透過最小的ASCII值的增減來使字串中的所有字元相同

The ASCII (American Standard Code for Information Interchange) system is often used in programming to manipulate characters. In this article, we will be examining an interesting problem where we need to make all 的of a string same by the minimum number of increments or decrements of ASCII values of characters. We will provide a detailed explanation of the problem, propose an efficient #solution in C , and analyze its complexity.

#理解問題

Given a string consisting of lowercase English letters, our task is to make all characters in the string the same by changing their ASCII values. The catch is that we need to do this using the smallest

我們可以透過遞增或遞減字元的ASCII值來進行操作，每次遞增或遞減都算是一次操作。目標是找到使字串中所有字元相同所需的最小操作次數。

方法

To solve this problem, we need to find the character that appears most frequently in the string. The reason is that it would require fewer operations to change all other characters to this most common character.##common character.#.

首先，我們將統計字串中每個字元的頻率。然後，我們將找到頻率最高的字元。將所有字元與此字元相同所需的操作次數將是最頻繁字元的ASCII值與所有其他字元的ASCII值之間的差異的總和。

C Solution

Example

的中文翻譯為：

範例

以下是解決問題的C 程式碼 -

#include<bits/stdc++.h>
using namespace std;

int minOperations(string str) {
   int freq[26] = {0};
   
   for (char c : str) {
      freq[c - 'a']++;
   }
   
   int max_freq = *max_element(freq, freq+26);
   int total_chars = str.length();
   return total_chars - max_freq;
}

int main() {
   string str;
   cout << "Enter the string: ";
   cin >> str;
   cout << "Minimum operations: " << minOperations(str) << endl;
   return 0;
}

Output

#

Enter the string: Minimum operations: 0

Code Explanation

Consider the string "abcdd". The character 'd' appears twice, more than any other character. Therefore, we should change all other characters to 'd'. The ASCII value of 'd' is 100. The ASCII valueues of 'a', 'b', and 'c' are 97, 98, and 99, 'b', and 'c' are 97, 98, and 99, respectively. So, the minimum number of operations will be (100-97) (100-98) (100-99) = 3 2 1 = 6. However, since we need to minimize the number of operations, we will instead decrement the ASCII values of 'a', 'b', and 'c'. In this case, the minimum number of operations will be (97 -97) (98-97) (99-97) = 0 1 2 = 3.

Conclusion

在本文中，我們看到如何在C 中解決涉及ASCII值和字串操作的獨特問題。

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