Count of numbers in the range [L, R] which satisfy the given conditions

Given a range [L, R], the task is to find the count of numbers from this range that satisfy the below conditions:

  1. All the digit in the number are distinct.
  2. All the digits are less than or equal to 5.

Examples:

Input: L = 4, R = 13
Output: 5
4, 5, 10, 12 and 13 are the only
valid numbers in the range [4, 13].



Input: L = 100, R = 1000
Output: 100

Approach: The question seems simple if the range is small because in that case, all the numbers from the range can be iterated and checked whether they are valid or not. But since the range could be large, it can be observed all the digits of a valid number has to be distinct and from the range [0, 5] which suggests that the maximum number cannot exceed 543210.

Now instead of checking for every number, the next valid number in the series can be generated from the previously generated numbers. The idea is similar to the approach discussed here.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Maximum possible valid number
#define MAX 543210
  
// To store all the required number
// from the range [1, MAX]
vector<string> ans;
  
// Function that returns true if x
// satisfies the given conditions
bool isValidNum(string x)
{
  
    // To store the digits of x
    map<int, int> mp;
  
    for (int i = 0; i < x.length(); i++) {
  
        // If current digit appears more than once
        if (mp.find(x[i] - '0') != mp.end()) {
            return false;
        }
  
        // If current digit is greater than 5
        else if (x[i] - '0' > 5) {
            return false;
        }
  
        // Put the digit in the map
        else {
            mp[x[i] - '0'] = 1;
        }
    }
    return true;
}
  
// Function to generate all the required
// numbers in the range [1, MAX]
void generate()
{
  
    // Insert first 5 valid numbers
    queue<string> q;
    q.push("1");
    q.push("2");
    q.push("3");
    q.push("4");
    q.push("5");
  
    bool flag = true;
  
    // Inserting 0 externally because 0 cannot
    // be the leading digit in any number
    ans.push_back("0");
  
    while (!q.empty()) {
        string x = q.front();
        q.pop();
  
        // If x satisfies the given conditions
        if (isValidNum(x)) {
            ans.push_back(x);
        }
  
        // Cannot append anymore digit as
        // adding a digit will repeat one of
        // the already present digits
        if (x.length() == 6)
            continue;
  
        // Append all the valid digits one by
        // one and push the new generated
        // number to the queue
        for (int i = 0; i <= 5; i++) {
            string z = to_string(i);
  
            // Append the digit
            string temp = x + z;
  
            // Push the newly generated
            // number to the queue
            q.push(temp);
        }
    }
}
  
// Function to copmpare two strings
// which represent a numerical value
bool comp(string a, string b)
{
    if (a.size() == b.size())
        return a < b;
    else
        return a.size() < b.size();
}
  
// Function to return the count of
// valid numbers in the range [l, r]
int findcount(string l, string r)
{
  
    // Generate all the valid numbers
    // in the range [1, MAX]
    generate();
  
    // To store the count of numbers
    // in the range [l, r]
    int count = 0;
  
    // For every valid number in
    // the range [1, MAX]
    for (int i = 0; i < ans.size(); i++) {
  
        string a = ans[i];
  
        // If current number is within
        // the required range
        if (comp(l, a) && comp(a, r)) {
            count++;
        }
  
        // If number is equal to either l or r
        else if (a == l || a == r) {
            count++;
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
  
    string l = "1", r = "1000";
  
    cout << findcount(l, r);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Maximum possible valid number
static int MAX = 543210;
  
// To store all the required number
// from the range [1, MAX]
static Vector<String> ans = new Vector<String>();
  
// Function that returns true if x
// satisfies the given conditions
static boolean isValidNum(String x)
{
  
    // To store the digits of x
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();
  
    for (int i = 0; i < x.length(); i++)
    {
  
        // If current digit appears more than once
        if (mp.containsKey(x.charAt(i) - '0')) 
        {
            return false;
        }
  
        // If current digit is greater than 5
        else if (x.charAt(i) - '0' > 5)
        {
            return false;
        }
  
