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This approach uses a hash map to store the first and last occurrence of each character as you traverse the string. For each character, calculate the distance between its occurrences and keep track of the maximum distance found. This efficiently provides the length of the longest substring between two identical characters.
Time Complexity: O(n) where n is the length of the string.
Space Complexity: O(1) since the hash map size is constant (fixed at 256 for all possible lowercase characters).
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int MaxLengthBetweenEqualCharacters(string s) {
6 Dictionary<char, int> firstOccurrence = new Dictionary<char, int>();
7 int maxLen = -1;
8 for (int i = 0; i < s.Length; i++) {
9 if (!firstOccurrence.ContainsKey(s[i])) {
10 firstOccurrence[s[i]] = i;
11 } else {
12 maxLen = Math.Max(maxLen, i - firstOccurrence[s[i]] - 1);
13 }
14 }
15 return maxLen;
16 }
17
18 public static void Main(string[] args) {
19 Solution sol = new Solution();
20 Console.WriteLine(sol.MaxLengthBetweenEqualCharacters("abca"));
21 }
22}This C# solution uses a Dictionary to track first indexes of characters in the string, aiding to compute longest substring length between identical characters.
This approach involves two pass-throughs of the string. The first pass records each character's first occurrence, while the second calculates potential maximum lengths using each character's last occurrence.
Time Complexity: O(n).
Space Complexity: O(1).
1#include <unordered_map>
using namespace std;
int maxLengthBetweenEqualCharacters(string s) {
unordered_map<char, int> firstOccurrence;
unordered_map<char, int> lastOccurrence;
int maxLen = -1;
for (int i = 0; i < s.length(); i++) {
if (firstOccurrence.find(s[i]) == firstOccurrence.end()) {
firstOccurrence[s[i]] = i;
}
lastOccurrence[s[i]] = i;
}
for (auto pair : firstOccurrence) {
maxLen = max(maxLen, lastOccurrence[pair.first] - pair.second - 1);
}
return maxLen;
}
int main() {
string s = "abca";
int result = maxLengthBetweenEqualCharacters(s);
cout << result << endl;
return 0;
}C++ implementation makes use of two unordered maps for recording first and last occurrences, facilitating the calculation of the longest substring length between identical characters.