Minimum number of substrings the given string can be splitted into that satisfy the given conditions
Given a string str and a string array arr[], the task is to find the minimum count of substrings this can be split into such that every substring is present in the given string array arr[].
Examples:
Input: str = “111112”, arr[] = {“11”, “111”, “11111”, “2”}
Output: 2
“11111” and “2” are the required substrings.
“11”, “111” and “2” can also be a valid answer
but it is not the minimum.Input: str = “123456”, arr[] = {“1”, “12345”, “2345”, “56”, “23”, “456”}
Output: 3
Approach: Iterate the string from index 0 and build the prefix string, if the prefix string exists in the given array (a set can be used to check this) then call the function again from the next position in the string and keep track of the overall minimum count of substrings.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Set to store all the strings // from the given array unordered_set<string> uSet; // To store the required count int minCnt = INT_MAX; // Recursive function to find the count of // substrings that can be splitted starting // from the index start such that all // the substrings are present in the map void findSubStr(string str, int cnt, int start) { // All the chosen substrings // are present in the map if (start == str.length()) { // Update the minimum count // of substrings minCnt = min(cnt, minCnt); } // Starting from the substrings of length 1 // that start with the given index for ( int len = 1; len <= (str.length() - start); len++) { // Get the substring string subStr = str.substr(start, len); // If the substring is present in the set if (uSet.find(subStr) != uSet.end()) { // Recursive call for the // rest of the string findSubStr(str, cnt + 1, start + len); } } } // Function that inserts all the strings // from the given array in a set and calls // the recursive function to find the // minimum count of substrings str can be // splitted into that satisfy the given condition void findMinSubStr(string arr[], int n, string str) { // Insert all the strings from // the given array in a set for ( int i = 0; i < n; i++) uSet.insert(arr[i]); // Find the required count findSubStr(str, 0, 0); } // Driver code int main() { string str = "123456" ; string arr[] = { "1" , "12345" , "2345" , "56" , "23" , "456" }; int n = sizeof (arr) / sizeof (string); findMinSubStr(arr, n, str); cout << minCnt; return 0; } |
Java
//Java implementation of above approach import java.util.*; class GFG { // Set to store all the Strings // from the given array static Set<String> uSet = new HashSet<String>(); // To store the required count static int minCnt = Integer.MAX_VALUE; // Recursive function to find the count of // subStrings that can be splitted starting // from the index start such that all // the subStrings are present in the map static void findSubStr(String str, int cnt, int start) { // All the chosen subStrings // are present in the map if (start == str.length()) { // Update the minimum count // of subStrings minCnt = Math.min(cnt, minCnt); } // Starting from the subStrings of length 1 // that start with the given index for ( int len = 1 ; len <= (str.length() - start); len++) { // Get the subString String subStr = str.substring(start, start + len); // If the subString is present in the set if (uSet.contains(subStr)) { // Recursive call for the // rest of the String findSubStr(str, cnt + 1 , start + len); } } } // Function that inserts all the Strings // from the given array in a set and calls // the recursive function to find the // minimum count of subStrings str can be // splitted into that satisfy the given condition static void findMinSubStr(String arr[], int n, String str) { // Insert all the Strings from // the given array in a set for ( int i = 0 ; i < n; i++) uSet.add(arr[i]); // Find the required count findSubStr(str, 0 , 0 ); } // Driver code public static void main(String args[]) { String str = "123456" ; String arr[] = { "1" , "12345" , "2345" , "56" , "23" , "456" }; int n = arr.length; findMinSubStr(arr, n, str); System.out.print(minCnt); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach import sys # Set to store all the strings # from the given array uSet = set (); # To store the required count minCnt = sys.maxsize; # Recursive function to find the count of # substrings that can be splitted starting # from the index start such that all # the substrings are present in the map def findSubStr(string, cnt, start) : global minCnt; # All the chosen substrings # are present in the map if (start = = len (string)) : # Update the minimum count # of substrings minCnt = min (cnt, minCnt); # Starting from the substrings of length 1 # that start with the given index for length in range ( 1 , len (string) - start + 1 ) : # Get the substring subStr = string[start : start + length]; # If the substring is present in the set if subStr in uSet : # Recursive call for the # rest of the string findSubStr(string, cnt + 1 , start + length); # Function that inserts all the strings # from the given array in a set and calls # the recursive function to find the # minimum count of substrings str can be # splitted into that satisfy the given condition def findMinSubStr(arr, n, string) : # Insert all the strings from # the given array in a set for i in range (n) : uSet.add(arr[i]); # Find the required count findSubStr(string, 0 , 0 ); # Driver code if __name__ = = "__main__" : string = "123456" ; arr = [ "1" , "12345" , "2345" , "56" , "23" , "456" ]; n = len (arr); findMinSubStr(arr, n, string); print (minCnt); # This code is contributed by AnkitRai01 |
C#
// C# implementation of above approach using System; using System.Collections.Generic; class GFG { // Set to store all the Strings // from the given array static HashSet<String> uSet = new HashSet<String>(); // To store the required count static int minCnt = int .MaxValue; // Recursive function to find the count of // subStrings that can be splitted starting // from the index start such that all // the subStrings are present in the map static void findSubStr(String str, int cnt, int start) { // All the chosen subStrings // are present in the map if (start == str.Length) { // Update the minimum count // of subStrings minCnt = Math.Min(cnt, minCnt); } // Starting from the subStrings of length 1 // that start with the given index for ( int len = 1; len <= (str.Length - start); len++) { // Get the subString String subStr = str.Substring(start, len); // If the subString is present in the set if (uSet.Contains(subStr)) { // Recursive call for the // rest of the String findSubStr(str, cnt + 1, start + len); } } } // Function that inserts all the Strings // from the given array in a set and calls // the recursive function to find the // minimum count of subStrings str can be // splitted into that satisfy the given condition static void findMinSubStr(String []arr, int n, String str) { // Insert all the Strings from // the given array in a set for ( int i = 0; i < n; i++) uSet.Add(arr[i]); // Find the required count findSubStr(str, 0, 0); } // Driver code public static void Main(String []args) { String str = "123456" ; String []arr = { "1" , "12345" , "2345" , "56" , "23" , "456" }; int n = arr.Length; findMinSubStr(arr, n, str); Console.WriteLine(minCnt); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach // Set to store all the strings // from the given array var uSet = new Set(); // To store the required count var minCnt = 1000000000; // Recursive function to find the count of // substrings that can be splitted starting // from the index start such that all // the substrings are present in the map function findSubStr( str, cnt, start) { // All the chosen substrings // are present in the map if (start == str.length) { // Update the minimum count // of substrings minCnt = Math.min(cnt, minCnt); } // Starting from the substrings of length 1 // that start with the given index for ( var len = 1; len <= (str.length - start); len++) { // Get the substring var subStr = str.substring(start, start+len); // If the substring is present in the set if (uSet.has(subStr)) { // Recursive call for the // rest of the string findSubStr(str, cnt + 1, start + len); } } } // Function that inserts all the strings // from the given array in a set and calls // the recursive function to find the // minimum count of substrings str can be // splitted into that satisfy the given condition function findMinSubStr(arr, n, str) { // Insert all the strings from // the given array in a set for ( var i = 0; i < n; i++) uSet.add(arr[i]); // Find the required count findSubStr(str, 0, 0); } // Driver code var str = "123456" ; var arr = [ "1" , "12345" , "2345" , "56" , "23" , "456" ]; var n = arr.length; findMinSubStr(arr, n, str); document.write( minCnt); </script> |
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