Given an integer N in the range [0, 109], the task is to find the maximum product of two integers that are formed by dividing any permutation of digits of integer N into two parts.
Example:
Input: N = 123
Output: 63
Explanation: The number of ways of dividing N = 123 into 2 integers are {12, 3}, {21, 3}, {13, 2}, {31, 2}, {23, 1} and {32, 1}. The product of {21, 3} is 63 which is the maximum over all possibilities.Input: N = 101
Output: 10
Approach: The given problem can be solved by using basic permutation and combination with the help of the following observations:
- The total number of ways to divide the given integer into two parts can be calculated as (Number of possible permutations) * (Ways to divide a permutation) => 9! * 8 => 2903040. Therefore, all possible separation can be iterated.
- Leading zeroes in any permutation do not affect the answer as well.
Therefore, all the permutations of the digits of the current integer can be iterated using the next permutation function, and for each permutation maintain the maximum product of all the possible ways to divide the permutation in a variable, which is the required answer.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to find maximum product // of integers formed by dividing any // permutation of N into two parts. int maxProduct(string N)
{ // Stores the maximum product
int ans = 0;
// Sort the digits in the string
sort(N.begin(), N.end());
// Iterating over all permutations
do {
// Loop to iterate over all
// possible partitions
for ( int i = 1; i < N.size(); i++) {
int l = 0, r = 0;
// Calculate the left partition
for ( int j = 0; j < i; j++)
l = l * 10 + N[j] - '0' ;
// Calculate the right partition
for ( int j = i; j < N.size(); j++)
r = r * 10 + N[j] - '0' ;
// Update answer
ans = max(ans, l * r);
}
} while (next_permutation(N.begin(), N.end()));
// Return answer
return ans;
} // Driver code int main()
{ int N = 101;
cout << maxProduct(to_string(N));
return 0;
} |
// Java implementation of the above approach import java.util.*;
class GFG{
static boolean next_permutation( char [] p) {
for ( int a = p.length - 2 ; a >= 0 ; --a)
if (p[a] < p[a + 1 ])
for ( int b = p.length - 1 ;; --b)
if (p[b] > p[a]) {
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1 ; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
return false ;
}
// Function to find maximum product
// of integers formed by dividing any
// permutation of N into two parts.
static int maxProduct(String a)
{
// Stores the maximum product
int ans = 0 ;
// Sort the digits in the String
char []N = a.toCharArray();
Arrays.sort(N);
// Iterating over all permutations
do {
// Loop to iterate over all
// possible partitions
for ( int i = 1 ; i < N.length; i++) {
int l = 0 , r = 0 ;
// Calculate the left partition
for ( int j = 0 ; j < i; j++)
l = l * 10 + N[j] - '0' ;
// Calculate the right partition
for ( int j = i; j < N.length; j++)
r = r * 10 + N[j] - '0' ;
// Update answer
ans = Math.max(ans, l * r);
}
} while (next_permutation(N));
// Return answer
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 101 ;
System.out.print(maxProduct(String.valueOf(N)));
}
} // This code is contributed by umadevi9616 |
# Python3 implementation of the above approach # Function for next permutation def next_permutation(arr):
# Find the length of the array
n = len (arr)
# Start from the right most digit and
# find the first digit that is smaller
# than the digit next to it.
k = n - 2
while k > = 0 :
if arr[k] < arr[k + 1 ]:
break
k - = 1
# Reverse the list if the digit that
# is smaller than the digit next to
# it is not found.
if k < 0 :
arr = arr[:: - 1 ]
else :
# Find the first greatest element
# than arr[k] from the end of the list
for l in range (n - 1 , k, - 1 ):
if arr[l] > arr[k]:
break
# Swap the elements at arr[k] and arr[l
arr[l], arr[k] = arr[k], arr[l]
# Reverse the list from k + 1 to the end
# to find the most nearest greater number
# to the given input number
arr[k + 1 :] = reversed (arr[k + 1 :])
return arr
# Function to find maximum product # of integers formed by dividing any # permutation of N into two parts. def maxProduct(N):
# Stores the maximum product
ans = 0
# Sort the digits in the string
Nstr = sorted (N)
Instr = sorted (N)
# Iterating over all permutations
while next_permutation(Nstr) ! = Instr:
# Loop to iterate over all
# possible partitions
for i in range ( len (Nstr)):
l = 0
r = 0
# Calculate the left partition
for j in range ( 0 , i):
l = l * 10 + ord (N[j]) - ord ( '0' )
# Calculate the right partition
for j in range (i, len (Nstr)):
r = r * 10 + ord (N[j]) - ord ( '0' )
# Update answer
ans = max (ans, l * r)
# Return answe
return ans
# Driver code N = 101
print (maxProduct( str (N)))
# This code is contributed by Potta Lokesh |
// C# implementation of the above approach using System;
class GFG {
static bool next_permutation( char [] p)
{
for ( int a = p.Length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for ( int b = p.Length - 1;; --b)
if (p[b] > p[a]) {
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1; a < b;
++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
return false ;
}
// Function to find maximum product
// of integers formed by dividing any
// permutation of N into two parts.
static int maxProduct( string a)
{
// Stores the maximum product
int ans = 0;
// Sort the digits in the String
char [] N = a.ToCharArray();
Array.Sort(N);
// Iterating over all permutations
do {
// Loop to iterate over all
// possible partitions
for ( int i = 1; i < N.Length; i++) {
int l = 0, r = 0;
// Calculate the left partition
for ( int j = 0; j < i; j++)
l = l * 10 + N[j] - '0' ;
// Calculate the right partition
for ( int j = i; j < N.Length; j++)
r = r * 10 + N[j] - '0' ;
// Update answer
ans = Math.Max(ans, l * r);
}
} while (next_permutation(N));
// Return answer
return ans;
}
// Driver code
public static void Main( string [] args)
{
int N = 101;
Console.Write(maxProduct(N.ToString()));
}
} // This code is contributed by ukasp. |
<script> // javascript implementation of the above approach function next_permutation( p) {
for ( var a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (b = p.length - 1;; --b)
if (p[b] > p[a]) {
var t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true ;
}
return false ;
}
// Function to find maximum product
// of integers formed by dividing any
// permutation of N into two parts.
function maxProduct( a) {
// Stores the maximum product
var ans = 0;
// Sort the digits in the String
N = a.split( "" );
N.sort();
// Iterating over all permutations
do {
// Loop to iterate over all
// possible partitions
for ( var i = 1; i < N.length; i++) {
var l = 0, r = 0;
// Calculate the left partition
for ( var j = 0; j < i; j++)
l = l * 10 + parseInt(N[j])- '0' ;
// Calculate the right partition
for ( var j = i; j < N.length; j++)
r = r * 10 + parseInt(N[j])- '0' ;
// Update answer
ans = Math.max(ans, l * r);
}
} while (next_permutation(N));
// Return answer
return ans;
}
// Driver code
var N = "101" ;
document.write(maxProduct(N));
// This code is contributed by Rajput-Ji </script> |
10
Time Complexity: O((log N)! * log N)
Auxiliary Space: O(1)