与给定字符串的汉明距离恰好为 K 的字典最小字符串
给定一个长度为N的小写字符串A和一个整数K ，找到与 A 长度相同的字典上最小的字符串B ，使得 A 和 B 之间的汉明距离恰好为 K。
例子：
Input : A = "pqrs", k = 1.
Output : aqrs
We can differ by at most one
character. So we put 'a' in the
beginning to make the result
lexicographically smallest.
Input : A = "pqrs", k = 2.
Output : aars我们从左到右开始，如果字符A的当前位置的字符串是'a'，那么我们分配字符串B的当前位置的字符'a'。这个位置不会影响汉明距离。如果A中这个位置的字符不等于'a'，那么我们也将分配字符串B字符'a'的当前位置，现在这将有助于汉明距离，这最多可以完成k次，因为汉明距离有等于K，如果已经这样做了K次，我们将B的这个位置分配与A相同的字符。
如果经过上一步，A和B之间的汉明距离为K，我们就完成了，否则我们必须对B进行更多的更改。现在我们将在B中从右到左开始，如果当前位置的字符等于A的对应字符，将B的字符更改为'b'，因此将汉明距离增加1，我们将这样做直到汉明距离等于K。
下面是这种方法的实现：
C++
// CPP program to find Lexicographically
// smallest string whose hamming distance
// from given string is exactly K
#include
using namespace std;
// function to find Lexicographically
// smallest string with hamming distance k
void findString(string str, int n, int k)
{
// If k is 0, output input string
if (k == 0) {
cout << str << endl;
return;
}
// Copying input string into output
// string
string str2 = str;
int p = 0;
// Traverse all the character of the
// string
for (int i = 0; i < n; i++) {
// If current character is not 'a'
if (str2[i] != 'a') {
// copy character 'a' to
// output string
str2[i] = 'a';
p++;
// If hamming distance became k,
// break;
if (p == k)
break;
}
}
// If k is less than p
if (p < k) {
// Traversing string in reverse
// order
for (int i = n - 1; i >= 0; i--)
if (str[i] == 'a') {
str2[i] = 'b';
p++;
if (p == k)
break;
}
}
cout << str2 << endl;
}
// Driven Program
int main()
{
string str = "pqrs";
int n = str.length();
int k = 2;
findString(str, n, k);
return 0;
} Java
// Java program to find Lexicographically
// smallest string whose hamming distance
// from given string is exactly K
class GFG {
// function to find Lexicographically
// smallest string with hamming distance k
static void findString(String str, int n, int k)
{
// If k is 0, output input string
if (k == 0) {
System.out.println(str);;
return;
}
// Copying input string into output
// string
String str2 = str;
int p = 0;
// Traverse all the character of the
// string
for (int i = 0; i < n; i++) {
// If current character is not 'a'
if (str2.charAt(i) != 'a') {
// copy character 'a' to
// output string
str2 = str2.substring(0,i)+'a'+str2.substring(i+1);
//str2[i] = 'a';
p++;
// If hamming distance became k,
// break;
if (p == k)
break;
}
}
// If k is less than p
if (p < k) {
// Traversing string in reverse
// order
for (int i = n - 1; i >= 0; i--)
if (str.charAt(i) == 'a') {
str2 = str2.substring(0,i)+'b'+str2.substring(i+1);
p++;
if (p == k)
break;
}
}
System.out.println(str2);
}
// Driven Program
public static void main(String[] args) {
String str = "pqrs";
int n = str.length();
int k = 2;
findString(str, n, k);
}
}
// This code is contributed by 29AjayKumarPython 3
# Python 3 program to find Lexicographically
# smallest string whose hamming distance
# from the given string is exactly K
# function to find Lexicographically
# smallest string with hamming distance k
def findString(str, n, k):
# If k is 0, output input string
if (k == 0):
print(str)
return
# Copying input string into output
# string
str2 = str
p = 0
# Traverse all the character of the
# string
for i in range(0, n, 1):
# If current character is not 'a'
if (str2[i] != 'a'):
# copy character 'a' to
# output string
str2 = str2.replace(str2[i], 'a')
p += 1
# If hamming distance became k,
# break;
if (p == k):
break
# If k is less than p
if (p < k):
# Traversing string in reverse
# order
i = n - 1
while(i >= 0):
if (str[i] == 'a'):
str2 = str2.replace(str2[i], 'b')
p += 1
if (p == k):
break
i -= 1
print(str2)
# Driver Code
if __name__ == '__main__':
str = "pqrs"
n = len(str)
k = 2
findString(str, n, k)
# This code is contributed by
# Surendra_GangwarC#
// C# program to find Lexicographically
// smallest string whose hamming distance
// from given string is exactly K
using System;
class GFG
{
// function to find Lexicographically
// smallest string with hamming distance k
static void findString(String str,
int n, int k)
{
// If k is 0, output input string
if (k == 0)
{
Console.Write(str);;
return;
}
// Copying input string into
// output string
String str2 = str;
int p = 0;
// Traverse all the character
// of the string
for (int i = 0; i < n; i++)
{
// If current character is not 'a'
if (str2[i] != 'a')
{
// copy character 'a' to
// output string
str2 = str2.Substring(0, i) + 'a' +
str2.Substring(i + 1);
//str2[i] = 'a';
p++;
// If hamming distance became k,
// break;
if (p == k)
break;
}
}
// If k is less than p
if (p < k)
{
// Traversing string in reverse
// order
for (int i = n - 1; i >= 0; i--)
if (str[i] == 'a')
{
str2 = str2.Substring(0, i) + 'b' +
str2.Substring(i + 1);
p++;
if (p == k)
break;
}
}
Console.Write(str2);
}
// Driver Code
public static void Main()
{
String str = "pqrs";
int n = str.Length;
int k = 2;
findString(str, n, k);
}
}
// This code is contributed by 29AjayKumarPHP
= 0; $i--)
if ($str[$i] == 'a')
{
$str2[$i] = 'b';
$p++;
if ($p == $k)
break;
}
}
echo $str2 . "\n";
}
// Driver Code
$str = "pqrs";
$n = strlen($str);
$k = 2;
findString($str, $n, $k);
// This code is contributed
// by Akanksha Rai
?>Javascript
输出：
aars