Counting Words With a Given Prefix

<?php
/**
 * @param String[] $words
 * @param String $pref
 * @return Integer
 */
function countWordsWithPrefix($words, $pref) {
    $count = 0;
    foreach ($words as $word) {
        if (strpos($word, $pref) === 0) {
            $count++;
        }
    }
    return $count;
}

// Example Usage
$words1 = ["pay", "attention", "practice", "attend"];
$pref1 = "at";
echo countWordsWithPrefix($words1, $pref1); // Output: 2

$words2 = ["leetcode", "win", "loops", "success"];
$pref2 = "code";
echo countWordsWithPrefix($words2, $pref2); // Output: 0
?>

2185. Counting Words With a Given Prefix

Difficulty: Easy

Topics: Array, String, String Matching

Given an array of strings words and a string pref, return the number of strings in words that contain pref as a prefix.

A prefix of a string s is any leading contiguous substring of s.

Example 1:

• Input: words = ["pay","attention","practice","attend"], pref = "at"
• Output: 2
• Explanation: The 2 strings that contain "at" as a prefix are: "attention" and "attend".

Example 2:

• Input: words = ["leetcode","win","loops","success"], pref = "code"
• Output: 0
• Explanation: There are no strings that contain "code" as a prefix.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 20
• 1 <= pref.length <= 20
• words[i] and pref consist of lowercase English letters.

Improved Solution (using strpos):

The provided solution uses substr which is less efficient than strpos for this specific task. strpos directly checks for the prefix at the beginning of the string, avoiding unnecessary substring creation.

This improved PHP solution uses strpos:

<?php
function countWordsWithPrefix(array $words, string $pref): int {
    $count = 0;
    foreach ($words as $word) {
        if (strpos($word, $pref) === 0) { // Check if pref is at the beginning (index 0)
            $count++;
        }
    }
    return $count;
}
?>

Time Complexity: O(n*m) in the worst case, where n is the number of words and m is the length of the prefix. However, on average, it will be faster than the original substr solution.

Space Complexity: O(1) - Constant extra space is used.

This revised answer provides a more efficient solution and maintains the clarity of the explanation. The image remains unchanged as it's relevant to the problem statement.

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