        // Put the digit in the map
        else
        {
            mp.put(x.charAt(i) - '0', 1);
        }
    }
    return true;
}
  
// Function to generate all the required
// numbers in the range [1, MAX]
static void generate()
{
  
    // Insert first 5 valid numbers
    Queue<String> q = new LinkedList<String>();
    q.add("1");
    q.add("2");
    q.add("3");
    q.add("4");
    q.add("5");
  
    boolean flag = true;
  
    // Inserting 0 externally because 0 cannot
    // be the leading digit in any number
    ans.add("0");
  
    while (!q.isEmpty())
    {
        String x = q.peek();
        q.remove();
  
        // If x satisfies the given conditions
        if (isValidNum(x)) 
        {
            ans.add(x);
        }
  
        // Cannot append anymore digit as
        // adding a digit will repeat one of
        // the already present digits
        if (x.length() == 6)
            continue;
  
        // Append all the valid digits one by
        // one and push the new generated
        // number to the queue
        for (int i = 0; i <= 5; i++) 
        {
            String z = String.valueOf(i);
  
            // Append the digit
            String temp = x + z;
  
            // Push the newly generated
            // number to the queue
            q.add(temp);
        }
    }
}
  
// Function to copmpare two Strings
// which represent a numerical value
static boolean comp(String a, String b)
{
    if (a.length()== b.length())
    {
        int i = a.compareTo(b);
      
        return i < 0 ? true : false;
    }
    else
        return a.length() < b.length();
}
  
// Function to return the count of
// valid numbers in the range [l, r]
static int findcount(String l, String r)
{
  
    // Generate all the valid numbers
    // in the range [1, MAX]
    generate();
  
    // To store the count of numbers
    // in the range [l, r]
    int count = 0;
  
    // For every valid number in
    // the range [1, MAX]
    for (int i = 0; i < ans.size(); i++) 
    {
  
        String a = ans.get(i);
  
        // If current number is within
        // the required range
        if (comp(l, a) && comp(a, r)) 
        {
            count++;
        }
  
        // If number is equal to either l or r
        else if (a == l || a == r) 
        {
            count++;
        }
    }
    return count;
}
  
// Driver code
public static void main (String[] args)
{
    String l = "1", r = "1000";
  
    System.out.println(findcount(l, r));
}
}
  
// This code is contributed by PrinciRaj1992

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
from collections import deque
  
# Maximum possible valid number
MAX = 543210
  
# To store all the required number
# from the range [1, MAX]
ans = []
  
# Function that returns true if x
# satisfies the given conditions
def isValidNum(x):
  
    # To store the digits of x
    mp = dict()
  
    for i in range(len(x)):
  
        # If current digit appears more than once
        if (ord(x[i]) - ord('0') in mp.keys()):
            return False
  
        # If current digit is greater than 5
        elif (ord(x[i]) - ord('0') > 5):
            return False
  
        # Put the digit in the map
        else:
            mp[ord(x[i]) - ord('0')] = 1
  
    return True
  
# Function to generate all the required
# numbers in the range [1, MAX]
def generate():
  
    # Insert first 5 valid numbers
    q = deque()
    q.append("1")
    q.append("2")
    q.append("3")
    q.append("4")
    q.append("5")
  
    flag = True
  
    # Inserting 0 externally because 0 cannot
    # be the leading digit in any number
    ans.append("0")
  
    while (len(q) > 0):
        x = q.popleft()
  
        # If x satisfies the given conditions
        if (isValidNum(x)):
            ans.append(x)
  
        # Cannot append anymore digit as
        # adding a digit will repeat one of
        # the already present digits
        if (len(x) == 6):
            continue
  
        # Append all the valid digits one by
        # one and append the new generated
        # number to the queue
        for i in range(6):
            z = str(i)
  
            # Append the digit
            temp = x + z
  
            # Push the newly generated
            # number to the queue
            q.append(temp)
  
# Function to copmpare two strings
# which represent a numerical value
def comp(a, b):
    if (len(a) == len(b)):
        if a < b:
            return True
    else:
        return len(a) < len(b)
  
