Given two integers N and K, the task is to find the Kth lexicographical ordered string of length N, which has distinct products of each of the substrings, return -1 if such a string does not exist.
Example:
Input: N = 3, K = 4
Output: 238
Explanation: The valid strings series is “234”, “235”, “237”, “238”. The 4th string in the series is “238” and the product of all its substrings {2, 3, 8, 6, 24, 48} is distinct.Input: N = 1, K = 11
Output: -1
Explanation: There are only 10 valid strings of length 1: “0”, “1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”.
Approach: The problem can be solved using recursion. Recursively check all possible numeric strings of length N and find Kth valid numeric string with distinct products of each substring.
Follow the steps to solve the problem:
- Notice when any two characters repeat in the string, it will not be a valid string as two substrings of length 1 will same product.
- All strings with length N > 10 will have no answer because at least one character will be repeated.
- For strings of length > 1, strings that will have ‘1’ or ‘0’ in them will not be valid as any number multiplied with these characters will be the same or converted to 0.
- Maintain a recursive function to calculate the total possible strings with the character ‘0’ -> ‘1’.
- Keep track of all the products till now and if the product is repeated then exit the function immediately.
- Start from 0->9 one by one and if a valid string is obtained then decrease K. For, any string S if K becomes 0, this string will be the final 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 the required stringvoid getString(int curlen, string& ans, string& s, int N, int& K, vector<int>& prod){ // If the current length is // equal to n exit from here only if (curlen == N) { K--; if (K == 0) ans = s; return; } char ch; int ok, t, i; // Iterate for all the characters for (ch = '2'; ch <= '9'; ch++) { s += ch; ok = 1; t = 1; for (i = curlen; i >= 0; i--) { t *= s[i] - 48; // Check if the product // is present before if (prod[t]) ok = 0; prod[t]++; } // If the current string is good // then recurse for // the next character if (ok) getString(curlen + 1, ans, s, N, K, prod); t = 1; // Decrease all the products back // to their original state for (i = curlen; i >= 0; i--) { t *= s[i] - 48; prod[t]--; } // Erase the last character s.erase(s.length() - 1); }}// Function to calculate// kth ordered valid stringstring kthValidString(int N, int K){ // Check for the base cases if (N > 10) { return "-1"; } if (N == 1) { // There are atmost 10 // valid strings for n = 1 if (K > 10) { return "-1"; } string s = ""; K--; s += (K + '0'); return s; } string ans = "-1"; string s = ""; // Vector to keep a check on // number of occurrences of products vector<int> prod(10005, 0); // Recursively construct the strings getString(0, ans, s, N, K, prod); return ans;}// Driver Codeint main(){ int N = 3, K = 4; cout << kthValidString(N, K);} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static String ans,s; static int K;// Function to find the required Stringstatic void getString(int curlen, int N, int[] prod){ // If the current length is // equal to n exit from here only if (curlen == N) { K--; if (K == 0) ans = s; return; } char ch; int ok, t, i; // Iterate for all the characters for (ch = '2'; ch <= '9'; ch++) { s += ch; ok = 1; t = 1; for (i = curlen ; i >= 0 && s.length()>i; i--) { t *= s.charAt(i) - 48; // Check if the product // is present before if (prod[t]!=0) ok = 0; prod[t]++; } // If the current String is good // then recurse for // the next character if (ok!=0) getString(curlen + 1, N, prod); t = 1; // Decrease all the products back // to their original state for (i = curlen; i >= 0&& s.length()>i; i--) { t *= s.charAt(i) - 48; prod[t]--; } // Erase the last character if(s.length()>0) s=s.substring(0,s.length() - 1); }}// Function to calculate// kth ordered valid Stringstatic String kthValidString(int N){ // Check for the base cases if (N > 10) { return "-1"; } if (N == 1) { // There are atmost 10 // valid Strings for n = 1 if (K > 10) { return "-1"; } String s = ""; K--; s += (K + '0'); return s; } ans = "-1"; s = ""; // Vector to keep a check on // number of occurrences of products int []prod = new int[10005]; // Recursively construct the Strings getString(0, N, prod); return ans;}// Driver Codepublic static void main(String[] args){ int N = 3; K = 4; System.out.print(kthValidString(N));}}// This code contributed by shikhasingrajput |
