3006. Find Beautiful Indices in the Given Array I

Description

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.

Example 1:

Input: s = "isawsquirrelnearmysquirrelhouseohmy", a = "my", b = "squirrel", k = 15
Output: [16,33]
Explanation: There are 2 beautiful indices: [16,33].
- The index 16 is beautiful as s[16..17] == "my" and there exists an index 4 with s[4..11] == "squirrel" and |16 - 4| <= 15.
- The index 33 is beautiful as s[33..34] == "my" and there exists an index 18 with s[18..25] == "squirrel" and |33 - 18| <= 15.
Thus we return [16,33] as the result.

Example 2:

Input: s = "abcd", a = "a", b = "a", k = 4
Output: [0]
Explanation: There is 1 beautiful index: [0].
- The index 0 is beautiful as s[0..0] == "a" and there exists an index 0 with s[0..0] == "a" and |0 - 0| <= 4.
Thus we return [0] as the result.

Constraints:

1 <= k <= s.length <= 10⁵
1 <= a.length, b.length <= 10
s, a, and b contain only lowercase English letters.

Solutions

Solution 1

Python Code

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class Solution:
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        def build_prefix_function(pattern):
            prefix_function = [0] * len(pattern)
            j = 0
            for i in range(1, len(pattern)):
                while j > 0 and pattern[i] != pattern[j]:
                    j = prefix_function[j - 1]
                if pattern[i] == pattern[j]:
                    j += 1
                prefix_function[i] = j
            return prefix_function

        def kmp_search(pattern, text, prefix_function):
            occurrences = []
            j = 0
            for i in range(len(text)):
                while j > 0 and text[i] != pattern[j]:
                    j = prefix_function[j - 1]
                if text[i] == pattern[j]:
                    j += 1
                if j == len(pattern):
                    occurrences.append(i - j + 1)
                    j = prefix_function[j - 1]
            return occurrences

        prefix_a = build_prefix_function(a)
        prefix_b = build_prefix_function(b)

        resa = kmp_search(a, s, prefix_a)
        resb = kmp_search(b, s, prefix_b)

        res = []
        print(resa, resb)
        i = 0
        j = 0
        while i < len(resa):
            while j < len(resb):
                if abs(resb[j] - resa[i]) <= k:
                    res.append(resa[i])
                    break
                elif j + 1 < len(resb) and abs(resb[j + 1] - resa[i]) < abs(
                    resb[j] - resa[i]
                ):
                    j += 1
                else:
                    break
            i += 1
        return res

Java Code

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public class Solution {
    public void computeLPS(String pattern, int[] lps) {
        int M = pattern.length();
        int len = 0;

        lps[0] = 0;

        int i = 1;
        while (i < M) {
            if (pattern.charAt(i) == pattern.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    lps[i] = 0;
                    i++;
                }
            }
        }
    }

    public List<Integer> KMP_codestorywithMIK(String pat, String txt) {
        int N = txt.length();
        int M = pat.length();
        List<Integer> result = new ArrayList<>();

        int[] lps = new int[M];
        computeLPS(pat, lps);

        int i = 0; // Index for text
        int j = 0; // Index for pattern

        while (i < N) {
            if (pat.charAt(j) == txt.charAt(i)) {
                i++;
                j++;
            }

            if (j == M) {
                result.add(i - j); // Pattern found at index i-j+1 (If you have to return 1 Based
                                   // indexing, that's why added + 1)
                j = lps[j - 1];
            } else if (i < N && pat.charAt(j) != txt.charAt(i)) {
                if (j != 0) {
                    j = lps[j - 1];
                } else {
                    i++;
                }
            }
        }

        return result;
    }

    private int lowerBound(List<Integer> list, int target) {
        int left = 0, right = list.size() - 1, result = list.size();

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (list.get(mid) >= target) {
                result = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return result;
    }

    public List<Integer> beautifulIndices(String s, String a, String b, int k) {
        int n = s.length();

        List<Integer> i_indices = KMP_codestorywithMIK(a, s);
        List<Integer> j_indices = KMP_codestorywithMIK(b, s);

