给定一个字符串S由N个字母和两个正整数L型和U，代表转换任何字符为小写和分别大写的成本，该任务是修改字符串s，使得在字符串S字母表的每次出现是任一小写或大写，并且将原始字符串转换为修改后的字符串的成本必须最小。
例子：
Input: S = “aabbAA”, L = 1, U = 1
Output: AAbbAA
Explanation:
The cost of converting “aabbAA” to “AAbbAA” is 1*2 = 2, which is minimum among all the possible combinations of conversions.
Input: S = “aApbBp”, L = 1, U = 2
Output: aapbbp
方法：可以通过使用贪婪方法解决给定的问题，该方法是检查将字符的所有出现转换为小写的代价是否小于大写，然后将所有出现转换为小写的代价。否则，将所有出现的都转换为大写。请按照以下步骤解决问题：
- 将所有小写和大写字符的频率分别存储在数组lowerFreq [26]和upperFreq [26]中。
- 初始化另一个数组，将result [26]设置为0 ，其中result [i] = 1表示第i个字符必须小写，否则它必须大写。
- 使用变量i遍历[ 0，25 ]范围，并执行以下步骤：
- 存储成本以将每次出现的字符i分别转换为变量costLower和costUpper的小写和大写。
- 乘以的成本为1个字符转换为小写的，即，L以其大写的频率，即，upperFreq [I]找到costLower的值。同样，通过将U乘以lowerFreq [i]来找到costUpper的值。
- 如果costLower
- 使用变量i遍历字符串S并执行以下步骤：
- 将S [i]的索引存储在变量X中。
- 如果result [i]的值等于1 ，则将S [i]转换为其小写。否则，将S [i]转换为大写。
- 完成上述步骤后，将字符串S打印为修改后的字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum cost
// to convert all distinct characters
// to either uppercase or lowercase
void minimumCost(string s, int L, int U)
{
// Store the size of the string
int N = s.size();
string ans = "";
// Stores the frequency of lowercase
// & uppercase characters respectively
int lowerFreq[26] = { 0 };
int upperFreq[26] = { 0 };
// Traverse the string S
for (int i = 0; i < N; i++) {
// Update uppercase
// frequency of s[i]
if (isupper(s[i]))
upperFreq[s[i] - 'A']++;
// Otherwise, update lowercase
// frequency of s[i]
else
lowerFreq[s[i] - 'a']++;
}
// Stores if the i-th character
//should be lowercase or not
int result[26] = { 0 };
// Iterate over the range [0, 25]
for (int i = 0; i < 26; i++) {
// If the character is present
// in the string
if (lowerFreq[i] != 0
|| upperFreq[i] != 0) {
// Store the cost to convert
// every occurence of i to
// uppercase and lowercase
int costToUpper = U * lowerFreq[i];
int costToLower = L * upperFreq[i];
// Update result[i] to 1 if
// lowercase cost is less
if (costToLower < costToUpper) {
result[i] = 1;
}
}
}
// Traverse the string S
for (int i = 0; i < N; i++) {
// Store the index
// of the character
int index = 0;
if (islower(s[i]))
index = s[i] - 'a';
else
index = s[i] - 'A';
// Convert the current character
// to uppercase or lowercase
// according to the condition
if (result[index] == 1) {
// Update s[i]
s[i] = tolower(s[i]);
}
else {
// Update s[i]
s[i] = toupper(s[i]);
}
}
// Print the modified string
cout << s;
}
// Driver Code
int main()
{
string S = "aabbAA";
int L = 1, U = 1;
minimumCost(S, L, U);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find the minimum cost
// to convert all distinct characters
// to either uppercase or lowercase
static void minimumCost(String str, int L, int U)
{
// Store the size of the string
int N = str.length();
char s[] = str.toCharArray();
String ans = "";
// Stores the frequency of lowercase
// & uppercase characters respectively
int lowerFreq[] = new int[26];
int upperFreq[] = new int[26];
// Traverse the string S
for(int i = 0; i < N; i++)
{
// Update uppercase
// frequency of s[i]
if (Character.isUpperCase(s[i]))
upperFreq[s[i] - 'A']++;
// Otherwise, update lowercase
// frequency of s[i]
else
lowerFreq[s[i] - 'a']++;
}
// Stores if the i-th character
// should be lowercase or not
int result[] = new int[26];
// Iterate over the range [0, 25]
for(int i = 0; i < 26; i++)
{
// If the character is present
// in the string
if (lowerFreq[i] != 0 || upperFreq[i] != 0)
{
// Store the cost to convert
// every occurence of i to
// uppercase and lowercase
int costToUpper = U * lowerFreq[i];
int costToLower = L * upperFreq[i];
// Update result[i] to 1 if
// lowercase cost is less
if (costToLower < costToUpper)
{
result[i] = 1;
}
}
}
// Traverse the string S
for(int i = 0; i < N; i++)
{
// Store the index
// of the character
int index = 0;
if (Character.isLowerCase(s[i]))
index = s[i] - 'a';
else
index = s[i] - 'A';
// Convert the current character
// to uppercase or lowercase
// according to the condition
if (result[index] == 1)
{
// Update s[i]
s[i] = Character.toLowerCase(s[i]);
}
else
{
// Update s[i]
s[i] = Character.toUpperCase(s[i]);
}
}
// Print the modified string
System.out.println(new String(s));
}
// Driver Code
public static void main(String[] args)
{
String S = "aabbAA";
int L = 1, U = 1;
minimumCost(S, L, U);
}
}
// This code is contributed by Kingash输出：
AAbbAA时间复杂度： O(N)
辅助空间： O(26)