通过用字母替换字符来修改字符串，字母与字符的距离等于其频率
原文:https://www . geeksforgeeks . org/通过字母替换字符来修改字符串，字母与字符的距离等于其频率/
给定一个由 N 个小写字母组成的字符串 S ，任务是通过用字母替换每个字符来修改字符串 S ，该字母与字符的圆形距离等于字符在 S 中的出现频率。
示例:
输入: S =【极客】 T3】输出:hgglt T6】说明:T8】对字符串 S 做了如下修改:
- 字符串中“g”的频率是 1。因此，“g”被“h”代替。
- 字符串中“e”的频率是 2。因此，“e”被“g”所取代。
- 字符串中“e”的频率是 2。因此，“e”被“g”所取代。
- 字符串中“k”的频率是 1。因此，“k”被转换为“k”+1 =“l”。
- 字符串中“s”的频率为 1。因此，“s”被转换为“s”+1 =“t”。
因此，修改后的字符串 S 为“hgglt”。
输入: S =【爵士】 输出:【kbbb】
方法:给定的问题可以通过维护频率数组来解决，该数组存储字符串中每个字符的出现次数。按照以下步骤解决问题:
- 初始化一个数组，freq【26】，最初所有元素为 0 ，存储字符串每个字符的频率。
- 遍历给定的字符串 S ，将数组freq【】中每个字符S【I】的频率增加 1 。
- 遍历字符串 S 使用变量 i 并执行以下步骤:
- 将要添加到 S[i] 的值存储在变量中，将添加为 (freq[i] % 26) 。
- 如果将的值加到 S[i] 后， S[i] 没有超过字符 z ，则更新 S[i] 为 S[i] +加。
- 否则，将的值更新为(S[I]+add–z)，然后将 S[i] 设置为(a+add–1)。
- 完成上述步骤后，打印修改后的字符串 S 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
void addFrequencyToCharacter(string s)
{
// Stores frequency of characters
int frequency[26] = { 0 };
// Stores length of the string
int n = s.size();
// Traverse the given string S
for (int i = 0; i < n; i++) {
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for (int i = 0; i < n; i++) {
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if (int(s[i]) + add <= int('z'))
s[i] = char(int(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else {
add = (int(s[i]) + add) - (int('z'));
s[i] = char(int('a') + add - 1);
}
}
// Print the modified string
cout << s;
}
// Driver Code
int main()
{
string str = "geeks";
addFrequencyToCharacter(str);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG{
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
static void addFrequencyToCharacter(char[] s)
{
// Stores frequency of characters
int frequency[] = new int[26];
// Stores length of the string
int n = s.length;
// Traverse the given string S
for(int i = 0; i < n; i++)
{
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for(int i = 0; i < n; i++)
{
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if ((int)(s[i]) + add <= (int)('z'))
s[i] = (char)((int)(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else
{
add = ((int)(s[i]) + add) - ((int)('z'));
s[i] = (char)((int)('a') + add - 1);
}
}
// Print the modified string
System.out.println(s);
}
// Driver Code
public static void main(String[] args)
{
String str = "geeks";
addFrequencyToCharacter(str.toCharArray());
}
}
// This code is contributed by AnkThon
Python 3
# Python3 program for the above approach
# Function to modify string by replacing
# characters by the alphabet present at
# distance equal to frequency of the string
def addFrequencyToCharacter(s):
# Stores frequency of characters
frequency = [0] * 26
# Stores length of the string
n = len(s)
# Traverse the given string S
for i in range(n):
# Increment frequency of
# current character by 1
frequency[ord(s[i]) - ord('a')] += 1
# Traverse the string
for i in range(n):
# Store the value to be added
# to the current character
add = frequency[ord(s[i]) - ord('a')] % 26
# Check if after adding the
# frequency, the character is
# less than 'z' or not
if (ord(s[i]) + add <= ord('z')):
s[i] = chr(ord(s[i]) + add)
# Otherwise, update the value of
# add so that s[i] doesn't exceed 'z'
else:
add = (ord(s[i]) + add) - (ord('z'))
s[i] = chr(ord('a') + add - 1)
# Print the modified string
print("".join(s))
# Driver Code
if __name__ == '__main__':
str = "geeks"
addFrequencyToCharacter([i for i in str])
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
static void addFrequencyToCharacter(char[] s)
{
// Stores frequency of characters
int[] frequency = new int[26];
// Stores length of the string
int n = s.Length;
// Traverse the given string S
for(int i = 0; i < n; i++)
{
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for(int i = 0; i < n; i++)
{
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if ((int)(s[i]) + add <= (int)('z'))
s[i] = (char)((int)(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else
{
add = ((int)(s[i]) + add) - ((int)('z'));
s[i] = (char)((int)('a') + add - 1);
}
}
// Print the modified string
Console.WriteLine(s);
}
// Driver Code
public static void Main(string[] args)
{
string str = "geeks";
addFrequencyToCharacter(str.ToCharArray());
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// JavaScript program for the above approach
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
function addFrequencyToCharacter(s) {
// Stores frequency of characters
var frequency = new Array(26).fill(0);
// Stores length of the string
var n = s.length;
// Traverse the given string S
for (var i = 0; i < n; i++) {
// Increment frequency of
// current character by 1
frequency[s[i].charCodeAt(0) - "a".charCodeAt(0)] += 1;
}
// Traverse the string
for (var i = 0; i < n; i++) {
// Store the value to be added
// to the current character
var add = frequency[s[i].charCodeAt(0) - "a".charCodeAt(0)] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if (s[i].charCodeAt(0) + add <= "z".charCodeAt(0))
s[i] = String.fromCharCode(s[i].charCodeAt(0) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else {
add = s[i].charCodeAt(0) + add - "z".charCodeAt(0);
s[i] = String.fromCharCode("a".charCodeAt(0) + add - 1);
}
}
// Print the modified string
document.write(s.join("") + "<br>");
}
// Driver Code
var str = "geeks";
addFrequencyToCharacter(str.split(""));
</script>
Output:
hgglt
时间复杂度:O(N) T5辅助空间:** O(1)
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