# Function to return the count of
# valid numbers in the range [l, r]
def findcount(l, r):
  
    # Generate all the valid numbers
    # in the range [1, MAX]
    generate()
  
    # To store the count of numbers
    # in the range [l, r]
    count = 0
  
    # For every valid number in
    # the range [1, MAX]
    for i in range(len(ans)):
  
        a = ans[i]
  
        # If current number is within
        # the required range
        if (comp(l, a) and comp(a, r)):
            count += 1
  
        # If number is equal to either l or r
        elif (a == l or a == r):
            count += 1
  
    return count
  
# Driver code
l = "1"
r = "1000"
  
print(findcount(l, r))
  
# This code is contributed by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
using System.Collections.Generic;     
      
class GFG
{
  
// Maximum possible valid number
static int MAX = 543210;
  
// To store all the required number
// from the range [1, MAX]
static List<String> ans = new List<String>();
  
// Function that returns true if x
// satisfies the given conditions
static bool isValidNum(String x)
{
  
    // To store the digits of x
    Dictionary<int, int> mp = new Dictionary<int, int>();
  
    for (int i = 0; i < x.Length; i++)
    {
  
        // If current digit appears more than once
        if (mp.ContainsKey(x[i] - '0')) 
        {
            return false;
        }
  
        // If current digit is greater than 5
        else if (x[i] - '0' > 5)
        {
            return false;
        }
  
        // Put the digit in the map
        else
        {
            mp.Add(x[i] - '0', 1);
        }
    }
    return true;
}
  
// Function to generate all the required
// numbers in the range [1, MAX]
static void generate()
{
  
    // Insert first 5 valid numbers
    Queue<String> q = new Queue<String>();
    q.Enqueue("1");
    q.Enqueue("2");
    q.Enqueue("3");
    q.Enqueue("4");
    q.Enqueue("5");
  
    bool flag = true;
  
    // Inserting 0 externally because 0 cannot
    // be the leading digit in any number
    ans.Add("0");
  
    while (q.Count!=0)
    {
        String x = q.Peek();
        q.Dequeue();
  
        // If x satisfies the given conditions
        if (isValidNum(x)) 
        {
            ans.Add(x);
        }
  
        // Cannot append anymore digit as
        // adding a digit will repeat one of
        // the already present digits
        if (x.Length == 6)
            continue;
  
        // Append all the valid digits one by
        // one and push the new generated
        // number to the queue
        for (int i = 0; i <= 5; i++) 
        {
            String z = i.ToString();
  
            // Append the digit
            String temp = x + z;
  
            // Push the newly generated
            // number to the queue
            q.Enqueue(temp);
        }
    }
}
  
// Function to copmpare two Strings
// which represent a numerical value
static bool comp(String a, String b)
{
    if (a.Length == b.Length)
    {
        int i = a.CompareTo(b);
      
        return i < 0 ? true : false;
    }
    else
        return a.Length < b.Length;
}
  
// Function to return the count of
// valid numbers in the range [l, r]
static int findcount(String l, String r)
{
  
    // Generate all the valid numbers
    // in the range [1, MAX]
    generate();
  
    // To store the count of numbers
    // in the range [l, r]
    int count = 0;
  
    // For every valid number in
    // the range [1, MAX]
    for (int i = 0; i < ans.Count; i++) 
    {
  
        String a = ans[i];
  
        // If current number is within
        // the required range
        if (comp(l, a) && comp(a, r)) 
        {
            count++;
        }
  
        // If number is equal to either l or r
        else if (a == l || a == r) 
        {
            count++;
        }
    }
    return count;
}
  
// Driver code
public static void Main (String[] args)
{
    String l = "1", r = "1000";
  
    Console.WriteLine(findcount(l, r));
}
}
  
// This code is contributed by Princi Singh

chevron_right


Output:

130


My Personal Notes arrow_drop_up

Competitive Programmer, Full Stack Developer, Technical Content Writer, Machine Learner

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.