Length of smallest substring of a given string which contains another string as subsequence | Set 2

04 Feb, 2022
Given two strings A and B, the task is to find the smallest substring of A having B as a subsequence.
Examples:
Input: A = “abcdefababaef”, B = “abf”Output: 5Explanation:Smallest substring of A having B as subsequence is abcdef.Therefore, the required length is 5.
Input: A = “abcdefababaef”, B = “aef”Output: 3
Approach: Follow the steps below to solve the problem:
Store all the indices of the characters of A which are also present in B in a Map CharacterIndex.
Traverse over all the characters of string B.
Check if the first character of string B is present in the string A or not:If found to be true, then initialize two variables firstVar and lastVar with the index of the first occurrence of B[0] in the string A.After updating the values, remove that character from the Map CharacterIndex.Otherwise, no further substring is possible.
If found to be true, then initialize two variables firstVar and lastVar with the index of the first occurrence of B[0] in the string A.After updating the values, remove that character from the Map CharacterIndex.Otherwise, no further substring is possible.
If found to be true, then initialize two variables firstVar and lastVar with the index of the first occurrence of B[0] in the string A.
After updating the values, remove that character from the Map CharacterIndex.
Otherwise, no further substring is possible.
For the remaining characters of B, check if the character is present in the string A or not. If found to be true, traverse through all the occurrences of that character in the string A and if the index of that character in the string A exceeds lastVar, then update the lastVar with that index. Otherwise, no further substring is possible.
If B is completely traversed, update answer with the difference between firstVar and the lastVar.
Print the final minimized answer.
Below is the implementation of the above approach:
C++
Java
Python3
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;  // Function to find the length of// smallest substring of a having// string b as a subsequenceint minLength(string a, string b){      // Stores the characters present    // in string b    map<char, int> Char;    for (int i = 0; i < b.length(); i++) {          Char[b[i]]++;    }      // Find index of characters of a    // that are also present in string b    map<char, vector<int> > CharacterIndex;      for (int i = 0; i < a.length(); i++) {          char x = a[i];          // If character is present in string b        if (Char.find(x) != Char.end()) {              // Store the index of character            CharacterIndex[x].push_back(i);        }    }      int len = INT_MAX;      // Flag is used to check if    // substring is possible    int flag;      while (true) {          // Assume that substring is        // possible        flag = 1;          // Stores first and last        // indices of the substring        // respectively        int firstVar, lastVar;          for (int i = 0; i < b.length(); i++) {              // For first character of string b            if (i == 0) {                  // If the first character of                // b is not present in a                if (CharacterIndex.find(b[i])                    == CharacterIndex.end()) {                      flag = 0;                    break;                }                  // If the first character of b                // is present in a                else {                      int x = *(                        CharacterIndex[b[i]].begin());                      // Remove the index from map                    CharacterIndex[b[i]].erase(                        CharacterIndex[b[i]].begin());                      // Update indices of                    // the substring                    firstVar = x;                    lastVar = x;                }            }              // For the remaining characters of b            else {                  int elementFound = 0;                for (auto e : CharacterIndex[b[i]]) {                      if (e > lastVar) {                          // If index possible for                        // current character                        elementFound = 1;                        lastVar = e;                        break;                    }                }                if (elementFound == 0) {                      // If no index is possible                    flag = 0;                    break;                }            }        }          if (flag == 0) {              // If no more substring            // is possible            break;        }          // Update the minimum length        // of substring        len = min(len,                  abs(lastVar - firstVar) + 1);    }      // Return the result    return len;}  // Driver Codeint main(){      // Given two string    string a = "abcdefababaef";    string b = "abf";      int len = minLength(a, b);    if (len != INT_MAX) {          cout << len << endl;    }    else {          cout << "Impossible" << endl;    }}
