Substring Finder - Problem

Given a main string and a substring, find all starting positions where the substring occurs in the main string.

Return an array of indices representing all positions where the substring begins. If the substring is not found, return an empty array.

Important: You must implement this without using built-in string search functions like indexOf(), find(), or contains().

Example: In string "abcabcabc", substring "abc" appears at positions [0, 3, 6].

Input & Output

Example 1 — Multiple Matches

$ Input: mainString = "abcabcabc", substring = "abc"

› Output: [0,3,6]

💡 Note: The pattern 'abc' appears at positions 0, 3, and 6 in the main string 'abcabcabc'

Example 2 — Single Match

$ Input: mainString = "hello world", substring = "world"

› Output: [6]

💡 Note: The pattern 'world' appears only once at position 6 in 'hello world'

Example 3 — No Match

$ Input: mainString = "programming", substring = "xyz"

› Output: []

💡 Note: The pattern 'xyz' does not appear anywhere in 'programming', so return empty array

Constraints

1 ≤ mainString.length ≤ 10⁴
1 ≤ substring.length ≤ 10²
mainString and substring consist of lowercase English letters

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32 f Facebook 28

The key insight is that after finding a match or mismatch, we can avoid redundant character comparisons using pattern preprocessing. The brute force approach checks every position (O(n×m)), while the KMP algorithm uses a failure function to achieve optimal O(n+m) time complexity.

Common Approaches

✓ Brute Force

⏱️ Time: O(n×m) Space: O(1)

For each position in the main string, check if the substring matches starting from that position. Compare character by character until either a mismatch is found or the entire substring is matched.

KMP Algorithm

⏱️ Time: O(n+m) Space: O(m)

Build a failure function (LPS array) that tells us how many characters we can skip when a mismatch occurs. This eliminates the need to restart matching from scratch after each failure.

Brute Force — Algorithm Steps

Step 1: Loop through each position i in mainString
Step 2: For each position, check if substring matches starting at i
Step 3: Compare characters one by one: mainString[i+j] vs substring[j]
Step 4: If all characters match, add i to result array

Visualization

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Step-by-Step Walkthrough

Start at position 0

Compare substring with mainString starting at index 0

Move to position 1

If no match, move to next position and repeat

Continue until end

Check all positions until mainString.length - substring.length

Code -

solution.c — C

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

int* solution(char* mainString, char* substring, int* returnSize) {
    int n = strlen(mainString);
    int m = strlen(substring);
    int* result = malloc(n * sizeof(int));
    *returnSize = 0;
    
    if (m == 0) {
        return result;
    }
    
    for (int i = 0; i <= n - m; i++) {
        int match = 1;
        for (int j = 0; j < m; j++) {
            if (mainString[i + j] != substring[j]) {
                match = 0;
                break;
            }
        }
        if (match) {
            result[(*returnSize)++] = i;
        }
    }
    
    return result;
}

int main() {
    char mainString[1001], substring[1001];
    fgets(mainString, sizeof(mainString), stdin);
    mainString[strcspn(mainString, "\n")] = 0;
    
    fgets(substring, sizeof(substring), stdin);
    substring[strcspn(substring, "\n")] = 0;
    
    int returnSize;
    int* result = solution(mainString, substring, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

For each of n positions, we might compare up to m characters

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, result array not counted

✓ Linear Space

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Substring Finder - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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