Maximize the number N by inserting given digit at any position

Given a positive integer N and a digit K, the task is to find the maximum value of the given number N by inserting the given digit K in it N.

Examples:

Input: N = 6673, K = 6
Output: 66763
Explanation:
All the numbers formed by inserting K at any position in N are {66673, 66763, 66736}. The maximum among all the formed number is 66763.

Input: N = 1234, K = 5
Output: 51234

Approach: The given problem can be solved by inserting K at that position where the next digit is smaller than K. Follow the steps below to solve the problem:

• Initialize two strings say S, by typecasting the given number N into a string and a string result as “” to store the maximum possible number after inserting K in it.
• Initialize two variables, say L as the length of string S and i as 0.
• Traverse the string S until K is less than S[i] and append the character S[i] to the string result.
• Now append K to the result and then append all the remaining characters of the string to the result.
• After completing the above steps, print the string result and the maximum possible number.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum value
// of N after inserting the digit K
void maximizeNumber(int N, int K)
{
    // Convert it into N to string
    string s = to_string(N);
    int L = s.length();

    // Stores the maximum value of N
    // after inserting K
    string result;
    int i = 0;

    // Iterate till all digits that
    // are not less than K
    while ((i < L) && (K <= (s[i] - '0'))) {

        // Add the current digit to
        // the string result
        result.push_back(s[i]);
        ++i;
    }

    // Add digit 'K' to result
    result.push_back(char(K + '0'));

    // Iterate through all remaining
    // characters
    while (i < L) {

        // Add current digit to result
        result.push_back(s[i]);
        ++i;
    }

    // Print the maximum number formed
    cout << result;
}

// Driver Code
int main()
{
    int N = 6673, K = 6;
    maximizeNumber(N, K);

    return 0;
}

Java

// Java program for the above approach

class GFG {

    // Function to find the maximum value
    // of N after inserting the digit K
public static void maximizeNumber(int N, int K)
{
    // Convert it into N to string
    String s = Integer.toString(N);
    int L = s.length();

    // Stores the maximum value of N
    // after inserting K
    String result =  "";
    int i = 0;

    // Iterate till all digits that
    // are not less than K
    while ((i < L) && (K <= ((int)s.charAt(i) - (int)'0'))) {

        // Add the current digit to
        // the string result
        result += (s.charAt(i));
        ++i;
    }

    // Add digit 'K' to result
    result += ((char)(K + (int)'0'));

    // Iterate through all remaining
    // characters
    while (i < L) {

        // Add current digit to result
        result += (s.charAt(i));
        ++i;
    }

    // Print the maximum number formed
    System.out.println(result);
}

    // Driver Code
    public static void main (String args[]) {
        int N = 6673, K = 6;
        maximizeNumber(N, K);
    }

}

// This code is contributed by _saurabh_Jaiswal.

Python3

# Python 3 program for the above approach

# Function to find the maximum value
# of N after inserting the digit K
def maximizeNumber(N, K):

    # Convert it into N to string
    s = str(N)
    L = len(s)

    # Stores the maximum value of N
    # after inserting K
    result = ""
    i = 0

    # Iterate till all digits that
    # are not less than K
    while ((i < L) and (K <= (ord(s[i]) - ord('0')))):

        # Add the current digit to
        # the string result
        result += (s[i])
        i += 1

    # Add digit 'K' to result
    result += (chr(K + ord('0')))

    # Iterate through all remaining
    # characters
    while (i < L):

        # Add current digit to result
        result += (s[i])
        i += 1

    # Print the maximum number formed
    print(result)


# Driver Code
if __name__ == "__main__":

    N = 6673
    K = 6
    maximizeNumber(N, K)

    # This code is contributed by ukasp.

C#

// C# program for above approach
using System;

class GFG{

    // Function to find the maximum value
    // of N after inserting the digit K
public static void maximizeNumber(int N, int K)
{
    // Convert it into N to string
    String s = N.ToString();
    int L = s.Length;

    // Stores the maximum value of N
    // after inserting K
    string result =  "";
    int i = 0;

    // Iterate till all digits that
    // are not less than K
    while ((i < L) && (K <= ((int)s[i]- (int)'0'))) {

        // Add the current digit to
        // the string result
        result += (s[i]);
        ++i;
    }

    // Add digit 'K' to result
    result += ((char)(K + (int)'0'));

    // Iterate through all remaining
    // characters
    while (i < L) {

        // Add current digit to result
        result += (s[i]);
        ++i;
    }

    // Print the maximum number formed
     Console.Write(result);
}
// Driver Code
static void Main()
{

    int N = 6673, K = 6;
    maximizeNumber(N, K);
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// Javascript program for the above approach

// Function to find the maximum value
// of N after inserting the digit K
function maximizeNumber(N, K)
{

    // Convert it into N to string
    let s = String(N);
    let L = s.length;

    // Stores the maximum value of N
    // after inserting K
    let result = [];
    let i = 0;

    // Iterate till all digits that
    // are not less than K
    while ((i < L) && (K <= (s[i].charCodeAt(0) - '0'.charCodeAt(0)))) {

        // Add the current digit to
        // the string result
        result.push(s[i]);
        ++i;
    }

    // Add digit 'K' to result
    result.push(String.fromCharCode(K + '0'.charCodeAt(0)));

    // Iterate through all remaining
    // characters
    while (i < L) {

        // Add current digit to result
        result.push(s[i]);
        ++i;
    }

    // Print the maximum number formed
    document.write(result.join(""));
}

// Driver Code
let N = 6673, K = 6;
maximizeNumber(N, K);

// This code is contributed by gfgking.
</script>

Output:

66763

Time Complexity: O(N)
Auxiliary Space: O(N)