Check If String Is a Prefix of Array

Check If String Is a Prefix of Array - Problem

Given a string s and an array of strings words, determine whether s is a prefix string of words.

A string s is a prefix string of words if s can be made by concatenating the first k strings in words for some positive k no larger than words.length.

Return true if s is a prefix string of words, or false otherwise.

Input & Output

Example 1 — Perfect Match

$ Input: s = "iloveleetcode", words = ["i", "love", "leetcode", "apples"]

› Output: true

💡 Note: Concatenating first 3 words: "i" + "love" + "leetcode" = "iloveleetcode" which equals s exactly

Example 2 — No Valid Prefix

$ Input: s = "iloveleetcode", words = ["apples", "i", "love", "leetcode"]

› Output: false

💡 Note: First word "apples" doesn't match start of s="iloveleetcode", so no valid prefix exists

Example 3 — Partial Match Only

$ Input: s = "ccccccccc", words = ["c", "cc"]

› Output: false

💡 Note: Concatenating all words gives "c" + "cc" = "ccc", but s has length 9, so s cannot be formed as a prefix

Constraints

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

Visualization

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Asked in

f Facebook 15 G Google 12 a Amazon 8

The optimal approach uses two pointers to match characters without creating intermediate strings. The key insight is that we can check character-by-character while iterating through words, achieving O(m) time and O(1) space complexity where m is the length of the target string.

Common Approaches

✓ Brute Force - Try All Combinations

⏱️ Time: O(n × m) Space: O(m)

For each possible k from 1 to words.length, concatenate the first k words and check if it equals s. This explores all possible prefix combinations.

Optimized - Build and Check Incrementally

⏱️ Time: O(n × m) Space: O(m)

Instead of rebuilding the entire concatenation for each k, we build it incrementally by appending each word. We check if we have an exact match after each addition and can exit early.

Two Pointers - Character by Character Matching

⏱️ Time: O(m) Space: O(1)

Use one pointer for the target string s and another to track position within current word in words array. This avoids creating intermediate concatenated strings and provides optimal space efficiency.

Brute Force - Try All Combinations — Algorithm Steps

Iterate k from 1 to words.length
For each k, concatenate words[0] to words[k-1]
Check if concatenation equals s

Visualization

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Step-by-Step Walkthrough

Try k=1

Concatenate first 1 word

Try k=2

Concatenate first 2 words

Check Match

Compare with target string

Code -

solution.c — C

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

int solution(char* s, char words[][101], int wordsSize) {
    for (int k = 1; k <= wordsSize; k++) {
        char concatenated[10001] = "";
        for (int i = 0; i < k; i++) {
            strcat(concatenated, words[i]);
        }
        if (strcmp(concatenated, s) == 0) {
            return 1;
        }
    }
    return 0;
}

int main() {
    char s[10001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char line[20001];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    char words[1000][101];
    int wordsSize = 0;
    
    char* token = strtok(line + 1, ",");
    while (token != NULL && wordsSize < 1000) {
        while (*token == ' ') token++;
        if (*token == '"') token++;
        char* end = token + strlen(token) - 1;
        while (*end == ' ' || *end == '"' || *end == ']') end--;
        *(end + 1) = '\0';
        strcpy(words[wordsSize], token);
        wordsSize++;
        token = strtok(NULL, ",");
    }
    
    int result = solution(s, words, wordsSize);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

n iterations, each concatenating up to m total characters

✓ Linear Growth

Space Complexity

O(m)

Temporary strings store up to m characters total

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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