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In this method, we need to count the frequency of each character in strings s and t. We then calculate the number of changes required in t by comparing these frequencies. Specifically, for each character, if the frequency count in s is greater than in t, those are the extra characters needed in t. The total number of these extra characters across all characters gives the result.
Time Complexity: O(n), where n is the length of the strings (since they are of equal length).
Space Complexity: O(1), since the space required does not depend on the input size.
1using System;
2
3class Solution {
4 public int MinSteps(string s, string t) {
5 int[] count = new int[26];
6 int steps = 0;
7
8 foreach (char c in s) {
9 count[c - 'a']++;
10 }
11
12 foreach (char c in t) {
13 count[c - 'a']--;
14 }
15
16 foreach (int freq in count) {
17 if (freq > 0) {
18 steps += freq;
19 }
20 }
21
22 return steps;
23 }
24
25 static void Main() {
26 Solution sol = new Solution();
27 Console.WriteLine(sol.MinSteps("leetcode", "practice"));
28 }
29}The C# code follows the same structure as previous implementations, utilizing an array to track letter frequencies and adjusting based on t's characters.