通过最小的ASCII值的增减来使字符串中的所有字符相同

王林转载: 2023-08-26 12:53:12 1017浏览

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 characters 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.

理解问题

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 number of changes.

我们可以通过递增或递减字符的ASCII值来进行操作，每次递增或递减都算作一次操作。目标是找到使字符串中所有字符相同所需的最小操作次数。

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.

首先，我们将统计字符串中每个字符的频率。然后，我们将找到频率最高的字符。将所有字符与此字符相同所需的操作次数将是最频繁字符的ASCII值与所有其他字符的ASCII值之间的差值的总和。

C++ Solution

Example

的中文翻译为：

示例

以下是解决问题的C++代码 -

#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 values of 'a', '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.