Find Beautiful Indices in the Given Array II

Find Beautiful Indices in the Given Array II - Problem

You are given a 0-indexed string s, a string a, a string b, and an integer k.

An index i is beautiful if:

0 <= i <= s.length - a.length
s[i..(i + a.length - 1)] == a
There exists an index j such that:
- 0 <= j <= s.length - b.length
- s[j..(j + b.length - 1)] == b
- |j - i| <= k

Return the array that contains beautiful indices in sorted order from smallest to largest.

Input & Output

Example 1 — Basic Case

$ Input: s = "isawsquirrelnearmylaptop", a = "my", b = "squirrel", k = 15

› Output: [16]

💡 Note: Pattern "my" found at index 16. Pattern "squirrel" found at index 8. Distance |8-16| = 8 ≤ 15, so index 16 is beautiful.

Example 2 — Multiple Matches

$ Input: s = "abcd", a = "a", b = "a", k = 4

› Output: [0]

💡 Note: Pattern "a" found at index 0. Same pattern "a" also found at index 0. Distance |0-0| = 0 ≤ 4, so index 0 is beautiful.

Example 3 — No Beautiful Indices

$ Input: s = "abcd", a = "a", b = "d", k = 1

› Output: []

💡 Note: Pattern "a" found at index 0. Pattern "d" found at index 3. Distance |3-0| = 3 > 1, so no beautiful indices exist.

Constraints

1 ≤ s.length ≤ 5 × 10⁵
1 ≤ a.length, b.length ≤ s.length
0 ≤ k ≤ s.length
s, a, and b consist only of lowercase English letters

Visualization

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Understanding the Visualization

Input

String s with patterns a, b and distance constraint k

Process

Find pattern matches and check distance constraints

Output

Indices where pattern a occurs and pattern b is within distance k

Key Takeaway

🎯 Key Insight: Separate pattern finding from distance checking for efficiency

Asked in

G Google 25 f Meta 18 a Amazon 15

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n³)	O(1)	Check every position for pattern a, then check all positions for pattern b within distance k
Two Pass with Lists	O(n × (m + p) + \|a_pos\| × \|b_pos\|)	O(\|a_pos\| + \|b_pos\|)	Find all occurrences of both patterns first, then check distance constraints efficiently
Binary Search Optimization	O(n × (m + p) + \|a_pos\| × log\|b_pos\|)	O(\|a_pos\| + \|b_pos\|)	Use binary search to efficiently find closest pattern b positions for each pattern a position

Brute Force — Algorithm Steps

Step 1: For each position i, check if pattern a matches
Step 2: If match found, check all positions j for pattern b
Step 3: If pattern b found and |j-i| <= k, mark i as beautiful

Visualization

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Step-by-Step Walkthrough

Find Pattern A

Check each position for pattern a matches

Search Pattern B

For each match of a, search all positions for pattern b

Check Distance

Verify |j-i| <= k constraint

Code -

solution.c — C

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

int* solution(char* s, char* a, char* b, int k, int* returnSize) {
    int* beautiful = malloc(1000 * sizeof(int));
    *returnSize = 0;
    int n = strlen(s);
    int lenA = strlen(a);
    int lenB = strlen(b);
    
    for (int i = 0; i <= n - lenA; i++) {
        // Check if pattern a matches at position i
        if (strncmp(s + i, a, lenA) == 0) {
            // Check if there exists a j where pattern b matches and |j-i| <= k
            int found = 0;
            for (int j = 0; j <= n - lenB; j++) {
                if (strncmp(s + j, b, lenB) == 0 && abs(j - i) <= k) {
                    found = 1;
                    break;
                }
            }
            if (found) {
                beautiful[(*returnSize)++] = i;
            }
        }
    }
    
    return beautiful;
}

int main() {
    char s[100001], a[100001], b[100001];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(a, sizeof(a), stdin);
    a[strcspn(a, "\n")] = 0;
    fgets(b, sizeof(b), stdin);
    b[strcspn(b, "\n")] = 0;
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(s, a, b, k, &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(n³)

For each position (n), we check pattern a (length m), then for each match we check all positions for pattern b (n×m)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for indices and temporary storage

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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