Convert a number of length N such that it contains any one digit at least ‘K’ times

Given the value ‘N’ which is the length of a number ‘A’. Your task is to convert the digits such that at least ‘K’ times in the number ‘A’ any digit exists. In order to replace one of ‘N’ digits, you also need to calculate the cost, which is the absolute difference between the old digit and the new one. The task is to print the minimum cost it took to convert the initial number to the final number and also print the final number.
Note: If there are several such numbers, then print the lexicographically minimum one.

Examples:

Input: N = 6, K = 5, A = 898196
Output: 4, 888188
Number = “898196”, the second digit “9” will be replaced by “8” costs |9 – 8| = 1 . Replacing the fifth digit with an “8” will cost the same. Replacing the fifth digit cost |6 – 8| = 2. As a result, 4 will be the total cost and the final number will be “888188”.

Input: N = 16, K = 14, A = 6124258626539246
Output: 22, 4444448444449444

Approach:

1. Initialize a number ‘A’ of length ‘N’.
2. Initialize a PAIR STL to store the minimum cost and Number.
3. Store the number as a string in the temp variable.
4. Using two for loop checks all the digits with a difference of ‘j’ and replace them with ‘i’, break if the cost is achieved.
5. Replace the minimum cost with the previous one.
6. Finally, print the minimum cost and the final number.

Below is the implementation of the above approach:

C++

 // C++ program to illustrate// the above problem#include using namespace std; // function to calculate the minimum// value and the final numberint finalNumber(int n, int k, string a){    // modtemp = modified temp string    int modtemp;     // store the count of numbers changed to k    int co;     // temporary temp string    string temp;     // To store the minimum cost and no    pair ans = make_pair(INT_MAX, "");     for (int i = 0; i < 10; i++) {        // 'i' will replace the digits of N's to        // generate a number with k same digits         // store the main str in temp str for modification        temp = a;         // To store the temporary value of the modified number        modtemp = 0;         // Initial count for the given number to replace 'i'        co = count(a.begin(), a.end(), i + '0');         // 'j' manages the difference 'i' and 'j'        for (int j = 1; j < 10; j++) {             // For the elements ahead of 'i' index            if (i + j < 10) {                 // Checks all elements with difference 'j'                // and replaces them with 'i'                for (int p = 0; p < n; p++) {                     // Break if count is achieved                    if (co >= k)                        break;                     if (i + '0' == temp[p] - j) {                         // Replaces all elements with difference                        // 'j' and with 'i'                        temp[p] = i + '0';                        modtemp += j;                        co++;                    }                }            }            // For the elements before 'i' index            if (i - j >= 0) {                for (int p = n - 1; p >= 0; p--) {                    if (co >= k)                        break;                     if (i + '0' == temp[p] + j) {                        temp[p] = i + '0';                        modtemp += j;                        co++;                    }                }            }        }         // replace the minimum cost with the previous one        ans = min(ans, make_pair(modtemp, temp));    }    // print the minimum cost and the final number    cout << ans.first << endl         << ans.second << endl;} // Driver codeint main(){    // initialize number length and k    int n = 5, k = 4;     // initialize the number    string a = "21122";     finalNumber(n, k, a);     return 0;}

Java

 // Java implementation of the approachimport java.util.*;class GFG{    static class pair    {        int first;        String second;        static pair make_pair(int first, String second)        {            pair p = new pair();            p.first = first;            p.second = second;            return p;        }    }     // count for the given characterstatic int count(String a,char c){    int co = 0;    for(int i = 0; i < a.length(); i++)    if(a.charAt(i) == c)        co++;    return co;} // function to calculate the minimum// value and the final numberstatic int finalNumber(int n, int k, String a){    // modtemp = modified temp String    int modtemp;     // store the count of numbers changed to k    int co;     // temporary temp String    char temp[] = new char[a.length()];     // To store the minimum cost and no    pair ans = pair.make_pair(Integer.MAX_VALUE, "");     for (int i = 0; i < 10; i++)     {        // 'i' will replace the digits of N's to        // generate a number with k same digits         // store the main str in temp str for modification        temp = a.toCharArray();         // To store the temporary value of the modified number        modtemp = 0;         // Initial count for the given number to replace 'i'        co = count(a, (char)(i + '0'));         // 'j' manages the difference 'i' and 'j'        for (int j = 1; j < 10; j++)         {             // For the elements ahead of 'i' index            if (i + j < 10)             {                 // Checks all elements with difference 'j'                // and replaces them with 'i'                for (int p = 0; p < n; p++)                 {                     // Break if count is achieved                    if (co >= k)                        break;                     if (i + '0' == temp[p] - j)                     {                         // Replaces all elements with difference                        // 'j' and with 'i'                        temp[p] = (char)(i + '0');                        modtemp += j;                        co++;                    }                }            }                         // For the elements before 'i' index            if (i - j >= 0)             {                for (int p = n - 1; p >= 0; p--)                 {                    if (co >= k)                        break;                     if (i + '0' == temp[p] + j)                     {                        temp[p] = (char)(i + '0');                        modtemp += j;                        co++;                    }                }            }        }         // replace the minimum cost with the previous one        if(ans.first > modtemp)        ans = pair.make_pair(modtemp, new String(temp));    }         // print the minimum cost and the final number    System.out.print( ans.first + "\n"                    + ans.second + "\n");         return -1;} // Driver codepublic static void main(String args[]){    // initialize number length and k    int n = 5, k = 4;     // initialize the number    String a = "21122";     finalNumber(n, k, a);}} // This code is contributed by Arnab Kundu

