Find all distinct three digit numbers from given array of digits

Last Updated : 10 Aug, 2022

Given an array containing digits[], where each element is a single digit integer. The array may contain duplicates. The task is to find all the unique integers that follow the given requirements:

The integer consists of the concatenation of three elements from digits in any arbitrary order.
The integer does not have leading zeros.

Examples:

Input: digits[] = {2, 1, 3, 0}
Output: {102, 103, 120, 123, 130, 132, 201, 203, 210, 213, 230, 231, 301, 302, 310, 312, 320, 321}
Explanation: The above are the three digit numbers formed.

Input: digits[] = {3, 7, 5}
Output: [357, 375, 537, 573, 735, 753 ]

Approach: This problem can be solved by using Frequency Map. Find count of all elements in given digits array. Follow the steps below to solve the given problem.

Check for all numbers between 100 to 999 whether they can be formed by the digits present in the digits vector.
Use 2 maps for the same. If the number can be made, then add it to the answer.
In the end, return the answer.

Below is the implementation of the above approach:

C++

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

// Function to find all the unique
// 3 digit number that can be formed
// from the given digits
vector<int> find3DigitNumbers(vector<int>&
                                  digits)
{
    // Generating frequency map
    // of the given digits
    vector<int> count(10, 0);
    for (auto& d : digits)
        count[d]++;

    vector<int> res;

    for (int num = 100; num < 999; num++) {

        // Generating frequency map
        // of the current number
        vector<int> currCount(10, 0);
        int temp = num;

        while (temp) {
            currCount[temp % 10]++;
            temp /= 10;
        }

        // Checking if the number
        // can be generated or not
        bool flag = true;

        for (int i = 0; i < 10; i++) {
            if (currCount[i] > count[i]) {
                flag = false;
                break;
            }
        }

        if (flag) {
            res.push_back(num);
        }
    }
    return res;
}

// Function to print answer
void printAnswer(vector<int>& v1)
{
    for (int i = 0; i < v1.size(); i++) {
        cout << v1[i] << " ";
    }
    cout << endl;
}

// Driver code
int main()
{
    vector<int> v1 = { 2, 1, 3, 0 };

    // Function Call
    vector<int> ans = find3DigitNumbers(v1);

    // Printing answer
    printAnswer(ans);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find all the unique
  // 3 digit number that can be formed
  // from the given digits
  static void find3DigitNumbers(int digits[], List<Integer> res)
  {
    // Generating frequency map
    // of the given digits
    int count[] = new int[10];;
    for (int i = 0; i < digits.length; i++)
      count[digits[i]]++;

    for (int num = 100; num < 999; num++) {

      // Generating frequency map
      // of the current number
      int currCount[] = new int[10];
      int temp = num;

      while (temp > 0) {
        currCount[temp % 10]++;
        temp /= 10;
      }

      // Checking if the number
      // can be generated or not
      Boolean flag = true;

      for (int i = 0; i < 10; i++) {
        if (currCount[i] > count[i]) {
          flag = false;
          break;
        }
      }

      if (flag == true) {
        res.add(num);
      }
    }
  }

  // Function to print answer
  static void printAnswer(List<Integer> res)
  {
    for(int i = 0;  i < res.size(); i++)
      System.out.print(res.get(i) + " ");
  }

  // Driver code
  public static void main (String[] args) 
  {

    int arr[] = { 2, 1, 3, 0 };

    List<Integer> ans=new ArrayList<Integer>(); 
    // Function Call
    find3DigitNumbers(arr, ans);

    // Printing answer
    printAnswer(ans);
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach

# Function to find all the unique
# 3 digit number that can be formed
# from the given digits
def find3DigitNumbers(digits):
  
    # Generating frequency map
    # of the given digits
    count = [0] * 10
    for d in digits:
        count[d] += 1

    res = []

    for num in range(100, 999):

        # Generating frequency map
        # of the current number
        currCount = [0] * 10
        temp = num

        while (temp):
            currCount[temp % 10] += 1
            temp = temp // 10

        # Checking if the number
        # can be generated or not
        flag = True

        for i in range(10):
            if (currCount[i] > count[i]):
                flag = False
                break

        if (flag):
            res.append(num)

    return res

# Function to print answer
def printAnswer(v1):
    for i in range(len(v1)):
        print(v1[i], end=" ")

    print('')

# Driver code
v1 = [2, 1, 3, 0]

# Function Call
ans = find3DigitNumbers(v1)

# Printing answer
printAnswer(ans)


# This code is contributed by Saurabh Jaiswal

// C# program for the above approach
using System;
using System.Collections;
class GFG {

  // Function to find all the unique
  // 3 digit number that can be formed
  // from the given digits
  static void find3DigitNumbers(int []digits, ArrayList res)
  {
    // Generating frequency map
    // of the given digits
    int []count = new int[10];;
    for (int i = 0; i < digits.Length; i++)
      count[digits[i]]++;

    for (int num = 100; num < 999; num++) {

      // Generating frequency map
      // of the current number
      int []currCount = new int[10];
      int temp = num;

      while (temp > 0) {
        currCount[temp % 10]++;
        temp /= 10;
      }

      // Checking if the number
      // can be generated or not
      bool flag = true;

      for (int i = 0; i < 10; i++) {
        if (currCount[i] > count[i]) {
          flag = false;
          break;
        }
      }

      if (flag == true) {
        res.Add(num);
      }
    }
  }

  // Function to print answer
  static void printAnswer(ArrayList res)
  {
    for(int i = 0;  i < res.Count; i++)
      Console.Write(res[i] + " ");
  }

  // Driver code
  public static void Main () 
  {

    int []arr = { 2, 1, 3, 0 };

    ArrayList ans=new ArrayList(); 
    // Function Call
    find3DigitNumbers(arr, ans);

    // Printing answer
    printAnswer(ans);
  }
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

 <script>
        // JavaScript code for the above approach 

        // Function to find all the unique
        // 3 digit number that can be formed
        // from the given digits
        function find3DigitNumbers(
            digits) {
            // Generating frequency map
            // of the given digits
            let count = new Array(10).fill(0);
            for (let d of digits)
                count[d]++;

            let res = [];

            for (let num = 100; num < 999; num++) {

                // Generating frequency map
                // of the current number
                let currCount = new Array(10).fill(0);
                let temp = num;

                while (temp) {
                    currCount[temp % 10]++;
                    temp = Math.floor(temp / 10);
                }

                // Checking if the number
                // can be generated or not
                let flag = true;

                for (let i = 0; i < 10; i++) {
                    if (currCount[i] > count[i]) {
                        flag = false;
                        break;
                    }
                }

                if (flag) {
                    res.push(num);
                }
            }
            return res;
        }

        // Function to print answer
        function printAnswer(v1) {
            for (let i = 0; i < v1.length; i++) {
                document.write(v1[i] + " ");
            }
            document.write('<br>')
        }

        // Driver code

        let v1 = [2, 1, 3, 0];

        // Function Call
        let ans = find3DigitNumbers(v1);

        // Printing answer
        printAnswer(ans);


       // This code is contributed by Potta Lokesh
    </script>

Output

102 103 120 123 130 132 201 203 210 213 230 231 301 302 310 312 320 321

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

Find all distinct three digit numbers from given array of digits

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Find all distinct three digit numbers from given array of digits

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