Given two strings A and B, the task is to find the smallest substring of A having B as a subsequence. If there are multiple such substrings in A, return the substring that has the smallest starting index.
Examples :
Input : A = “abcedbaced” B = “bed”
Output : “bced”
Explanation : The substring A[2 : 5] is the shortest substring that contains the string ‘B’ as a subsequence.Input : A = “abcdbad” B = “bd”
Output : “bcd”
Explanation : Although both the substrings A[2:4] and A[5:7] have the same length, the substring which has the smallest starting index is “bcd” so it is the final answer.
Naive Approach: The simplest approach to solve the given problem is to check for string B occurring as a subsequence in every substring of A.
Among such substrings, the answer will be the one shortest in length and which has the smallest index.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Dynamic Programming because the above problem has Overlapping subproblems and an Optimal substructure. The subproblems can be stored in dp[][] table memoization where dp[i][j] denotes that in the substring A(0…i) there exists a subsequence corresponding to B(0…j) which begins at index dp[i][j]. Follow the steps below to solve the problem:
- Initialize a global multidimensional array dp[100][100] with all values as -1 that stores the result of each dp state.
- Each column in the dp table represents a character in string B.
- Initialize the first column of the dp array, i.e., store the nearest occurrence of the first character of string B in every prefix of string A.
- Fill the remaining columns of the dp table. For a particular character in string B at position ‘j’, check if there exists a matching character in string A.
- If A[i] == B[j], then starting index of the required substring in the string A is equal to the starting index when the first j – 1 characters of string B are matched with the first i – 1 characters of string A.
- If A[i] != B[j], then starting index of the required substring in the string A is equal to the starting index when the first j characters of string B are matched with the first i – 1 characters of string A.
- Check if any of the values in the last column of dp table is greater than or equal to 0. For a particular index i in string A, if dp[i][b – 1] >= 0, then the required substring which has B as a subsequence is A[dp[i][b-1] : i]. Update the answer with the shortest possible substring.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Stores the dp-statesint dp[100][100];// Function to find the smallest substring in string A which// contains string B as a subsequence.string smallestSubstring(string& A, string& B){ // Size of string A int a = A.size(); // Size of string B int b = B.size(); // Initializing the first column of dp array. // Storing the occurrence of first character of string B // in first (i + 1) characters of string A. for (int i = 0; i < a; ++i) { // If the current character of string A does not // match the first character of string B. if (i > 0 and A[i] != B[0]) { dp[i][0] = dp[i - 1][0]; } // If the current character of string A is equal to // the first character of string B. if (A[i] == B[0]) { dp[i][0] = i; } } // Iterating through remaining characters of string B. for (int j = 1; j < b; ++j) { // Checking if any character in string A matches // with the current character of string B. for (int i = 1; i < a; ++i) { // If there is a match, then starting index of // required substring in string 'A' is equal to // the starting index when first 'j - 1' // characters of string 'B' matched with first // 'i - 1' characters of string 'A'. if (A[i] == B[j]) { dp[i][j] = dp[i - 1][j - 1]; } // Else, starting index of required substring in // string 'A' is equal to the starting index // when first 'j' characters of string 'B' // matched with first 'i - 1' characters of // string 'A'. else { dp[i][j] = dp[i - 1][j]; } } } // String for storing final answer string answer = ""; // Length of smallest substring int best_length = 1e9; for (int i = 0; i < a; ++i) { // dp[i][b-1] is the index in string 'A', such that // the substring A(dp[i][b-1] : i) contains string // 'B' as a subsequence. if (dp[i][b - 1] != -1) { // Starting index of substring int start = dp[i][b - 1]; // Ending index of substring int end = i; // Length of substring int current_length = end - start + 1; // if current length is lesser than the best // length update the answer. if (current_length < best_length) { best_length = current_length; // Update the answer answer = A.substr(start, best_length); } } } // Return the smallest substring return answer;}// This function is initializing dp with -1// and printing the resultvoid smallestSubstringUtil(string& A, string& B){ // Initialize dp array with -1 memset(dp, -1, sizeof dp); // Function call cout << smallestSubstring(A, B) << endl;}// Driver codeint main(){ // Input strings string A = "abcedbaced"; string B = "bed"; // Function Call smallestSubstringUtil(A, B); return 0;} |
Java
