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3104. Find Longest Self-Contained Substring 🔒
Description
Given a string s, your task is to find the length of the longest self-contained substring of s.
A substring t of a string s is called self-contained if t != s and for every character in t, it doesn't exist in the rest of s.
Return the length of the longest self-contained substring of s if it exists, otherwise, return -1.
Example 1:
Input: s = "abba"
Output: 2
Explanation:
Let's check the substring "bb". You can see that no other "b" is outside of this substring. Hence the answer is 2.
Example 2:
Input: s = "abab"
Output: -1
Explanation:
Every substring we choose does not satisfy the described property (there is some character which is inside and outside of that substring). So the answer would be -1.
Example 3:
Input: s = "abacd"
Output: 4
Explanation:
Let's check the substring "abac". There is only one character outside of this substring and that is "d". There is no "d" inside the chosen substring, so it satisfies the condition and the answer is 4.
Constraints:
2 <= s.length <= 5 * 104sconsists only of lowercase English letters.
Solutions
Solution 1: Enumeration
We notice that the start of a substring that meets the conditions must be the position where a character appears for the first time.
Therefore, we can use two arrays or hash tables first and last to record the positions where each character appears for the first time and the last time, respectively.
Next, we enumerate each character c. Suppose the position where c first appears is $i$, and the position where it last appears is $mx$. Then we can start traversing from $i$. For each position $j$, we find the position $a$ where $s[j]$ first appears and the position $b$ where it last appears. If $a < i$, it means that $s[j]$ is on the left of $c$, which does not meet the enumeration conditions, and we can exit the loop directly. Otherwise, we update $mx = \max(mx, b)$. If $mx = j$ and $j - i + 1 < n$, we update the answer to $ans = \max(ans, j - i + 1)$.
Finally, return the answer.
The time complexity is $O(n \times |\Sigma|)$, and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $s$; and $|\Sigma|$ is the size of the character set. In this problem, the character set is lowercase letters, so $|\Sigma| = 26$.
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class Solution { public int maxSubstringLength(String s) { int[] first = new int[26]; int[] last = new int[26]; Arrays.fill(first, -1); int n = s.length(); for (int i = 0; i < n; ++i) { int j = s.charAt(i) - 'a'; if (first[j] == -1) { first[j] = i; } last[j] = i; } int ans = -1; for (int k = 0; k < 26; ++k) { int i = first[k]; if (i == -1) { continue; } int mx = last[k]; for (int j = i; j < n; ++j) { int a = first[s.charAt(j) - 'a']; int b = last[s.charAt(j) - 'a']; if (a < i) { break; } mx = Math.max(mx, b); if (mx == j && j - i + 1 < n) { ans = Math.max(ans, j - i + 1); } } } return ans; } } -
class Solution { public: int maxSubstringLength(string s) { vector<int> first(26, -1); vector<int> last(26); int n = s.length(); for (int i = 0; i < n; ++i) { int j = s[i] - 'a'; if (first[j] == -1) { first[j] = i; } last[j] = i; } int ans = -1; for (int k = 0; k < 26; ++k) { int i = first[k]; if (i == -1) { continue; } int mx = last[k]; for (int j = i; j < n; ++j) { int a = first[s[j] - 'a']; int b = last[s[j] - 'a']; if (a < i) { break; } mx = max(mx, b); if (mx == j && j - i + 1 < n) { ans = max(ans, j - i + 1); } } } return ans; } }; -
class Solution: def maxSubstringLength(self, s: str) -> int: first, last = {}, {} for i, c in enumerate(s): if c not in first: first[c] = i last[c] = i ans, n = -1, len(s) for c, i in first.items(): mx = last[c] for j in range(i, n): a, b = first[s[j]], last[s[j]] if a < i: break mx = max(mx, b) if mx == j and j - i + 1 < n: ans = max(ans, j - i + 1) return ans -
func maxSubstringLength(s string) int { first := [26]int{} last := [26]int{} for i := range first { first[i] = -1 } n := len(s) for i, c := range s { j := int(c - 'a') if first[j] == -1 { first[j] = i } last[j] = i } ans := -1 for k := 0; k < 26; k++ { i := first[k] if i == -1 { continue } mx := last[k] for j := i; j < n; j++ { a, b := first[s[j]-'a'], last[s[j]-'a'] if a < i { break } mx = max(mx, b) if mx == j && j-i+1 < n { ans = max(ans, j-i+1) } } } return ans } -
function maxSubstringLength(s: string): number { const first: number[] = Array(26).fill(-1); const last: number[] = Array(26).fill(0); const n = s.length; for (let i = 0; i < n; ++i) { const j = s.charCodeAt(i) - 97; if (first[j] === -1) { first[j] = i; } last[j] = i; } let ans = -1; for (let k = 0; k < 26; ++k) { const i = first[k]; if (i === -1) { continue; } let mx = last[k]; for (let j = i; j < n; ++j) { const a = first[s.charCodeAt(j) - 97]; if (a < i) { break; } const b = last[s.charCodeAt(j) - 97]; mx = Math.max(mx, b); if (mx === j && j - i + 1 < n) { ans = Math.max(ans, j - i + 1); } } } return ans; }