K distant string

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

// CPP program to find the k distant string
#include <bits/stdc++.h>
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 Kumari

C#

// 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.

Javascript

<script>

// JavaScript program to find the k distant string


// function to find the
// lost string
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));

            // 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 = String.fromCharCode(maybe);
            }
        }

        // decrease the remaining
        // distance
        k -= best_distance;
        str = str.replace(str[i],best_letter);

    }

    // we found a correct
    // string only if "k == 0"
    if (k == 0)
        return str;
    else
        return "No";
}

// driver function

let str = "bear";
let k = 26;
document.write(findKDistantString(str, k),"</br>");

str = "af";
k = 7;

document.write(findKDistantString(str, k),"</br>");


// This code is contributed by shinjanpatra

</script>

Output

zcar
hf

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.