// Java implementation of above approachimport java.io.*;import java.util.*;class GFG { // Stores the dp-states static int[][] dp = new int[100][100]; // Function to find the smallest subString in String A which // contains String B as a subsequence. static String smallestSubstring(String A, String B) { // Size of String A int a = A.length(); // Size of String B int b = B.length(); // Initializing the first column of dp array. // Storing the occurrence of first character of String B // in first (i + 1) characters of String A. for (int i = 0; i < a; ++i) { // If the current character of String A does not // match the first character of String B. if (i > 0 && A.charAt(i) != B.charAt(0)) { dp[i][0] = dp[i - 1][0]; } // If the current character of String A is equal to // the first character of String B. if (A.charAt(i) == B.charAt(0)) { dp[i][0] = i; } } // Iterating through remaining characters of String B. for (int j = 1; j < b; ++j) { // Checking if any character in String A matches // with the current character of String B. for (int i = 1; i < a; ++i) { // If there is a match, then starting index of // required subString in String 'A' is equal to // the starting index when first 'j - 1' // characters of String 'B' matched with first // 'i - 1' characters of String 'A'. if (A.charAt(i) == B.charAt(j)) { dp[i][j] = dp[i - 1][j - 1]; } // Else, starting index of required subString in // String 'A' is equal to the starting index // when first 'j' characters of String 'B' // matched with first 'i - 1' characters of // String 'A'. else { dp[i][j] = dp[i - 1][j]; } } } // String for storing final answer String answer = ""; // Length of smallest substring int best_length = 100000000; for (int i = 0; i < a; ++i) { // dp[i][b-1] is the index in String 'A', such that // the subString A(dp[i][b-1] : i) contains string // 'B' as a subsequence. if (dp[i][b - 1] != -1) { // Starting index of substring int start = dp[i][b - 1]; // Ending index of substring int end = i; // Length of substring int current_length = end - start + 1; // if current length is lesser than the best // length update the answer. if (current_length < best_length) { best_length = current_length; // Update the answer answer = A.substring(start, best_length+1); } } } // Return the smallest substring return answer; } // This function is initializing dp with -1 // and printing the result static void smallestSubstringUtil(String A, String B) { // Initialize dp array with -1 for(int i=0;i<100;i++) { for(int j=0;j<100;j++) { dp[i][j] = -1; } } // Function call System.out.print( smallestSubstring(A, B) ); } // Driver Code public static void main(String[] args) { // Input strings String A = "abcedbaced"; String B = "bed"; // Function Call smallestSubstringUtil(A, B); }}// This code is contributed by sanjoy_62. |
Python3
# python3 program for the above approach# Stores the dp-statesdp = [[-1 for _ in range(100)] for _ in range(100)]# Function to find the smallest substring in string A which# contains string B as a subsequence.def smallestSubstring(A, B): # Size of string A a = len(A) # Size of string B b = len(B) # Initializing the first column of dp array. # Storing the occurrence of first character of string B # in first (i + 1) characters of string A. for i in range(0, a): # If the current character of string A does not # match the first character of string B. if (i > 0 and A[i] != B[0]): dp[i][0] = dp[i - 1][0] # If the current character of string A is equal to # the first character of string B. if (A[i] == B[0]): dp[i][0] = i # Iterating through remaining characters of string B. for j in range(1, b): # Checking if any character in string A matches # with the current character of string B. for i in range(1, a): # If there is a match, then starting index of # required substring in string 'A' is equal to # the starting index when first 'j - 1' # characters of string 'B' matched with first # 'i - 1' characters of string 'A'. if (A[i] == B[j]): dp[i][j] = dp[i - 1][j - 1] # Else, starting index of required substring in # string 'A' is equal to the starting index # when first 'j' characters of string 'B' # matched with first 'i - 1' characters of # string 'A'. else: dp[i][j] = dp[i - 1][j] # String for storing final answer answer = "" # Length of smallest substring best_length = 1e9 for i in range(0, a): # dp[i][b-1] is the index in string 'A', such that # the substring A(dp[i][b-1] : i) contains string # 'B' as a subsequence. if (dp[i][b - 1] != -1): # Starting index of substring start = dp[i][b - 1] # Ending index of substring end = i # Length of substring current_length = end - start + 1 # if current length is lesser than the best # length update the answer. if (current_length < best_length): best_length = current_length # Update the answer answer = A[start: start + best_length] # Return the smallest substring return answer# This function is initializing dp with -1# and printing the resultdef smallestSubstringUtil(A, B): # Initialize dp array with -1 # Function call print(smallestSubstring(A, B))# Driver codeif __name__ == "__main__": # Input strings A = "abcedbaced" B = "bed" # Function Call smallestSubstringUtil(A, B)# This code is contributed by rakeshsahni |
C#
