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++ 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 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.
# 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.
<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)