01. Non-Repeating Character

The problem can be found at the following link: Problem Link

✨ LeetCode Problem of the Day (POTD) Continued ✨

• As part of the LeetCode Problem of the Day (POTD) series, here is the solution for December 1, 2024. 🎯

• My latest solution is now live: 1346. Check If N and Its Double Exist

Problem DescriptionGiven a string s consisting of lowercase Latin letters, return the first non-repeating character in s. If there is no non-repeating character, return '$'.Note: When you return '$', the driver code will output -1.Examples:Input: s = "geeksforgeeks" Output: 'f'Explanation: In the given string, 'f' is the first character in the string which does not repeat.Input: s = "racecar" Output: 'e'Explanation: In the given string, 'e' is the only character in the string which does not repeat.Input: s = "aabbccc" Output: -1Explanation: All the characters in the given string are repeating.Constraints:$1 <= s.size() <= 10^5$My ApproachFrequency Array Method:Since the input consists only of lowercase Latin letters, a frequency array of size 26 is sufficient to count occurrences of each character.Traverse the string once to update the frequency of each character in the array.Traverse the string a second time to identify the first character with a frequency of 1.Steps:Initialize a frequency array freq[26] and set all elements to 0.For each character in the string, increment its corresponding frequency in the array.Iterate through the string again, checking for the first character with a frequency of 1.If no such character exists, return '$'.Time and Auxiliary Space ComplexityExpected Time Complexity: O(n), where n is the size of the string. The algorithm requires two linear passes through the string.Expected Auxiliary Space Complexity: O(1), as the frequency array uses a fixed amount of additional space (26 elements).Code (C)char nonRepeatingChar(char s[]) { int freq[26] = {0}; for (int i = 0; s[i] != '\0'; i++) { freq[s[i] - 'a']++; } for (int i = 0; s[i] != '\0'; i++) { if (freq[s[i] - 'a'] == 1) { return s[i]; } } return '$';}Code (Cpp)class Solution {public: char nonRepeatingChar(string &s) { int freq[26] = {0}; for (char c : s) { freq[c - 'a']++; } for (char c : s) { if (freq[c - 'a'] == 1) { return c; } } return '$'; }};Code (Java)class Solution { static char nonRepeatingChar(String s) { int[] freq = new int[26]; for (char c : s.toCharArray()) { freq[c - 'a']++; } for (char c : s.toCharArray()) { if (freq[c - 'a'] == 1) { return c; } } return '$'; }}Code (Python)class Solution: def nonRepeatingChar(self, s): freq = [0] * 26 for c in s: freq[ord(c) - ord('a')] += 1 for c in s: if freq[ord(c) - ord('a')] == 1: return c return '$'Contribution and Support

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