Check If String Is a Prefix of Array - Practice Coding Problems

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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Understanding the Visualization

Input

String s and array of words

Process

Try concatenating first k words for each k

Output

Return true if any concatenation equals s

Key Takeaway

🎯 Key Insight: A string is a prefix of an array if it exactly matches the concatenation of the first k words for some k

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

Approach	Time	Space	Notes
✓ Two Pointers - Character by Character Matching	O(m)	O(1)	Use two pointers to match characters without building intermediate strings
Optimized - Build and Check Incrementally	O(n × m)	O(m)	Build the concatenated string incrementally and check for exact match
Brute Force - Try All Combinations	O(n × m)	O(m)	Generate all possible concatenations and check if any matches s

Two Pointers - Character by Character Matching — Algorithm Steps

Initialize pointer i for string s
For each word in words, match characters using inner pointer j
If all characters match and we've consumed entire s, return true
If s is consumed but more words remain, return false

Visualization

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Step-by-Step Walkthrough

Initialize

Set pointer i=0 for target string

Match Characters

Compare each character and advance pointer

Check Complete

Return true when entire string matched

Code -

solution.c — C

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

int solution(char* s, char words[][101], int wordsSize) {
    int i = 0; // pointer for string s
    int sLen = strlen(s);
    
    for (int w = 0; w < wordsSize; w++) {
        for (int j = 0; words[w][j] != '\0'; j++) {
            if (i >= sLen || s[i] != words[w][j]) {
                return 0;
            }
            i++;
        }
        
        // Check if we've matched entire string s
        if (i == sLen) {
            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(m)

Single pass through target string s of length m, each character checked once

✓ Linear Growth

Space Complexity

O(1)

Only uses a few integer pointers, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Pointers - Character by Character Matching — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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