Counting Words With a Given Prefix

Counting Words With a Given Prefix - Problem

You are 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.

Input & Output

Example 1 — Basic Case

$ Input: words = ["pay","attention","practice","attend"], pref = "at"

› Output: 2

💡 Note: Only "attention" and "attend" start with "at", so return 2

Example 2 — No Matches

$ Input: words = ["leetcode","win","loops","success"], pref = "code"

› Output: 0

💡 Note: None of the words start with "code" as a prefix

Example 3 — All Match

$ Input: words = ["car","care","careful"], pref = "car"

› Output: 3

💡 Note: All three words start with "car"

Constraints

1 ≤ words.length ≤ 1000
1 ≤ words[i].length, pref.length ≤ 100
words[i] and pref consist of lowercase English letters

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use built-in string methods like startsWith() to efficiently check if each word begins with the given prefix. Best approach uses built-in string prefix checking with Time: O(n × m), Space: O(1).

Common Approaches

✓ Built-in String Prefix Check

⏱️ Time: O(n × m) Space: O(1)

Leverage the built-in startsWith() or similar methods to efficiently check if each word begins with the given prefix. This is cleaner and more readable than manual character comparison.

Character-by-Character Comparison

⏱️ Time: O(n × m) Space: O(1)

For each word, iterate through characters and compare them one by one with the prefix characters. Count words where all prefix characters match at the beginning.

Built-in String Prefix Check — Algorithm Steps

Iterate through each word in the array
Use built-in method to check if word starts with prefix
Count and return total matches

Visualization

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Step-by-Step Walkthrough

Use Built-in Method

Call startsWith() or similar for each word

Efficient Check

Built-in method handles character comparison

Count Results

Increment counter for matches

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char words[][100], int wordCount, char* pref) {
    int count = 0;
    int prefLen = strlen(pref);
    
    for (int i = 0; i < wordCount; i++) {
        if (strncmp(words[i], pref, prefLen) == 0) {
            count++;
        }
    }
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple JSON parsing for array
    char words[100][100];
    int wordCount = 0;
    char* token = strtok(line, "[],\"");
    while (token != NULL && wordCount < 100) {
        if (strlen(token) > 0 && strcmp(token, " ") != 0) {
            strcpy(words[wordCount], token);
            wordCount++;
        }
        token = strtok(NULL, "[],\"");
    }
    
    char pref[100];
    fgets(pref, sizeof(pref), stdin);
    pref[strcspn(pref, "\n")] = 0;
    
    int result = solution(words, wordCount, pref);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

n words × m average prefix length, but optimized by built-in string methods

✓ Linear Growth

Space Complexity

O(1)

Only using counter variable, no extra data structures needed

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Built-in String Prefix Check — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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