        List<Integer> result = new ArrayList<>();

        for (int i : i_indices) {

            int left_limit = Math.max(0, i - k); // To avoid out of bound -> I used max(0, i-k)
            int right_limit
                = Math.min(n - 1, i + k); // To avoid out of bound -> I used min(n-1, i+k)

            int lowerBoundIndex = lowerBound(j_indices, left_limit);

            if (lowerBoundIndex < j_indices.size()
                && j_indices.get(lowerBoundIndex) <= right_limit) {
                result.add(i);
            }
        }

        return result;
    }
}

C++ Code

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class Solution {
public:
    vector<int> beautifulIndices(string s, string patternA, string patternB, int k) {
        vector<int> beautifulIndicesA = kmpSearch(s, patternA);
        vector<int> beautifulIndicesB = kmpSearch(s, patternB);

        sort(beautifulIndicesB.begin(), beautifulIndicesB.end());

        vector<int> result;
        for (int indexA : beautifulIndicesA) {
            int left = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA - k) - beautifulIndicesB.begin();
            int right = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA + k + patternB.length()) - beautifulIndicesB.begin();

            left = (left >= 0) ? left : -(left + 1);
            right = (right >= 0) ? right : -(right + 1);

            for (int indexB = left; indexB < right; indexB++) {
                if (abs(beautifulIndicesB[indexB] - indexA) <= k) {
                    result.push_back(indexA);
                    break;
                }
            }
        }

        return result;
    }

private:
    vector<int> kmpSearch(string text, string pattern) {
        vector<int> indices;
        vector<int> pi = computePrefixFunction(pattern);

        int q = 0;
        for (int i = 0; i < text.length(); i++) {
            while (q > 0 && pattern[q] != text[i]) {
                q = pi[q - 1];
            }
            if (pattern[q] == text[i]) {
                q++;
            }
            if (q == pattern.length()) {
                indices.push_back(i - q + 1);
                q = pi[q - 1];
            }
        }

        return indices;
    }

    vector<int> computePrefixFunction(string pattern) {
        int m = pattern.length();
        vector<int> pi(m, 0);
        int k = 0;
        for (int q = 1; q < m; q++) {
            while (k > 0 && pattern[k] != pattern[q]) {
                k = pi[k - 1];
            }
            if (pattern[k] == pattern[q]) {
                k++;
            }
            pi[q] = k;
        }
        return pi;
    }
};

Go Code

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func beautifulIndices(s string, a string, b string, k int) []int {

	s_len := len(s)
	a_len := len(a)
	b_len := len(b)

	final := make([]int, 0)
	lps_a := make([]int, a_len)
	lps_b := make([]int, b_len)
	a_index := make([]int, 0)
	b_index := make([]int, 0)

	var pat func(lps []int, s_l int, pattern string)

	pat = func(lps []int, s_l int, pattern string) {

		l := 0
		lps[0] = 0
		i := 1

		for i < s_l {
			if pattern[i] == pattern[l] {
				l++
				lps[i] = l
				i++
			} else {
				if l != 0 {
					l = lps[l-1]
				} else {
					lps[i] = l
					i++
				}
			}
		}
	}

	pat(lps_a, a_len, a)
	pat(lps_b, b_len, b)

	var kmp func(pat string, pat_l int, lps []int, index *[]int)

	kmp = func(pat string, pat_l int, lps []int, index *[]int) {
		i := 0
		j := 0
		for s_len-i >= pat_l-j {
			if s[i] == pat[j] {
				i++
				j++
			}
			if j == pat_l {
				*index = append(*index, i-pat_l)
				j = lps[j-1]
			} else if s[i] != pat[j] {
				if j != 0 {
					j = lps[j-1]
				} else {
					i++
				}
			}
		}
	}

	kmp(a, a_len, lps_a, &a_index)
	kmp(b, b_len, lps_b, &b_index)

	// fmt.Println(a_index, b_index)

	i := 0
	j := 0

	for i < len(a_index) && j < len(b_index) {
		if a_index[i]+k >= b_index[j] && a_index[i]-k <= b_index[j] {
			final = append(final, a_index[i])
			i++
		} else if a_index[i]-k > b_index[j] {
			j++
		} else {
			i++
		}
	}

	return final
}

3006. Find Beautiful Indices in the Given Array I#

Description#

Solutions#

Solution 1#

3006. Find Beautiful Indices in the Given Array I

Description

Solutions

Solution 1