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:
- Initialize a number ‘A’ of length ‘N’.
- Initialize a PAIR STL to store the minimum cost and Number.
- Store the number as a string in the temp variable.
- Using two for loop checks all the digits with a difference of ‘j’ and replace them with ‘i’, break if the cost is achieved.
- Replace the minimum cost with the previous one.
- 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 <bits/stdc++.h> using namespace std; // function to calculate the minimum // value and the final number 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 string temp; // To store the minimum cost and no pair< int , string> 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 code int 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 approach import 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 character static 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 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 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 code public 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 problem import sys # function to calculate the # minimum value and the final # number def 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 code if __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 problem using 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< int , string > ans = new Tuple< int , string >(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< int , string >(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
<script> // Javascript implementation of the approach // count for the given character function count(a,c) { let co = 0; for (let i = 0; i < a.length; i++) if (a[i] == c) co++; return co; } // function to calculate the minimum // value and the final number function finalNumber(n,k,a) { // modtemp = modified temp String let modtemp; // store the count of numbers changed to k let co; // temporary temp String let temp = new Array(a.length); // To store the minimum cost and no let ans = [Number.MAX_VALUE, "" ]; for (let 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.split( "" ); // To store the temporary value of the modified number modtemp = 0; // Initial count for the given number to replace 'i' co = count(a, String.fromCharCode(i + '0 '.charCodeAt(0))); // ' j ' manages the difference ' i ' and ' j ' for (let 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 (let p = 0; p < n; p++) { // Break if count is achieved if (co >= k) break; if (i + ' 0 '.charCodeAt(0) == temp[p].charCodeAt(0) - j) { // Replaces all elements with difference // ' j ' and with ' i ' temp[p] = String.fromCharCode(i + ' 0 '.charCodeAt(0)); modtemp += j; co++; } } } // For the elements before ' i ' index if (i - j >= 0) { for (let p = n - 1; p >= 0; p--) { if (co >= k) break; if (i + ' 0 '.charCodeAt(0) == temp[p].charCodeAt(0) + j) { temp[p] = String.fromCharCode(i + ' 0'.charCodeAt(0)); modtemp += j; co++; } } } } // replace the minimum cost with the previous one if (ans[0] > modtemp) ans = [modtemp, temp.join( "" )]; } // print the minimum cost and the final number document.write( ans[0] + "<br>" + ans[1] + "<br>" ); return -1; } // Driver code // initialize number length and k let n = 5, k = 4; // initialize the number let a = "21122" ; finalNumber(n, k, a); // This code is contributed by avanitrachhadiya2155 </script> |
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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