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# Minimum number of substrings the given string can be splitted into that satisfy the given conditions

• Difficulty Level : Medium
• Last Updated : 27 May, 2021

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:
“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:

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 ``using` `namespace` `std;` `// Set to store all the strings``// from the given array``unordered_set 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 uSet = ``new` `HashSet();` `// 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 uSet = ``new` `HashSet();` `// 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

 ``

Output:

`3`

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