给定一个长度为n的字符串和一个非负整数k。查找给出K字符串的遥远的字符串。
两个字母之间的距离是它们在字母表中的位置之间的差。例如：
- dist(c，e)= dist(e，c)= 2。
- dist(a，z)= dist(z，a)= 25。
通过使用此概念，两个字符串之间的距离是相应字母的距离之和。例如 ：
- dist(af，hf)= dist(a，h)+ dist(f，f)= 7 + 0 = 7。
给定一个字符串和一个距离k。任务是找到一个字符串，使结果字符串与给定字符串的距离为k。如果无法使用k个远距字符串，则打印“否”。
注意：可能存在多种解决方案。我们急需找到其中一个。
例子 ：
Input : bear
k = 26
Output : zcar
Here, dist(bear, zcar) =
dist(b, z) + dist(e, c) +
+ dist(a, a) + dist(r, r)
= 24 + 2 + 0 + 0
= 26
Input : af
k = 7
Output : hf
Here, dist(af, hf) = dist(a, h) + dist(f, f)
= 7 + 0
= 7
Input : hey
k = 1000
Output : No
Explanation :
No such string exists.
如果给定的所需距离太大，则无法解决。考虑给定字符串的最大可能距离是多少。还是更有用的东西-如何构造丢失的字符串以最大化距离?分别对待每个字母，并用距离最远的字母替换。例如，我们应将“ c”替换为“ z”，并将“ y”替换为“ a”。更准确地说，对于字母表的前13个字母，最远的字母是“ z”，对于其他字母，它是“ a”。
该方法很简单，遍历给定字符串的字母并贪婪地更改它们。 “贪婪”一词意味着在更改字母时，不必关心接下来的字母。通常，必须有距离较远的字母，因为否则可能没有解决方案。对于给定字符串的每个字母，请将其更改为最远的字母，除非总距离太大。更改字母后，请减少剩余的所需距离。因此，对于给定字符串的每个字母，仅考虑不超过剩余距离的字母，并从中选择距离最远的字母。
CPP和Java实施：
CPP
// CPP program to find the k distant string
#include
using namespace std;
// function to find the
// lost string
string findKDistantString(string str, int k)
{
int n = str.length();
for (int i = 0; i < n; ++i) {
char best_letter = str[i];
int best_distance = 0;
for (char maybe = 'a';
maybe <= 'z'; ++maybe)
{
int distance = abs(maybe - str[i]);
// check if "distance <= k",
// so that, the total distance
// will not exceed among
// letters with "distance <= k"
if (distance <= k && distance >
best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
// decrease the remaining
// distance
k -= best_distance;
str[i] = best_letter;
}
assert(k >= 0);
// we found a correct
// string only if "k == 0"
if (k > 0)
return "No";
else
return str;
}
// driver function
int main()
{
string str = "bear";
int k = 26;
cout << findKDistantString(str, k) << endl;
str = "af";
k = 7;
cout << findKDistantString(str, k) << endl;
return 0;
} Java
// Java program to find k distant string
import java.util.*;
import java.lang.*;
public class GfG {
// function to find
// the lost string
public static String findKDistantString
(String str1, int k)
{
int n = str1.length();
char[] str = str1.toCharArray();
for (int i = 0; i < n; ++i) {
char best_letter = str[i];
int best_distance = 0;
for (char maybe = 'a';
maybe <= 'z'; ++maybe)
{
int distance =
Math.abs(maybe - str[i]);
// Check if "distance <= k"
// so that it should not
// exceed the total distance
// among letters with "distance
// <= k" we choose the most
// distant one
if (distance <= k && distance
> best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
// we decrease the remaining
// distance
k -= best_distance;
str[i] = best_letter;
}
assert(k >= 0);
// Correct string only
// if "k == 0"
if (k > 0)
return "No";
else
return (new String(str));
}
public static void main(String argc[])
{
String str = "bear";
int k = 26;
System.out.println(findKDistantString(str, k));
str = "af";
k = 7;
System.out.println(findKDistantString(str, k));
}
}Python3
# Python implementation to check if
# both halves of the string have
# at least one different character
MAX = 26
# Function which break string into two halves
# Counts frequency of characters in each half
# Compares the two counter array and returns
# true if these counter arrays differ
def function(st):
global MAX
l = len(st)
# Declaration and initialization
# of counter array
counter1, counter2 = [0]*MAX, [0]*MAX
for i in range(l//2):
counter1[ord(st[i]) - ord('a')] += 1
for i in range(l//2, l):
counter2[ord(st[i]) - ord('a')] += 1
for i in range(MAX):
if (counter2[i] != counter1[i]):
return True
return False
# Driver function
st = "abcasdsabcae"
if function(st): print("Yes, both halves differ by at least one character")
else: print("No, both halves do not differ at all")
# This code is contributed by Ansu KumariC#
// C# program to find k distant string
using System;
class GfG {
// function to find the lost string
public static String findKDistantString
(string str1, int k)
{
int n = str1.Length;
char []str = str1.ToCharArray();
for (int i = 0; i < n; ++i) {
char best_letter = str[i];
int best_distance = 0;
for (char maybe = 'a';
maybe <= 'z'; ++maybe)
{
int distance =
Math.Abs(maybe - str[i]);
// Check if "distance <= k"
// so that it should not
// exceed the total distance
// among letters with "distance
// <= k" we choose the most
// distant one
if (distance <= k && distance
> best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
// we decrease the remaining
// distance
k -= best_distance;
str[i] = best_letter;
}
//(k >= 0);
// Correct string only
// if "k == 0"
if (k > 0)
return "No";
else
return (new string(str));
}
// Driver code
public static void Main()
{
string str = "bear";
int k = 26;
Console.WriteLine(
findKDistantString(str, k));
str = "af";
k = 7;
Console.Write(
findKDistantString(str, k));
}
}
// This code is contributed by Nitin millal.输出：
zcar
hf