// C# program of the above approachusing System;using System.Linq; using System.Collections.Generic;class GFG{ // Stores the dp-states static int[,] dp = new int[100, 100]; // Function to find the smallest substring in string A which // contains string B as a subsequence. static string smallestSubstring(string A, string B) { // Size of string A int a = A.Length; // Size of string B int b = B.Length; // Initializing the first column of dp array. // Storing the occurrence of first character of string B // in first (i + 1) characters of string A. for (int i = 0; i < a; ++i) { // If the current character of string A does not // match the first character of string B. if (i > 0 && A[i] != B[0]) { dp[i,0] = dp[i - 1, 0]; } // If the current character of string A is equal to // the first character of string B. if (A[i] == B[0]) { dp[i,0] = i; } } // Iterating through remaining characters of string B. for (int j = 1; j < b; ++j) { // Checking if any character in string A matches // with the current character of string B. for (int i = 1; i < a; ++i) { // If there is a match, then starting index of // required substring in string 'A' is equal to // the starting index when first 'j - 1' // characters of string 'B' matched with first // 'i - 1' characters of string 'A'. if (A[i] == B[j]) { dp[i,j] = dp[i - 1,j - 1]; } // Else, starting index of required substring in // string 'A' is equal to the starting index // when first 'j' characters of string 'B' // matched with first 'i - 1' characters of // string 'A'. else { dp[i,j] = dp[i - 1,j]; } } } // String for storing final answer string answer = ""; // Length of smallest substring int best_length = 100000000; for (int i = 0; i < a; ++i) { // dp[i][b-1] is the index in string 'A', such that // the substring A(dp[i][b-1] : i) contains string // 'B' as a subsequence. if (dp[i,b - 1] != -1) { // Starting index of substring int start = dp[i,b - 1]; // Ending index of substring int end = i; // Length of substring int current_length = end - start + 1; // if current length is lesser than the best // length update the answer. if (current_length < best_length) { best_length = current_length; // Update the answer answer = A.Substring(start, best_length); } } } // Return the smallest substring return answer; } // This function is initializing dp with -1 // and printing the result static void smallestSubstringUtil(string A, string B) { // Initialize dp array with -1 for(int i=0;i<100;i++) { for(int j=0;j<100;j++) { dp[i,j] = -1; } } // Function call Console.Write( smallestSubstring(A, B)); } // Driver Code public static void Main() { // Input strings string A = "abcedbaced"; string B = "bed"; // Function Call smallestSubstringUtil(A, B); }}// This code is contributed by sanjoy_62. |
Javascript
<script> // JavaScript code for the above approach // Stores the dp-states let dp = new Array(100); for (let i = 0; i < dp.length; i++) { dp[i] = new Array(100).fill(-1) } // Function to find the smallest substring in string A which // contains string B as a subsequence. function smallestSubstring(A, B) { // Size of string A let a = A.length; // Size of string B let b = B.length; // Initializing the first column of dp array. // Storing the occurrence of first character of string B // in first (i + 1) characters of string A. for (let i = 0; i < a; ++i) { // If the current character of string A does not // match the first character of string B. if (i > 0 && A[i] != B[0]) { dp[i][0] = dp[i - 1][0]; } // If the current character of string A is equal to // the first character of string B. if (A[i] == B[0]) { dp[i][0] = i; } } // Iterating through remaining characters of string B. for (let j = 1; j < b; ++j) { // Checking if any character in string A matches // with the current character of string B. for (let i = 1; i < a; ++i) { // If there is a match, then starting index of // required substring in string 'A' is equal to // the starting index when first 'j - 1' // characters of string 'B' matched with first // 'i - 1' characters of string 'A'. if (A[i] == B[j]) { dp[i][j] = dp[i - 1][j - 1]; } // Else, starting index of required substring in // string 'A' is equal to the starting index // when first 'j' characters of string 'B' // matched with first 'i - 1' characters of // string 'A'. else { dp[i][j] = dp[i - 1][j]; } } } // String for storing final answer let answer = ""; // Length of smallest substring let best_length = 1e9; for (let i = 0; i < a; ++i) { // dp[i][b-1] is the index in string 'A', such that // the substring A(dp[i][b-1] : i) contains string // 'B' as a subsequence. if (dp[i][b - 1] != -1) { // Starting index of substring let start = dp[i][b - 1]; // Ending index of substring let end = i; // Length of substring let current_length = end - start + 1; // if current length is lesser than the best // length update the answer. if (current_length < best_length) { best_length = current_length; // Update the answer answer = A.slice(start, start + best_length); } } } // Return the smallest substring return answer; } // This function is initializing dp with -1 // and printing the result function smallestSubstringUtil(A, B) { // Function call document.write(smallestSubstring(A, B) + "<br>"); } // Driver code // Input strings let A = "abcedbaced"; let B = "bed"; // Function Call smallestSubstringUtil(A, B); // This code is contributed by Potta Lokesh </script> |
Output
bced
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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