C#
// C# program for the above approachusing System;class GFG{ static String ans,s; static int K;// Function to find the required Stringstatic void getString(int curlen, int N, int[] prod){ // If the current length is // equal to n exit from here only if (curlen == N) { K--; if (K == 0) ans = s; return; } char ch; int ok, t, i; // Iterate for all the characters for(ch = '2'; ch <= '9'; ch++) { s += ch; ok = 1; t = 1; for(i = curlen ; i >= 0 && s.Length>i; i--) { t *= s[i] - 48; // Check if the product // is present before if (prod[t] != 0) ok = 0; prod[t]++; } // If the current String is good // then recurse for // the next character if (ok != 0) getString(curlen + 1, N, prod); t = 1; // Decrease all the products back // to their original state for(i = curlen; i >= 0 && s.Length>i; i--) { t *= s[i] - 48; prod[t]--; } // Erase the last character if (s.Length > 0) s = s.Substring(0,s.Length - 1); }}// Function to calculate// kth ordered valid Stringstatic String kthValidString(int N){ String s = ""; // Check for the base cases if (N > 10) { return "-1"; } if (N == 1) { // There are atmost 10 // valid Strings for n = 1 if (K > 10) { return "-1"; } s = ""; K--; s += (K + '0'); return s; } ans = "-1"; s = ""; // List to keep a check on // number of occurrences of products int []prod = new int[10005]; // Recursively construct the Strings getString(0, N, prod); return ans;}// Driver Codepublic static void Main(String[] args){ int N = 3; K = 4; Console.Write(kthValidString(N));}}// This code is contributed by 29AjayKumar |
Python3
# Java program for the above approachs = ''K = 0# Function to find the required Stringdef getString(curlen, N, prod): global s, K # If the current length is # equal to n exit from here only if (curlen == N): K -= 1 if (K == 0): ans = s return # Iterate for all the characters ch = '2' while(ch <= '9'): s = chr(ord(s)+ ord(ch)) ok = 1 t = 1 i = curlen ch = chr(ord(ch) + 1) while(i >= 0 and len(s)) : t *= ord(s[i]) - 48 # Check if the product # is present before if (prod[t] != 0): ok = 0 prod[t] += 1 i -= 1 # If the current String is good # then recurse for # the next character if (ok != 0): getString(curlen + 1, N, prod) t = 1 # Decrease all the products back # to their original state i = curlen while(i >= 0 and len(s)>i): i-=1 t *= ord(s[i]) - 48 prod[t] -= 1 # Erase the last character if(len(s)>0): s = s[0: len(s) - 1] # Function to calculate# kth ordered valid Stringdef kthValidString(N): # Check for the base cases if (N > 10): return "-1" if (N == 1): # There are atmost 10 # valid Strings for n = 1 if (K > 10): return "-1" s = "" K-=1 s += (K + '0') return s ans = "-1" s = "" # Vector to keep a check on # number of occurrences of products prod = [0]*10005 # Recursively construct the Strings getString(0, N, prod) return ans# Driver CodeN = 3K = 4print(kthValidString(N))# This code is contributed by shikhasingrajput |
Javascript
<script>// JavaScript program for the above approach var ans,s; var K; // Function to find the required Stringfunction getString(curlen, N, prod){ // If the current length is // equal to n exit from here only if (curlen == N) { K--; if (K == 0) ans = s; return; } var ch; var ok, t, i; // Iterate for all the characters for (ch = '2'; ch <= '9'; ch++) { s += ch; ok = 1; t = 1; for (i = curlen ; i >= 0 && s.length > i; i--) { t *= s.charAt(i) - 48; // Check if the product // is present before if (prod[t] != 0) ok = 0; prod[t]++; } // If the current String is good // then recurse for // the next character if (ok!=0) getString(curlen + 1, N, prod); t = 1; // Decrease all the products back // to their original state for (i = curlen; i >= 0&& s.length > i; i--) { t *= s.charAt(i) - 48; prod[t]--; } // Erase the last character if(s.length>0) s = s.substring(0,s.length - 1); }}// Function to calculate// kth ordered valid Stringfunction kthValidString(N){ // Check for the base cases if (N > 10) { return "-1"; } if (N == 1) { // There are atmost 10 // valid Strings for n = 1 if (K > 10) { return "-1"; } var s = ""; K--; s += (K + '0'); return s; } var ans = "1"; var s = ""; // Vector to keep a check on // number of occurrences of products var prod = new Array(10005); // Recursively construct the Strings getString(0, N, prod); return ans;}// Driver Code var N = 3; var K = 4; document.write(kthValidString(N));// This code is contributed by shivanisinghss2110</script> |
Output
238
Time Complexity: 
Auxiliary Space:
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!