Given two strings A and B, the problem is to find if string B will be a subsequence of string A if we remove the substring [A[i]..A[j]] from string A. Assume that there are Q queries giving the indices i and j and each query is independent of the other.
Examples:
Input : A = abcabcxy, B = acy
Q = 2
i = 2, j = 5
i = 3, j = 6
Output :
Yes
No
Explanation :
In the first query we remove A[2]..A[5],
getting acxy and acy is its subsequence.
In the second query we remove A[3]..A[6],
getting abxy but acy is not its subsequence.
A brute force approach is, for each query remove the required substring from A and check if B is a subsequence of A, but is inefficient because we have to modify the string A for each query and also check if string B is its subsequence.
A more efficient approach is to do preprocessing on the strings as we have to encounter multiple queries. We can store the number of characters of string B that matches till each index of string A in both the forward and backward directions, in two separate arrays. Finally we can say the answer is Yes, if the following equation holds, otherwise No:
forward[i-1] + backward[j+1] >= length(B).
This works because we are removing A[i]..A[j] from A and want to know the sum of number of characters of B that match in A
from A[1]..A[i-1] and A[j+1]..A[len], which is a subsequence if this sum is atleast the length of string B.
Following is the implementation of the above approach:
// CPP program for answering queries to check
// whether a string subsequence or not after
// removing a substring.
#include <bits/stdc++.h>
using namespace std;
// arrays to store results of preprocessing
int *fwd, *bwd;
// function to preprocess the strings
void preProcess(string a, string b)
{
int n = a.size();
// Allocate memory for fwd and bwd, and
// initialize it as 0.
fwd = new int[n]();
bwd = new int[n]();
int j = 0;
// store subsequence count in forward direction
for (int i = 1; i <= a.size(); i++) {
if (j < b.size() && a[i - 1] == b[j])
j++;
// store number of matches till now
fwd[i] = j;
}
j = 0;
// store subsequence count in backward direction
for (int i = a.size(); i >= 1; i--) {
if (j < b.size() &&
a[i - 1] == b[b.size() - j - 1])
j++;
// store number of matches till now
bwd[i] = j;
}
}
// function that gives the output
void query(string a, string b, int x, int y)
{
// length of remaining string A is less
// than B's length
if ((x - 1 + a.size() - y) < b.size()) {
cout << "No\n";
return;
}
if (fwd[x - 1] + bwd[y + 1] >= b.size())
cout << "Yes\n";
else
cout << "No\n";
}
// driver function
int main()
{
string a = "abcabcxy", b = "acy";
preProcess(a, b);
// two queries
int x = 2, y = 5;
query(a, b, x, y);
x = 3, y = 6;
query(a, b, x, y);
return 0;
}
// Java program for answering
// queries to check whether
// a String subsequence or
// not after removing a substring.
class GFG
{
// arrays to store results
// of preprocessing
static int[] fwd = new int[100];
static int[] bwd = new int[100];
// function to preprocess
// the strings
static void preProcess(String a,
String b)
{
int n = a.length();
// initialize it as 0.
int j = 0;
// store subsequence count
// in forward direction
for (int i = 1;
i <= a.length(); i++)
{
if (j < b.length() &&
a.charAt(i - 1) == b.charAt(j))
{
j++;
}
// store number of
// matches till now
fwd[i] = j;
}
j = 0;
// store subsequence count
// in backward direction
for (int i = a.length(); i >= 1; i--)
{
if (j < b.length() && a.charAt(i - 1) ==
b.charAt(b.length() - j - 1))
{
j++;
}
// store number of
// matches till now
bwd[i] = j;
}
}
// function that gives
// the output
static void query(String a, String b,
int x, int y)
{
// length of remaining
// String A is less
// than B's length
if ((x - 1 + a.length() - y) < b.length())
{
System.out.print("No\n");
return;
}
if (fwd[x - 1] + bwd[y + 1] >= b.length())
{
System.out.print("Yes\n");
}
else
{
System.out.print("No\n");
}
}
// Driver Code
public static void main(String[] args)
{
String a = "abcabcxy", b = "acy";
preProcess(a, b);
// two queries
int x = 2, y = 5;
query(a, b, x, y);
x = 3;
y = 6;
query(a, b, x, y);
