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> |
66763
Time Complexity: O(N)
Auxiliary Space: O(N)