// Java program to implement// the above approachimport java.util.ArrayList;import java.util.HashMap;  class GFG{  // Function to find the length of// smallest substring of a having// string b as a subsequencestatic int minLength(String a, String b){          // Stores the characters present    // in string b    HashMap<Character, Integer> Char = new HashMap<>();      for(int i = 0; i < b.length(); i++)    {        Char.put(b.charAt(i),                 Char.getOrDefault(b.charAt(i), 0) + 1);    }      // Find index of characters of a    // that are also present in string b    HashMap<Character, ArrayList<Integer>>        CharacterIndex = new HashMap<>();      for(int i = 0; i < a.length(); i++)     {        char x = a.charAt(i);          // If character is present in string b        if (Char.containsKey(x))         {            if (CharacterIndex.get(x) == null)             {                CharacterIndex.put(                    x, new ArrayList<Integer>());            }                          // Store the index of character            CharacterIndex.get(x).add(i);        }    }      int len = Integer.MAX_VALUE;      // Flag is used to check if    // substring is possible    int flag;      while (true)    {          // Assume that substring is        // possible        flag = 1;          // Stores first and last        // indices of the substring        // respectively        int firstVar = 0, lastVar = 0;          for(int i = 0; i < b.length(); i++)         {                          // For first character of string b            if (i == 0)             {                  // If the first character of                // b is not present in a                if (CharacterIndex.containsKey(i))                {                    flag = 0;                    break;                }                  // If the first character of b                // is present in a                else                {                    int x = CharacterIndex.get(b.charAt(i)).get(0);                      // Remove the index from map                    CharacterIndex.get(b.charAt(i)).remove(                        CharacterIndex.get(b.charAt(i)).get(0));                      // Update indices of                    // the substring                    firstVar = x;                    lastVar = x;                }            }              // For the remaining characters of b            else             {                int elementFound = 0;                for(var e :                    CharacterIndex.get(b.charAt(i)))                 {                    if (e > lastVar)                    {                                                  // If index possible for                        // current character                        elementFound = 1;                        lastVar = e;                        break;                    }                }                if (elementFound == 0)                {                                          // If no index is possible                    flag = 0;                    break;                }            }        }          if (flag == 0)        {              // If no more substring            // is possible            break;        }          // Update the minimum length        // of substring        len = Math.min(            len, Math.abs(lastVar - firstVar) + 1);    }      // Return the result    return len;}  // Driver codepublic static void main(String[] args){          // Given two string    String a = "abcdefababaef";    String b = "abf";      int len = minLength(a, b);    if (len != Integer.MAX_VALUE)     {        System.out.println(len);    }    else     {        System.out.println("Impossible");    }}}  // This code is contributed by sk944795
# Python3 program to implement# the above approachimport sys  # Function to find the length of# smallest substring of a having# string b as a subsequencedef minLength(a, b):          # Stores the characters present    # in string b    Char = {}    for i in range(len(b)):        Char[b[i]] = Char.get(b[i], 0) + 1      # Find index of characters of a    # that are also present in string b    CharacterIndex = {}      for i in range(len(a)):        x = a[i]          # If character is present in string b        if (x in Char):                          # Store the index of character            CharacterIndex[x] = CharacterIndex.get(x, [])            CharacterIndex[x].append(i)      l = sys.maxsize      # Flag is used to check if    # substring is possible    while(True):                  # Assume that substring is        # possible        flag = 1          firstVar = 0        lastVar = 0                  # Stores first and last        # indices of the substring        # respectively        for i in range(len(b)):                          # For first character of string b            if (i == 0):                                  # If the first character of                # b is not present in a                if (b[i] not in CharacterIndex):                    flag = 0                    break                  # If the first character of b                # is present in a                else:                    x = CharacterIndex[b[i]][0]                      # Remove the index from map                    CharacterIndex[b[i]].remove(                    CharacterIndex[b[i]][0])                      # Update indices of                    # the substring                    firstVar = x                    lastVar = x              # For the remaining characters of b            else:                elementFound = 0                for e in CharacterIndex[b[i]]:                    if (e > lastVar):                                                  # If index possible for                        # current character                        elementFound = 1                        lastVar = e                        break                                      if (elementFound == 0):                                          # If no index is possible                    flag = 0                    break                          if (flag == 0):                          # If no more substring            # is possible            break          # Update the minimum length        # of substring        l = min(l, abs(lastVar - firstVar) + 1)      # Return the result    return l  # Driver Codeif __name__ == '__main__':          # Given two string    a = "abcdefababaef"    b = "abf"      l = minLength(a, b)    if (l != sys.maxsize):        print(l)    else:        print("Impossible")          # This code is contributed by SURENDRA_GANGWAR
5
 
Time Complexity: O(N2)Auxiliary Space: O(N)
 
SURENDRA_GANGWAR
sk944795
cpp-map
frequency-counting
sliding-window
subsequence
substring
Hash
Searching
Strings
sliding-window
Searching
Hash
Strings
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