Count Prefixes of a Given String

Count Prefixes of a Given String - Problem

Array String Easy

You are given a string array words and a string s, where words[i] and s comprise only of lowercase English letters.

Return the number of strings in words that are a prefix of s.

A prefix of a string is a substring that occurs at the beginning of the string. A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: words = ["a", "b", "c", "ab", "bc", "abc"], s = "abc"

› Output: 3

💡 Note: The words "a", "ab", and "abc" are prefixes of "abc". "b", "c", and "bc" do not start at the beginning of "abc".

Example 2 — No Matches

$ Input: words = ["a", "a"], s = "aa"

› Output: 2

💡 Note: Both instances of "a" are prefixes of "aa". The string "a" matches the beginning of "aa".

Example 3 — Empty Prefix Not Allowed

$ Input: words = ["hello", "world", "hi"], s = "hi"

› Output: 1

💡 Note: Only "hi" exactly matches the target string "hi" as a prefix. "hello" and "world" don't match the beginning.

Constraints

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

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is to check if each word matches the beginning characters of the target string. The optimal approach uses built-in string prefix methods like startsWith() for clean, readable code. Time: O(n×m), Space: O(1).

Common Approaches

✓ Sliding Window

⏱️ Time: N/A Space: N/A

Built-in String Methods

⏱️ Time: O(n × m) Space: O(1)

Instead of manually comparing characters, use the language's built-in string prefix checking methods like startsWith() in JavaScript/Java, or string slicing comparison in Python. This makes the code more readable and potentially optimized.

Character-by-Character Comparison

⏱️ Time: O(n × m) Space: O(1)

For each word in the array, compare it character by character with the beginning of string s. If all characters match and we don't run out of characters in s, then it's a valid prefix.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

long long solution(int* nums, int numsSize, int k) {
    // Find the maximum element in the array
    int maxElement = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] > maxElement) {
            maxElement = nums[i];
        }
    }
    
    long long count = 0;
    int left = 0;
    int maxCount = 0;
    
    // Use sliding window approach
    for (int right = 0; right < numsSize; right++) {
        // Expand window by including nums[right]
        if (nums[right] == maxElement) {
            maxCount++;
        }
        
        // Contract window from left while we have at least k max elements
        while (maxCount >= k) {
            // All subarrays from left to any position >= right are valid
            count += numsSize - right;
            
            // Contract from left
            if (nums[left] == maxElement) {
                maxCount--;
            }
            left++;
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]"); // Skip [
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    long long result = solution(nums, numsSize, k);
    printf("%lld\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

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