Given two integers N and K, the task is to find the string S of minimum length to contain all possible strings of size N as a substring. The characters of the string should be from integers ranging from 0 to K-1.

Examples:

Input: N = 2, K = 2
Output: 00110
Explanation: Allowed characters are from 0 to k-1 (i.e., 0 and 1). There are 4 string possible of size N=2 (i.e “00”, “01”,”10″,”11″) “00110” contains all possible string as a substring. It also has the minimum length.

Input: N = 2, K = 3
Output: 0010211220
Explanation: Allowed characters are from 0 to k-1 (i.e., 0, 1 and 2). There are total 9 strings possible of size N, given output string has the minimum length that contains all those strings as substring.

Approach: To solve the problem follow the below idea:

In this we find the smallest string of length N using digits from 0 to K-1, such that it contains all possible strings of size N as substrings. The solution utilizes a approach to iteratively build the string while ensuring that all permutations of size N are present as substrings.

• Initialize an empty string ans with N zeros.
• Decrement K by 1 to represent the range of digits from 0 to K-1.
• Create an unordered set visited to keep track of visited nodes and insert the initial string into it.
• Run a loop until all permutations are found, i.e., the size of the visited set equals (K+1)^N.
  • Extract the last N-1 digits from the current string ans and store them in the previous string.
  • Iterate from K to 0.
    • Create a new string currStr equal to the previous and push i into it.
    • If currStr is not in the visited set, insert it into the set, append the current digit to the current string ans, and break from the loop.
• Return the ans string.

Implementation of the above approach:

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

// Function to find the lexicographically smallest
// string of length N using K digits
string findString(int n, int k)
{
    // Initialize the result string with '0'
    string ans = "0";

    // Decrement k for 0-based indexing
    k -= 1;

    // Set to store visited strings
    unordered_set<string> visited;
    visited.insert("0");

    // Loop until all possible strings are generated
    while (visited.size() < pow((k + 1), n)) {
        // Get the substring for the previous n-1
        // characters
        string previous = ans.substr(ans.length() - n + 1);

        // Iterate from k to 0 to find the
        // lexicographically smallest character
        for (int i = k; i >= 0; i--) {
            string currStr = previous + to_string(i);

            // If the current string is not visited,
            // insert it and append to the result
            if (visited.find(currStr) == visited.end()) {
                visited.insert(currStr);
                ans += to_string(i);
                break;
            }
        }
    }

    return ans;
}

// Driver code
int main()
{
    // Set values for N and K
    int n = 2;
    int k = 2;

    // Call the findString method to obtain the result
    string result = findString(n, k);

    // Print the result
    cout << "Lexicographically smallest string: " << result
         << endl;

    return 0;
}

// This code is contributed by Susobhan Akhuli

import java.util.HashSet;
import java.util.Set;

public class GFG {

    // Function to find the lexicographically smallest
    // string of length N using K digits
    public static String findString(int n, int k) {
        // Initialize the result string with '0'
        StringBuilder ans = new StringBuilder("0");

        // Decrement k for 0-based indexing
        k -= 1;

        // Set to store visited strings
        Set<String> visited = new HashSet<>();
        visited.add("0");

        // Loop until all possible strings are generated
        while (visited.size() < Math.pow((k + 1), n)) {
            // Get the substring for the previous n-1 characters
            String previous = ans.substring(ans.length() - n + 1);

            // Iterate from k to 0 to find the lexicographically
            // smallest character
            for (int i = k; i >= 0; i--) {
                String currStr = previous + i;

                // If the current string is not visited,
                // insert it and append to the result
                if (!visited.contains(currStr)) {
                    visited.add(currStr);
                    ans.append(i);
                    break;
                }
            }
        }

        return ans.toString();
    }

    // Driver code
    public static void main(String[] args) {
        // Set values for N and K
        int n = 2;
        int k = 2;

        // Call the findString method to obtain the result
        String result = findString(n, k);

        // Print the result
        System.out.println("Lexicographically smallest string: " + result);
    }
}

using System;
using System.Collections.Generic;

class Program
{
    // Function to find the lexicographically smallest
    // string of length N using K digits
    static string FindString(int n, int k)
    {
        // Initialize the result string with '0'
        string ans = "0";

        // Decrement k for 0-based indexing
        k -= 1;

        // Set to store visited strings
        HashSet<string> visited = new HashSet<string>();
        visited.Add("0");

        // Loop until all possible strings are generated
        while (visited.Count < Math.Pow((k + 1), n))
        {
            // Get the substring for the previous n-1
            // characters
            string previous = ans.Substring(ans.Length - n + 1);

            // Iterate from k to 0 to find the
            // lexicographically smallest character
            for (int i = k; i >= 0; i--)
            {
                string currStr = previous + i.ToString();

                // If the current string is not visited,
                // insert it and append to the result
                if (!visited.Contains(currStr))
                {
                    visited.Add(currStr);
                    ans += i.ToString();
                    break;
                }
            }
        }

        return ans;
    }

    // Driver code
    static void Main()
    {
        // Set values for N and K
        int n = 2;
        int k = 2;

        // Call the FindString method to obtain the result
        string result = FindString(n, k);

        // Print the result
        Console.WriteLine("Lexicographically smallest string: " + result);
    }
}

function findString(n, k) {
    // Initialize the result string with '0'
    let ans = "0";

    // Decrement k for 0-based indexing
    k -= 1;

    // Set to store visited strings
    let visited = new Set();
    visited.add("0");

    // Loop until all possible strings are generated
    while (visited.size < Math.pow(k + 1, n)) {
        // Get the substring for the previous n-1 characters
        let previous = ans.substring(ans.length - n + 1);

        // Iterate from k to 0 to find the lexicographically
        // smallest character
        for (let i = k; i >= 0; i--) {
            let currStr = previous + i;

            // If the current string is not visited,
            // insert it and append to the result
            if (!visited.has(currStr)) {
                visited.add(currStr);
                ans += i;
                break;
            }
        }
    }

    return ans;
}

// Set values for N and K
let n = 2;
let k = 2;

// Call the findString method to obtain the result
let result = findString(n, k);

// Print the result
console.log("Lexicographically smallest string: " + result);

def findString(n, k):
    ans = "0"
    k -= 1
    visited = set()
    visited.add("0")

    while len(visited) < pow((k + 1), n):
        previous = ans[-n + 1:]
        for i in range(k, -1, -1):
            currStr = previous + str(i)
            if currStr not in visited:
                visited.add(currStr)
                ans += str(i)
                break

    return ans

# Driver code
n = 2
k = 2
result = findString(n, k)
print("Lexicographically smallest string:", result)

Lexicographically smallest string: 0110

Time Complexity: O(K^N N), As we are running the loop of size O(K^N N).
Auxiliary Space: O(K^N N), As we have the set to store the all the possible permutations.