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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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

✓ Binary Search Optimization

⏱️ Time: O(n log n) Space: O(n)

First find all positions of both patterns, then for each pattern a position, use binary search on the sorted pattern b positions to quickly find if any are within distance k. This avoids checking all b positions for each a position.

Brute Force

⏱️ Time: O(n³) Space: O(n)

For each position that matches pattern a, scan the entire string to find all positions that match pattern b and check if any are within distance k. This is straightforward but inefficient.

Two Pass with Arrays

⏱️ Time: O(n²) Space: O(n)

Make two passes through the string: first collect all positions where pattern a occurs, then collect all positions where pattern b occurs. Finally, for each pattern a position, check if any pattern b position is within distance k.

Binary Search Optimization — Algorithm Steps

Find and store all positions of pattern a and pattern b
Sort the positions of pattern b
For each a position, use binary search to find b positions within range [i-k, i+k]

Visualization

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Step-by-Step Walkthrough

Find All Patterns

Collect all positions of pattern a and pattern b

Binary Search Range

For each a position, binary search for b positions in range [i-k, i+k]

Quick Match

Binary search finds matches in O(log n) time

Code -

solution.c — C

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

int binarySearch(int* arr, int n, int target) {
    int left = 0, right = n;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
}

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;
    
    // Find all positions of pattern a
    for (int i = 0; i <= n - aLen; i++) {
        if (strncmp(s + i, a, aLen) == 0) {
            aPositions[aCount++] = i;
        }
    }
    
    // Find all positions of pattern b
    for (int j = 0; j <= n - bLen; j++) {
        if (strncmp(s + j, b, bLen) == 0) {
            bPositions[bCount++] = j;
        }
    }
    
    int* result = (int*)malloc(1000 * sizeof(int));
    int count = 0;
    
    for (int x = 0; x < aCount; x++) {
        int leftBound = aPositions[x] - k;
        int rightBound = aPositions[x] + k;
        
        int idx = binarySearch(bPositions, bCount, leftBound);
        
        if (idx < bCount && bPositions[idx] <= rightBound) {
            result[count++] = aPositions[x];
        }
    }
    
    free(aPositions);
    free(bPositions);
    *returnSize = count;
    return result;
}

int main() {
    char s[100001], 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(n log n)

O(n) to find patterns, O(m log m) to sort b positions, O(|a| log |b|) for binary searches

⚡ Linearithmic

Space Complexity

O(n)

Store positions of pattern a and b, plus result array

⚡ Linearithmic Space

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Common Approaches

Binary Search Optimization — Algorithm Steps

Visualization

Code -

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