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' appears at index 16. Pattern 'squirrel' appears at index 4. Distance |16-4| = 12 <= 15, so index 16 is beautiful.

Example 2 — Multiple Matches

$ Input: s = "abababab", a = "ab", b = "ba", k = 1

› Output: [0,2,4]

💡 Note: Pattern 'ab' at indices [0,2,4,6]. Pattern 'ba' at indices [1,3,5]. Valid pairs: (0,1), (2,1), (2,3), (4,3), (4,5), (6,5). Beautiful indices: [0,2,4,6]. But 6 is not within k=1 of any 'ba', so result is [0,2,4].

Example 3 — No Valid Pairs

$ Input: s = "abcd", a = "a", b = "d", k = 2

› Output: []

💡 Note: Pattern 'a' at index 0, pattern 'd' at index 3. Distance |3-0| = 3 > 2, so no beautiful indices exist.

Constraints

1 ≤ s.length ≤ 10⁵
1 ≤ a.length, b.length ≤ 10
0 ≤ k ≤ s.length
s, a, and b consist only of lowercase English letters.

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to first collect all pattern positions, then use binary search for efficient distance queries. The optimal Binary Search approach achieves O(n² + m₁ log m₂) time complexity by pre-sorting pattern b positions and performing range queries. Time: O(n² + m₁ log m₂), Space: O(m₁ + m₂).

Common Approaches

✓ Binary Search Optimization

⏱️ Time: O(n² + m₁ log m₂) Space: O(m₁ + m₂)

Pre-compute all positions of patterns a and b, then for each position of pattern a, use binary search to quickly find if there's any position of pattern b within distance k.

Brute Force

⏱️ Time: O(n³) Space: O(1)

Find all positions where pattern a occurs, then for each such position, search the entire string for pattern b and check if any occurrence is within distance k.

Two Pass with Lists

⏱️ Time: O(n² + m₁ × m₂) Space: O(m₁ + m₂)

Use two passes: first to collect all positions where patterns a and b occur, then for each position of pattern a, check if any position of pattern b is within distance k.

Binary Search Optimization — Algorithm Steps

Step 1: Find all positions where pattern a occurs
Step 2: Find all positions where pattern b occurs and sort them
Step 3: For each position of a, binary search for valid b positions
Step 4: Check if any position in range [pos_a-k, pos_a+k] exists

Visualization

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Step-by-Step Walkthrough

Find Patterns

Collect and sort all pattern positions

Binary Search

For each 'a', search range [pos-k, pos+k]

Range Check

Verify if any 'b' exists in valid range

Code -

solution.c — C

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

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

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

int binarySearchRight(int* arr, int size, int target) {
    int left = 0, right = size;
    while (left < right) {
        int mid = (left + right) / 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 lenA = strlen(a);
    int lenB = strlen(b);
    
    int* positionsA = (int*)malloc(1000 * sizeof(int));
    int* positionsB = (int*)malloc(1000 * sizeof(int));
    int countA = 0, countB = 0;
    
    // Find all positions of pattern a
    for (int i = 0; i <= n - lenA; i++) {
        if (strncmp(s + i, a, lenA) == 0) {
            positionsA[countA++] = i;
        }
    }
    
    // Find all positions of pattern b
    for (int i = 0; i <= n - lenB; i++) {
        if (strncmp(s + i, b, lenB) == 0) {
            positionsB[countB++] = i;
        }
    }
    
    // Sort positions_b for binary search
    qsort(positionsB, countB, sizeof(int), compare);
    
    int* result = (int*)malloc(1000 * sizeof(int));
    *returnSize = 0;
    
    // For each position of a, use binary search to find valid b positions
    for (int i = 0; i < countA; i++) {
        int posA = positionsA[i];
        int left = posA - k;
        int right = posA + k;
        
        int leftIdx = binarySearchLeft(positionsB, countB, left);
        int rightIdx = binarySearchRight(positionsB, countB, right);
        
        if (leftIdx < rightIdx) {
            result[(*returnSize)++] = posA;
        }
    }
    
    free(positionsA);
    free(positionsB);
    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 + m₁ log m₂)

O(n²) for pattern matching, O(m₂ log m₂) for sorting b positions, O(m₁ log m₂) for binary searches

⚠ Quadratic Growth

Space Complexity

O(m₁ + m₂)

Storing sorted positions of both patterns

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search Optimization — Algorithm Steps

Visualization

Code -

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