}
}
// This code is contributed by 29AjayKumar
# Python3 program for answering
# queries to check whether
# a String subsequence or
# not after removing a substring.
# arrays to store results
# of preprocessing
fwd = [0] * 100
bwd = [0] * 100
# function to preprocess
# the strings
def preProcess(a, b):
n = len(a)
# initialize it as 0.
j = 0
# store subsequence count
# in forward direction
for i in range(1, len(a) + 1):
if j < len(b) and a[i - 1] == b[j]:
j += 1
# store number of
# matches till now
fwd[i] = j
j = 0
# store subsequence count
# in backward direction
for i in range(len(a), 0, -1):
if (j < len(b) and
a[i - 1] == b[len(b) - j - 1]):
j += 1
# store number of
# matches till now
bwd[i] = j
# function that gives
# the output
def query(a, b, x, y):
# length of remaining
# String A is less
# than B's length
if (x - 1 + len(a) - y) < len(b):
print("No")
return
if (fwd[x - 1] + bwd[y + 1]) >= len(b):
print("Yes")
else:
print("No")
# Driver Code
if __name__ == "__main__":
a = "abcabcxy"
b = "acy"
preProcess(a, b)
x = 2
y = 5
query(a, b, x, y)
x = 3
y = 6
query(a, b, x, y)
# This code is contributed by
# sanjeev2552
// C# program for answering
// queries to check whether
// a string subsequence or
// not after removing a substring.
using System;
class GFG
{
// arrays to store results
// of preprocessing
static int []fwd = new int[100];
static int []bwd = new int[100];
// function to preprocess
// the strings
static void preProcess(string a,
string b)
{
int n = a.Length;
// initialize it as 0.
int j = 0;
// store subsequence count
// in forward direction
for (int i = 1;
i <= a.Length; i++)
{
if (j < b.Length &&
a[i - 1] == b[j])
j++;
// store number of
// matches till now
fwd[i] = j;
}
j = 0;
// store subsequence count
// in backward direction
for (int i = a.Length;
i >= 1; i--)
{
if (j < b.Length &&
a[i - 1] == b[b.Length - j - 1])
j++;
// store number of
// matches till now
bwd[i] = j;
}
}
// function that gives
// the output
static void query(string a, string b,
int x, int y)
{
// length of remaining
// string A is less
// than B's length
if ((x - 1 + a.Length - y) < b.Length)
{
Console.Write("No\n");
return;
}
if (fwd[x - 1] +
bwd[y + 1] >= b.Length)
Console.Write("Yes\n");
else
Console.Write("No\n");
}
// Driver Code
static void Main()
{
string a = "abcabcxy", b = "acy";
preProcess(a, b);
// two queries
int x = 2, y = 5;
query(a, b, x, y);
x = 3; y = 6;
query(a, b, x, y);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
<script>
// Javascript program for answering
// queries to check whether
// a String subsequence or
// not after removing a substring.
// arrays to store results
// of preprocessing
let fwd = new Array(100);
let bwd = new Array(100);
// function to preprocess
// the strings
function preProcess(a,b)
{
let n = a.length;
// initialize it as 0.
let j = 0;
// store subsequence count
// in forward direction
for (let i = 1;
i <= a.length; i++)
{
if (j < b.length &&
a[i - 1] == b[j])
{
j++;
}
// store number of
// matches till now
fwd[i] = j;
}
j = 0;
// store subsequence count
// in backward direction
for (let i = a.length; i >= 1; i--)
{
if (j < b.length && a[i-1] ==
b[b.length - j - 1])
{
j++;
}
// store number of
// matches till now
bwd[i] = j;
}
}
// function that gives
// the output
function query(a,b,x,y)
{
// length of remaining
// String A is less
// than B's length
if ((x - 1 + a.length - y) < b.length)
{
document.write("No<br>");
return;
}
if (fwd[x - 1] + bwd[y + 1] >= b.length)
{
document.write("Yes<br>");
}
else
{
document.write("No<br>");
}
}
// Driver Code
let a = "abcabcxy", b = "acy";
preProcess(a, b);
// two queries
let x = 2, y = 5;
query(a, b, x, y);
x = 3;
y = 6;
query(a, b, x, y);
// This code is contributed by rag2127
</script>
Output
Yes No
The time complexity of the above approach is O(n + q), where q is the number of queries and n is the length of string A.
Auxiliary Space: O(n)