K distant string
Last Updated :
18 Nov, 2022
Given a string of length n and a non-negative integer k. Find k distant string of given string. Distance between two letters is the difference between their positions in the alphabet. for example:
- dist(c, e) = dist(e, c) = 2.
- dist(a, z) = dist(z, a) = 25.
By using this concept, the distance between two strings is the sum of distances of corresponding letters. For example :
- dist(af, hf) = dist(a, h) + dist(f, f) = 7 + 0 = 7.
Given a string and a distance k. Task is to find a string such that distance of the result string is k from given string. If k distant string is not possible, then print “No”.
Note: There may exist multiple solutions. We need to find one of them.
Examples :
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.
There is no solution if the given required distance is too big. Think what is the maximum possible distance for the given string. Or the more useful thing — how to construct the lost string to maximize the distance? Treat each letter separately and replace it with the most distant letter.
For example, we should replace ‘c’ with ‘z’, and we should replace ‘y’ with ‘a’. To be more precise, for the first 13 letters of the alphabet the most distant letter is ‘z’, and for other letters, it is ‘a’. The approach is simple, iterate over letters of the given string and greedily change them. The word “greedily” means when changing a letter, don’t care about the next letters. Generally, there must be distant letters, because there may not be a solution otherwise. For each letter of the given string change it into the most distant letter, unless the total distance would be too big. As letters are changed, decrease the remaining required distance. So, for each letter of the given string consider only letters not exceeding the remaining distance, and among them choose the most distant one. CPP and JAVA Implementation:
Implementation:
CPP
#include <bits/stdc++.h>
using namespace std;
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]);
if (distance <= k && distance >
best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
k -= best_distance;
str[i] = best_letter;
}
assert (k >= 0);
if (k > 0)
return "No" ;
else
return str;
}
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
import java.util.*;
import java.lang.*;
public class GfG {
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]);
if (distance <= k && distance
> best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
k -= best_distance;
str[i] = best_letter;
}
assert (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
MAX = 26
def function(st):
global MAX
l = len (st)
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
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" )
|
C#
using System;
class GfG {
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]);
if (distance <= k && distance
> best_distance)
{
best_distance = distance;
best_letter = maybe;
}
}
k -= best_distance;
str[i] = best_letter;
}
if (k > 0)
return "No" ;
else
return ( new string (str));
}
public static void Main()
{
string str = "bear" ;
int k = 26;
Console.WriteLine(
findKDistantString(str, k));
str = "af" ;
k = 7;
Console.Write(
findKDistantString(str, k));
}
}
|
Javascript
<script>
function findKDistantString(str,k)
{
let n = str.length;
for (let i = 0; i < n; ++i) {
let best_letter = str[i];
let best_distance = 0;
for (let maybe = 97; maybe <= 122; ++maybe)
{
let distance = Math.abs(maybe - str.charCodeAt(i));
if (distance <= k && distance >
best_distance)
{
best_distance = distance;
best_letter = String.fromCharCode(maybe);
}
}
k -= best_distance;
str = str.replace(str[i],best_letter);
}
if (k == 0)
return str;
else
return "No" ;
}
let str = "bear" ;
let k = 26;
document.write(findKDistantString(str, k), "</br>" );
str = "af" ;
k = 7;
document.write(findKDistantString(str, k), "</br>" );
</script>
|
Time Complexity: O(n), where n is the size of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...