Python3

 # Python3 program to illustrate# the above problemimport sys # function to calculate the # minimum value and the final # numberdef finalNumber(n, k, a):     # To store the minimum     # cost and no    ans = [sys.maxsize, ""]     for i in range(10):               # 'i' will replace the         # digits of N's to generate         # a number with k same digits         # store the main str in temp         # str for modification        temp = a         # To store the temporary         # value of the modified number        modtemp = 0         # Initial count for the         # given number to replace 'i'        co = a.count(chr(i + ord('0')))         # 'j' manages the difference         # 'i' and 'j'        for j in range(1, 10):             # For the elements ahead             # of 'i' index            if (i + j < 10):                 # Checks all elements with                # difference 'j' and replaces                 # them with 'i'                for p in range(n):                     # Break if count is                     # achieved                    if (co >= k):                        break                     if (i + ord('0') ==                        ord(temp[p]) - j):                         # Replaces all elements                         # with difference 'j'                         # and with 'i'                        temp.replace(temp[p],                                     chr(i +                                         ord('0')), 1)                        modtemp += j                        co+= 1             # For the elements             # before 'i' index            if (i - j >= 0):                for p in range(n - 1,                               -1, -1):                    if (co >= k):                        break                    if (i + ord('0') ==                        ord(temp[p]) + j):                        temp.replace(temp[p],                                     chr(i +                                         ord('0')), 1)                        modtemp += j                        co += 1         # replace the minimum cost         # with the previous one        ans = min(ans, [modtemp,                         temp])     # print the minimum cost     # and the final number    print(ans[0])    print(ans[1]) # Driver codeif __name__ == "__main__":     # Initialize number     # length and k    n = 5    k = 4     # initialize the number    a = "21122"     finalNumber(n, k, a) # This code is contributed by Chitranayal

C#

 // C# program to illustrate// the above problemusing System;using System.Collections.Generic;class GFG {         // count for the given character    static int count(string a,char c)    {        int co = 0;        for(int i = 0; i < a.Length; i++)        if(a[i] == c)            co++;        return co;    }          // function to calculate the minimum    // value and the final number    static int finalNumber(int n, int k, string a)    {        // modtemp = modified temp String        int modtemp;              // store the count of numbers changed to k        int co;              // temporary temp String        char[] temp = new char[a.Length];              // To store the minimum cost and no        Tuple ans = new Tuple(Int32.MaxValue, "");               for (int i = 0; i < 10; i++)         {            // 'i' will replace the digits of N's to            // generate a number with k same digits                  // store the main str in temp str for modification            temp = a.ToCharArray();                  // To store the temporary value of the modified number            modtemp = 0;                  // Initial count for the given number to replace 'i'            co = count(a, (char)(i + '0'));                  // 'j' manages the difference 'i' and 'j'            for (int j = 1; j < 10; j++)             {                      // For the elements ahead of 'i' index                if (i + j < 10)                 {                          // Checks all elements with difference 'j'                    // and replaces them with 'i'                    for (int p = 0; p < n; p++)                     {                              // Break if count is achieved                        if (co >= k)                            break;                              if (i + '0' == temp[p] - j)                         {                                  // Replaces all elements with difference                            // 'j' and with 'i'                            temp[p] = (char)(i + '0');                            modtemp += j;                            co++;                        }                    }                }                                  // For the elements before 'i' index                if (i - j >= 0)                 {                    for (int p = n - 1; p >= 0; p--)                     {                        if (co >= k)                            break;                              if (i + '0' == temp[p] + j)                         {                            temp[p] = (char)(i + '0');                            modtemp += j;                            co++;                        }                    }                }            }                  // replace the minimum cost with the previous one            if(ans.Item1 > modtemp)            ans = new Tuple(modtemp, new string(temp));        }                  // print the minimum cost and the final number        Console.Write( ans.Item1 + "\n" + ans.Item2 + "\n");                  return -1;    }   static void Main()   {         // initialize number length and k    int n = 5, k = 4;         // initialize the number    string a = "21122";      finalNumber(n, k, a);  }} // This code is contributed by divyeshrabadiya07

Javascript



Output:
1
21222

Explanation: As on converting 1 to 2 just one time. 2 becomes k times in the number. So the cost is 2-1 = 1.

Time Complexity: O(10 * 10 * N), where N is the given length of the number.
Auxiliary Space: O(1) because constant space is used.

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