# Length of smallest substring of a given string which contains another string as subsequence

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: 5
Explanation:
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:
1. If found to be true, then initialize two variables firstVar and lastVar with the index of the first occurrence of B in the string A.
2. After updating the values, remove that character from the Map CharacterIndex.
3. 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++

 `// C++ program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the length of` `// smallest substring of a having` `// string b as a subsequence` `int` `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 Code` `int` `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;` `    ``}` `}`

## Python3

 `# Python3 program to implement` `# the above approach` `import` `sys`   `# Function to find the length of` `# smallest substring of a having` `# string b as a subsequence` `def` `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 Code` `if` `__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`

Output:

```5

```

Time Complexity: O(N2)
Auxiliary Space: O(N)

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