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Largest number by rearranging digits of a given positive or negative number

Last Updated : 24 Feb, 2022
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Given an integer N(positive or negative), the task is to find the maximum number that can be formed using all of the digits of this number.

Examples:

Input: -38290367
Output: -20336789
Explanation: As there is need to use all the digits, 0 cannot be the first digit because it becomes redundant at first position.

Input: 1203465
Output: 6543210

 

Approach: The digits in a number will range from 0-9, so the idea is to create a hash array of size 10 and store the count of every digit in the hashed array that occurs in the number. Follow the steps mentioned below to solve the problem: 

  • Then first check if the number is positive or negative
  • If the number is positive, then as explained in this article, traverse the hashed array from index 9 to 0 and calculate the number accordingly.
  • But if the number is negative, traverse the array from 0-9, such that:
    • There can be leading 0s. So, find the smallest non-zero digit present
    • Insert the smallest non-zero digit in the start, and then add all other digits from 0 to 9 in that order to the final answer
    • Then multiply the number by -1 to make it negative
  • Return the resultant number

Below is the implementation of the above approach

C++ 
// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the maximum number
long long printMaxNum(long long num)
{
    // Initialising hash array
    int hash[10] = { 0 };
    long long n = num < 0 ? num * -1 : num;
    long long ans = 0;
    while (n) {
        hash[n % 10]++;
        n = n / 10;
    }

    // If positive number
    if (num > 0) {
        for (int i = 9; i >= 0; i--)
            for (int j = 0; j < hash[i]; j++)
                ans = ans * 10 + i;
    }

    // If negative number
    else {

        // If 0 is present in the number
        if (hash[0] > 0) {
            for (int i = 1; i < 10; i++)
                if (hash[i] > 0) {
                    ans = i;
                    hash[i]--;
                    break;
                }
        }
        for (int i = 0; i < 10; i++)
            for (int j = 0; j < hash[i]; j++)
                ans = ans * 10 + i;
        ans = ans * -1;
    }
    return ans;
}

// Driver code
int main()
{
    int N = -38290367;

    // Function call
    cout << printMaxNum(N);
    return 0;
}
Java 
// Java program to implement the approach
class GFG 
{

  // Function to print the maximum number
  static long printMaxNum(long num)
  {

    // Initialising hash array
    int[] hash = new int[10];
    for (int i = 0; i < 10; i++) {
      hash[i] = 0;
    }
    long n = num < 0 ? num * -1 : num;
    long ans = 0;
    while (n > 0) {
      hash[(int)(n % 10)] += 1;
      n = n / 10;
    }

    // If positive number
    if (num > 0) {
      for (int i = 9; i >= 0; i--)
        for (int j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
    }

    // If negative number
    else {

      // If 0 is present in the number
      if (hash[0] > 0) {
        for (int i = 1; i < 10; i++)
          if (hash[i] > 0) {
            ans = i;
            hash[i]--;
            break;
          }
      }
      for (int i = 0; i < 10; i++)
        for (int j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
      ans = ans * -1;
    }
    return ans;
  }

  // Driver code
  public static void main(String args[])
  {
    int N = -38290367;

    // Function call
    System.out.println(printMaxNum(N));
  }
}

// This code is contributed by gfgking
Python3 
# Python program to implement the approach

# Function to print the maximum number
def printMaxNum(num):

    # Initialising hash array
    hash = []
    for i in range(0, 10):
        hash.append(0)

    if(num < 0):
        n = num * -1
    else:
        n = num

    ans = 0
    while (n != 0):
        hash[int(n % 10)] = hash[int(n % 10)] + 1
        n = n // 10

    # If positive number
    if (num > 0):
        for i in range(9, -1, -1):
            for j in range(0, hash[i]):
                ans = ans * 10 + i

    # If negative number
    else:

        # If 0 is present in the number
        if (hash[0] > 0):
            for i in range(1, 10):
                if (hash[i] > 0):
                    ans = i
                    hash[i] = hash[i]-1
                    break

        for i in range(0, 10):
            for j in range(0, hash[i]):
                ans = ans * 10 + i
        ans = ans * -1

    return ans

# Driver code
N = -38290367

# Function call
print(printMaxNum(N))

# This code is contributed by Taranpreet
C# 
// C# program to implement the approach
using System;
class GFG {
    // Function to print the maximum number
    static long printMaxNum(long num)
    {
        // Initialising hash array
        int[] hash = new int[10];
        for (int i = 0; i < 10; i++) {
            hash[i] = 0;
        }
        long n = num < 0 ? num * -1 : num;
        long ans = 0;
        while (n > 0) {
            hash[n % 10]++;
            n = n / 10;
        }

        // If positive number
        if (num > 0) {
            for (int i = 9; i >= 0; i--)
                for (int j = 0; j < hash[i]; j++)
                    ans = ans * 10 + i;
        }

        // If negative number
        else {

            // If 0 is present in the number
            if (hash[0] > 0) {
                for (int i = 1; i < 10; i++)
                    if (hash[i] > 0) {
                        ans = i;
                        hash[i]--;
                        break;
                    }
            }
            for (int i = 0; i < 10; i++)
                for (int j = 0; j < hash[i]; j++)
                    ans = ans * 10 + i;
            ans = ans * -1;
        }
        return ans;
    }

    // Driver code
    public static void Main()
    {
        int N = -38290367;

        // Function call
        Console.Write(printMaxNum(N));
    }
}

// This code is contributed by Samim Hossain Mondal.
JavaScript 
<script>
// javascript program to implement the approach

  // Function to print the maximum number
  function printMaxNum(num)
  {

    // Initialising hash array
    var hash = Array.from({length: 10}, (_, i) => 0);
    for (var i = 0; i < 10; i++) {
      hash[i] = 0;
    }
    var n = num < 0 ? num * -1 : num;
    var ans = 0;
    while (n > 0) {
      hash[parseInt(n % 10)] += 1;
      n = parseInt(n / 10);
    }

    // If positive number
    if (num > 0) {
      for (var i = 9; i >= 0; i--)
        for (var j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
    }

    // If negative number
    else {

      // If 0 is present in the number
      if (hash[0] > 0) {
        for (var i = 1; i < 10; i++)
          if (hash[i] > 0) {
            ans = i;
            hash[i]--;
            break;
          }
      }
      for (var i = 0; i < 10; i++)
        for (var j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
      ans = ans * -1;
    }
    return ans;
  }

  // Driver code
var N = -38290367;

// Function call
document.write(printMaxNum(N));

// This code is contributed by shikhasingrajput 
</script>

 
 


Output
-20336789


 

Time Complexity: O( length(N) )
Auxiliary Space: O(1)


 


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