// Author : Hao Chen | ||
// Date : 2021-07-22 | ||
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/***************************************************************************************************** | ||
* | ||
* There is a malfunctioning keyboard where some letter keys do not work. All other keys on the | ||
* keyboard work properly. | ||
* | ||
* Given a string text of words separated by a single space (no leading or trailing spaces) and a | ||
* string brokenLetters of all distinct letter keys that are broken, return the number of words in | ||
* text you can fully type using this keyboard. | ||
* | ||
* Example 1: | ||
* | ||
* Input: text = "hello world", brokenLetters = "ad" | ||
* Output: 1 | ||
* Explanation: We cannot type "world" because the 'd' key is broken. | ||
* | ||
* Example 2: | ||
* | ||
* Input: text = "leet code", brokenLetters = "lt" | ||
* Output: 1 | ||
* Explanation: We cannot type "leet" because the 'l' and 't' keys are broken. | ||
* | ||
* Example 3: | ||
* | ||
* Input: text = "leet code", brokenLetters = "e" | ||
* Output: 0 | ||
* Explanation: We cannot type either word because the 'e' key is broken. | ||
* | ||
* Constraints: | ||
* | ||
* 1 <= text.length <= 10^4 | ||
* 0 <= brokenLetters.length <= 26 | ||
* text consists of words separated by a single space without any leading or trailing spaces. | ||
* Each word only consists of lowercase English letters. | ||
* brokenLetters consists of distinct lowercase English letters. | ||
******************************************************************************************************/ | ||
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class Solution { | ||
public: | ||
int canBeTypedWords(string text, string brokenLetters) { | ||
vector<bool> borken(26, false); | ||
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for (auto ch : brokenLetters) { | ||
borken[ch - 'a'] = true; | ||
} | ||
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text += ' '; | ||
int cnt = 0; | ||
for (int i = 0; i < text.size(); i++ ) { | ||
if ( text[i] == ' ') continue; | ||
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bool skip = false; | ||
for (; text[i] != ' '; i++ ) { | ||
if (borken[text[i] - 'a'] == true ) skip = true; | ||
} | ||
if ( !skip ) cnt++; | ||
} | ||
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return cnt; | ||
} | ||
}; |