Find Beautiful Indices in the Given Array I

Find Beautiful Indices in the Given Array I - 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" occurs at index 16. Pattern "squirrel" occurs at index 5. Distance |16-5| = 11 ≤ 15, so index 16 is beautiful.

Example 2 — Multiple Matches

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

› Output: [0]

💡 Note: Pattern "a" occurs at index 0. The same pattern also serves as pattern b at the same position. Distance |0-0| = 0 ≤ 4, so index 0 is beautiful.

Example 3 — No Match

$ Input: s = "aaaa", a = "aa", b = "bb", k = 1

› Output: []

💡 Note: Pattern "aa" occurs at indices 0, 1, 2, but pattern "bb" never occurs in the string, so no indices are beautiful.

Constraints

1 ≤ s.length ≤ 10⁵
1 ≤ a.length, b.length ≤ 10³
0 ≤ k ≤ s.length
s, a, and b consist only of lowercase English letters

Visualization

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Understanding the Visualization

Input String

Given string s with patterns a and b, and maximum distance k

Find Patterns

Locate all occurrences of pattern a and pattern b in the string

Check Distances

For each pattern a position, verify if any pattern b position is within distance k

Key Takeaway

🎯 Key Insight: Use binary search to efficiently find pattern matches within the distance constraint

Asked in

G Google 12 M Microsoft 8 a Amazon 6

The key insight is to find all pattern positions first, then efficiently match them using binary search. The optimal approach uses binary search to quickly locate pattern b positions within distance k of each pattern a position. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Two Pass with Arrays	O(n²)	O(n)	First find all pattern a positions, then all pattern b positions, then match them up
Brute Force	O(n³)	O(n)	Check every position for pattern a, then check entire string for pattern b within distance k
Binary Search Optimization	O(n log n)	O(n)	Find pattern positions first, then use binary search to quickly locate nearby matches

Two Pass with Arrays — Algorithm Steps

First pass: find all positions where pattern a occurs
Second pass: find all positions where pattern b occurs
For each a position, check if any b position is within distance k

Visualization

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Step-by-Step Walkthrough

Collect Pattern A

First pass finds all positions of pattern a

Collect Pattern B

Second pass finds all positions of pattern b

Match Within Distance

For each a position, find b positions within distance k

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 n = strlen(s);
    int aLen = strlen(a);
    int bLen = strlen(b);
    int* aPositions = (int*)malloc(1000 * sizeof(int));
    int* bPositions = (int*)malloc(1000 * sizeof(int));
    int aCount = 0, bCount = 0;
    
    // First pass: find all positions of pattern a
    for (int i = 0; i <= n - aLen; i++) {
        if (strncmp(s + i, a, aLen) == 0) {
            aPositions[aCount++] = i;
        }
    }
    
    // Second pass: find all positions of pattern b
    for (int j = 0; j <= n - bLen; j++) {
        if (strncmp(s + j, b, bLen) == 0) {
            bPositions[bCount++] = j;
        }
    }
    
    // Find beautiful indices
    int* result = (int*)malloc(1000 * sizeof(int));
    int count = 0;
    
    for (int x = 0; x < aCount; x++) {
        for (int y = 0; y < bCount; y++) {
            if (abs(bPositions[y] - aPositions[x]) <= k) {
                result[count++] = aPositions[x];
                break;
            }
        }
    }
    
    free(aPositions);
    free(bPositions);
    *returnSize = count;
    return result;
}

int main() {
    char s[1001], a[1001], b[1001];
    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(n²)

Two O(n) passes to find patterns, then O(|a_positions| × |b_positions|) to match them up

⚠ Quadratic Growth

Space Complexity

O(n)

Store positions of pattern a and b, plus result array

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Pass with Arrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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