Given a range [L, R], the task is to find all the numbers from the range [L, R] whose digits are consecutive. Print all those numbers in increasing order.
Examples:
Input: L = 12, R = 34
Output: 12 23 34Input: L = 12, R = 25
Output: 12 23
Approach: The given problem can be solved by generating all possible numbers and store all those numbers that satisfy the given condition. After generating all the numbers print all the stored numbers in sorted order. Follow the steps below to solve the given problem:
- Initialize a variable, say num as “” that stores the string form of all possible numbers having consecutive digits and in increasing order.
- Iterate over the range [1, 9] using the variable i and perform the following steps:
- Update the string num to the string form of i and if this value lies over the range [L, R], then store this in the vector Ans[].
- Iterate over the range [1, 9] using the variable j add the character form of j to the string num, and if the integer form of the string num lies over the range [L, R], then store this in the vector Ans[].
- After completing the above steps, sort the vector Ans[] to print all the numbers generated.
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 the consecutive// digit numbers in the given rangevector<int> solve(int start, int end){ // Initialize num as empty string string num = ""; // Stores the resultant number vector<int> ans; // Iterate over the range [1, 9] for (int i = 1; i <= 9; i++) { num = to_string(i); int value = stoi(num); // Check if the current number // is within range if (value >= start and value <= end) { ans.push_back(value); } // Iterate on the digits starting // from i for (int j = i + 1; j <= 9; j++) { num += to_string(j); value = stoi(num); // Checking the consecutive // digits numbers starting // from i and ending at j // is within range or not if (value >= start and value <= end) { ans.push_back(value); } } } // Sort the numbers in the // increasing order sort(ans.begin(), ans.end()); return ans;}// Driver Codeint main(){ int L = 12, R = 87; vector<int> ans = solve(12, 87); // Print the required numbers for (auto& it : ans) cout << it << ' '; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the consecutive// digit numbers in the given rangestatic Vector<Integer> solve(int start, int end){ // Initialize num as empty String String num = ""; // Stores the resultant number Vector<Integer> ans = new Vector<>(); // Iterate over the range [1, 9] for(int i = 1; i <= 9; i++) { num = Integer.toString(i); int value = i; // Check if the current number // is within range if (value >= start && value <= end) { ans.add(value); } // Iterate on the digits starting // from i for(int j = i + 1; j <= 9; j++) { num += Integer.toString(j); value = Integer.valueOf(num); // Checking the consecutive // digits numbers starting // from i and ending at j // is within range or not if (value >= start && value <= end) { ans.add(value); } } } // Sort the numbers in the // increasing order Collections.sort(ans);; return ans;}// Driver Codepublic static void main(String[] args){ int L = 12, R = 87; Vector<Integer> ans = solve(L,R); // Print the required numbers for(int it : ans) System.out.print(it + " ");}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program for the above approach# Function to find the consecutive# digit numbers in the given rangedef solve(start, end): # Initialize num as empty string num = "" # Stores the resultant number ans = [] # Iterate over the range [1, 9] for i in range(1,10,1): num = str(i) value = int(num) # Check if the current number # is within range if (value >= start and value <= end): ans.append(value) # Iterate on the digits starting # from i for j in range(i + 1,10,1): num += str(j) value = int(num) # Checking the consecutive # digits numbers starting # from i and ending at j # is within range or not if (value >= start and value <= end): ans.append(value) # Sort the numbers in the # increasing order ans.sort() return ans# Driver Codeif __name__ == '__main__': L = 12 R = 87 ans = solve(12, 87) # Print the required numbers for it in ans: print(it,end = " ") # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class Program{ // Function to find the consecutive// digit numbers in the given rangestatic void solve(int start, int end){ // Initialize num as empty String string num = ""; // Stores the resultant number List<int> ans = new List<int>(); // Iterate over the range [1, 9] for(int i = 1; i <= 9; i++) { num = i.ToString(); int value = i; // Check if the current number // is within range if (value >= start && value <= end) { ans.Add(value); } // Iterate on the digits starting // from i for(int j = i + 1; j <= 9; j++) { num += j.ToString(); value = Int32.Parse(num); // Checking the consecutive // digits numbers starting // from i and ending at j // is within range or not if (value >= start && value <= end) { ans.Add(value); } } } // Sort the numbers in the // increasing order ans.Sort(); // Print the required numbers foreach(int it in ans) Console.Write(it + " ");} // Driver code static void Main() { int L = 12, R = 87; solve(L,R); }}// This code is contributed by SoumikMondal |
Javascript
<script>// javascript program for the above approach // Function to find the consecutive // digit numbers in the given range function solve(start , end) { // Initialize num as empty String var num = ""; // Stores the resultant number var ans = []; // Iterate over the range [1, 9] for (var i = 1; i <= 9; i++) { num = i.toString(); var value = i; // Check if the current number // is within range if (value >= start && value <= end) { ans.push(value); } // Iterate on the digits starting // from i for (j = i + 1; j <= 9; j++) { num += j.toString(); value = num; // Checking the consecutive // digits numbers starting // from i and ending at j // is within range or not if (value >= start && value <= end) { ans.push(value); } } } // Sort the numbers in the // increasing order return ans; } // Driver Code var L = 12, R = 87; var ans = solve(L, R); // Print the required numbers for (it of ans) document.write(it + " ");// This code is contributed by Rajput-Ji </script> |
Output:
12 23 34 45 56 67 78
Time Complexity: O(1)
Space Complexity: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
