Substring Matching Pattern

Substring Matching Pattern - Problem

You are given a string s and a pattern string p, where p contains exactly one '*' character. The '*' in p can be replaced with any sequence of zero or more characters.

Return true if p can be made a substring of s, and false otherwise.

Note: A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Pattern Match

$ Input: s = "hello", p = "l*e"

› Output: true

💡 Note: The pattern 'l*e' can match substring 'lle' by replacing '*' with 'l'. The substring 'lle' exists in 'hello' at positions 2-4.

Example 2 — No Match Possible

$ Input: s = "hello", p = "a*b"

› Output: false

💡 Note: There's no way to form a substring of 'hello' using pattern 'a*b' because 'hello' doesn't contain 'a' or 'b'.

Example 3 — Star as Empty String

$ Input: s = "abc", p = "a*c"

› Output: true

💡 Note: The pattern 'a*c' matches substring 'ac' by replacing '*' with empty string. 'ac' is a substring of 'abc'.

Constraints

1 ≤ s.length ≤ 1000
1 ≤ p.length ≤ 1000
p contains exactly one '*' character
s and p consist of lowercase English letters and '*'

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to split the pattern at the '*' character into prefix and suffix parts, then find positions where the prefix occurs and check if the suffix can follow at a valid distance. The optimal approach uses two-pass string matching with O(n×m) time complexity and O(k) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Two-Pass String Matching	O(n × m)	O(k)	Split pattern at '*', then find all prefix matches and check if any can extend to suffix matches
Brute Force Pattern Matching	O(n³)	O(m)	Try every possible replacement for '*' and check if any forms a substring of s

Two-Pass String Matching — Algorithm Steps

Split pattern p at '*' into prefix and suffix
Find all occurrences of prefix in string s
For each prefix occurrence, check if suffix can be matched after it
Return true if any valid prefix-suffix combination is found

Visualization

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Step-by-Step Walkthrough

Split Pattern

Split 'l*e' into prefix 'l' and suffix 'e'

Find Prefix

Find all 'l' occurrences in 'hello'

Check Suffix

For each 'l', check if 'e' can follow after some gap

Code -

solution.c — C

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

bool solution(char* s, char* p) {
    int sLen = strlen(s);
    int pLen = strlen(p);
    int starPos = -1;
    
    // Find '*' position
    for (int i = 0; i < pLen; i++) {
        if (p[i] == '*') {
            starPos = i;
            break;
        }
    }
    
    int prefixLen = starPos;
    int suffixLen = pLen - starPos - 1;
    
    // Special cases
    if (prefixLen == 0 && suffixLen == 0) return true;
    if (prefixLen == 0) return strstr(s, p + starPos + 1) != NULL;
    if (suffixLen == 0) return strstr(s, p) != NULL; // p without '*'
    
    // Find all prefix matches
    int prefixPositions[1000];
    int prefixCount = 0;
    
    for (int i = 0; i <= sLen - prefixLen; i++) {
        bool match = true;
        for (int j = 0; j < prefixLen; j++) {
            if (s[i + j] != p[j]) {
                match = false;
                break;
            }
        }
        if (match) {
            prefixPositions[prefixCount++] = i;
        }
    }
    
    // Check if any prefix match can be extended to suffix match
    for (int i = 0; i < prefixCount; i++) {
        int pos = prefixPositions[i];
        int startSearch = pos + prefixLen;
        
        // Look for suffix starting from after the prefix
        for (int j = startSearch; j <= sLen - suffixLen; j++) {
            bool match = true;
            for (int k = 0; k < suffixLen; k++) {
                if (s[j + k] != p[starPos + 1 + k]) {
                    match = false;
                    break;
                }
            }
            if (match) return true;
        }
    }
    
    return false;
}

int main() {
    char s[1001], p[1001];
    fgets(s, sizeof(s), stdin);
    fgets(p, sizeof(p), stdin);
    s[strcspn(s, "\n")] = 0;
    p[strcspn(p, "\n")] = 0;
    
    bool result = solution(s, p);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

Finding prefix takes O(n×m) time, then for each match we check suffix in O(n×m) time

✓ Linear Growth

Space Complexity

O(k)

Storage for up to k prefix match positions, where k ≤ n

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Two-Pass String Matching — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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