Maximize the number N by inserting given digit at any position

Last Updated : 05 Aug, 2021

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# 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:

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

Maximize a given unsigned number by swapping bits at it's extreme positions.

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