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Maximize product of integers formed by splitting digits of N into two parts in any permutation

  • Difficulty Level : Hard
  • Last Updated : 28 Feb, 2022

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++




// 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




// 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




# 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#




// 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.

Javascript




<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>
Output
10

Time Complexity: O((log N)! * log N) 
Auxiliary Space: O(1)

 


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