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

Given the value **‘N’** which is 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 needs to calculate the cost which is the absolute differences between the old digit and the new one. The task is to print the minimum cost took to convert the initial number to the final number 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 as “9” will be replaced to “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 string in the temp variable.
- Using two for loop checks all the digits with difference of ‘j’ and replace them with ‘i’, break if cost is achieved.
- Replace the minimum cost with the previous one.
- At last, print the minimum cost and the final number.

Below is the implementation of above approach :

`// 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() ` `{ ` ` ` `// initilaize number length and k ` ` ` `int` `n = 5, k = 4; ` ` ` ` ` `// initialize the number ` ` ` `string a = ` `"21122"` `; ` ` ` ` ` `finalNumber(n, k, a); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**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.

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