Check if string S1 can be formed using repeated insertions of another string S2
Last Updated :
09 Sep, 2021
Given two strings S1 and S2 consisting of unique characters, the task is to check S1 can be formed by repeated insertions of string S2.
Input: S1 = “aabb”, S2 = “ab”
Output: Yes
Explanation: the mentioned string can be obtained after series of moves:
- Insert string “ab” in an empty string. Current string will be “ab“
- insert “ab” after “a”. The final string will be “aabb”
Input: S1 = “ababcd”, S2 = “abc”
Output: No
Explanation: It is not possible to obtain above string with any series of moves.
Approach: Given problem can be solved using stack data structure. The idea is to insert the characters of S1 into the stack till the last character of S2 is found. Then pop the characters of length(S2) from Stack and compare with S2. If not the same, stop and return false. Else repeat the process till S1 becomes empty.
Below are the steps for the above approach:
- First, check the below cases and return false if found true:
- The count of unique characters in S1 must be the same as S2
- The length of string S1 must be a multiple of S2
- Maintain a stack for all the characters
- Iterate through the string S1 and push characters in a stack
- If the current character is the last character of the string S2 then match all the characters to the left in the stack
- If for any position the stack is empty or the character doesn’t matches then return False
- After the complete iteration on the string check if the stack is empty. If the stack is not empty then return false else return true
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool validInsertionstring(string S1, string S2)
{
int N = S1.length();
int M = S2.length();
stack<char> st;
for (int i = 0; i < N; i++) {
st.push(S1[i]);
if (S1[i] == S2[M - 1]) {
int idx = M - 1;
while (idx >= 0) {
if (st.empty()) {
return false;
}
char c = st.top();
st.pop();
if (c != S2[idx]) {
return false;
}
idx--;
}
}
}
if (!st.empty()) {
return false;
}
else {
return true;
}
}
int main()
{
string S1 = "aabb";
string S2 = "ab";
validInsertionstring(S1, S2) ? cout << "Yes\n"
: cout << "No\n";
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static boolean validInsertionstring(String S1,
String S2)
{
int N = S1.length();
int M = S2.length();
Stack<Character> st = new Stack<>();
for (int i = 0; i < N; i++) {
st.push(S1.charAt(i));
if (S1.charAt(i) == S2.charAt(M - 1)) {
int idx = M - 1;
while (idx >= 0) {
if (st.size() == 0) {
return false;
}
char c = st.peek();
st.pop();
if (c != S2.charAt(idx)) {
return false;
}
idx--;
}
}
}
if (st.size() > 0) {
return false;
}
else {
return true;
}
}
public static void main(String[] args)
{
String S1 = "aabb";
String S2 = "ab";
if (validInsertionstring(S1, S2) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
|
Python3
def validInsertionstring(S1, S2):
N = len(S1)
M = len(S2)
st = []
for i in range(N):
st.append(S1[i])
if (S1[i] == S2[M - 1]):
idx = M - 1
while (idx >= 0):
if (len(st) == 0):
return False
c = st[-1]
st.pop()
if (c != S2[idx]):
return False
idx-=1
if (len(st) != 0):
return False
else:
return True
S1 = "aabb"
S2 = "ab"
if validInsertionstring(S1, S2):
print("Yes")
else:
print("No")
|
C#
using System;
using System.Collections.Generic;
class GFG{
static bool validInsertionstring(string S1, string S2)
{
int N = S1.Length;
int M = S2.Length;
Stack<char> st = new Stack<char>();
for (int i = 0; i < N; i++) {
st.Push(S1[i]);
if (S1[i] == S2[M - 1]) {
int idx = M - 1;
while (idx >= 0) {
if (st.Count==0) {
return false;
}
char c = st.Peek();
st.Pop();
if (c != S2[idx]) {
return false;
}
idx--;
}
}
}
if (st.Count > 0) {
return false;
}
else {
return true;
}
}
public static void Main()
{
string S1 = "aabb";
string S2 = "ab";
if(validInsertionstring(S1, S2)==true)
Console.Write("Yes");
else
Console.Write("No");
}
}
|
Javascript
<script>
function validInsertionstring(S1, S2)
{
let N = S1.length;
let M = S2.length;
let st = [];
for (let i = 0; i < N; i++) {
st.push(S1[i]);
if (S1[i] == S2[M - 1]) {
let idx = M - 1;
while (idx >= 0) {
if (st.length == 0) {
return false;
}
let c = st[st.length - 1];
st.pop();
if (c != S2[idx]) {
return false;
}
idx--;
}
}
}
if (st.length != 0) {
return false;
}
else {
return true;
}
}
let S1 = "aabb";
let S2 = "ab";
validInsertionstring(S1, S2) ? document.write("Yes")
: document.write("No");
</script>
|
Time Complexity: O(N*M)
Auxiliary Space: O(N)
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