Check if a String can be converted to another by inserting character same as both neighbours
Last Updated :
21 Oct, 2022
Given 2 strings A and B . The task is to check if A can be converted into B by performing the following operation any number of times :
- Choose an index from 0 to N - 2 (suppose N is the length of A). If the character at ith index is equal to the character at (i + 1 )th index, then insert the same character between the ith and (i + 1)th index.
Examples:
Input: A = "abbaac", B = "abbbbaaac"
Output: YES
Explanation: A can be converted into B by performing the operations as follows:
- Insert b between 2nd and 3rd characters of A. ( A becomes abbbaac )
- Insert b between 2nd and 3rd characters of A.( A becomes abbbbaac )
- Insert a between 6th and 7th characters of A. ( A becomes abbbbaaac )
Input: A = "xyzz", B = "xyyzz"
Output: NO
Approach: The given problem can be solved by applying the concept of Run Length Encoding on given strings. For example, strings in 1st example can be encoded as follows :
A = {(a, 1), (b, 2), (a, 2), (c, 1)}
B = {(a, 1), (b, 4), (a, 3), (c, 1)}
Let's generalize the encoding of 2 strings as :
A = {(a1, x1), (a2, x2), (a3, x3).....(aN, xN)}
B = {(b1, y1), (b2, y2), (b3, y3)....(bM, yM)}
Now, A can be made equal to B, if the following 3 conditions are satisfied:
- N = M
- ai = bi, for i = 1, 2, 3....N
- Either (xi = yi ) or (xi < yi and xi ? 2) for i = 1, 2, 3....N
So, the problem can be solved easily by following the below steps:
- Apply Run Length Encoding on both strings.
- Check if the encodings of the strings satisfy above mentioned 3 conditions.
Below is the implementation for the above approach:
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
void runLengthEncoding(string S, vector<pair<char, int> >& V)
{
int count = 1;
for (int i = 1; i < S.length(); i++) {
if (S[i] != S[i - 1]) {
V.push_back({ S[i - 1], count });
count = 0;
}
count++;
}
V.push_back({ S.back(), count });
}
// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
bool isPossible(string A, string B)
{
// Declaring vectors V1 and V2 to store
// Run Length Encoding of Strings A and B
vector<pair<char, int> > V1, V2;
runLengthEncoding(A, V1);
runLengthEncoding(B, V2);
// Checking for 1st condition
if (V1.size() != V2.size()) {
return false;
}
for (int i = 0; i < V1.size(); i++) {
// Checking for second condition
if (V1[i].first != V2[i].first) {
return false;
}
// Checking for third condition
if (!(V1[i].second == V2[i].second || (V1[i].second < V2[i].second && V1[i].second >= 2))) {
return false;
}
}
// If all three conditions are
// satisfied, return true
return true;
}
// Driver Code
int main()
{
string A = "abbaac";
string B = "abbbbaaac";
// Function Call
bool answer = isPossible(A, B);
if (answer == 1) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
// Java code for the above approach
import java.util.*;
public class GFG
{
// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
static void runLengthEncoding(
String S, ArrayList<pair<Character, Integer> > V)
{
int count = 1;
for (int i = 1; i < S.length(); i++) {
if (S.charAt(i) != S.charAt(i - 1)) {
V.add(new pair<Character, Integer>(
S.charAt(i - 1), count));
count = 0;
}
count++;
}
V.add(new pair<Character, Integer>(
S.charAt(S.length() - 1), count));
}
// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
static boolean isPossible(String A, String B)
{
// Declaring vectors V1 and V2 to store
// Run Length Encoding of Strings A and B
ArrayList<pair<Character, Integer> > V1
= new ArrayList<pair<Character, Integer> >();
ArrayList<pair<Character, Integer> > V2
= new ArrayList<pair<Character, Integer> >();
runLengthEncoding(A, V1);
runLengthEncoding(B, V2);
// Checking for 1st condition
if (V1.size() != V2.size()) {
return false;
}
for (int i = 0; i < V1.size(); i++) {
// Checking for second condition
if (V1.get(i).first != V2.get(i).first) {
return false;
}
// Checking for third condition
if (!(V1.get(i).second == V2.get(i).second
|| (V1.get(i).second < V2.get(i).second
&& V1.get(i).second >= 2))) {
return false;
}
}
// If all three conditions are satisfied, return
// true
return true;
}
// Driver Code
public static void main(String[] args)
{
String A = "abbaac";
String B = "abbbbaaac";
// Function Call
boolean answer = isPossible(A, B);
if (answer == true) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
// Pair Class
class pair<T, U> {
T first;
U second;
public pair(T first, U second)
{
this.first = first;
this.second = second;
}
}
// This code is contributed by Tapesh(tapeshdua420)
# Python code for the above approach
# Function to perform Run Length Encoding
# on a string S and store that
# in a vector V
def runLengthEncoding(S, V):
count = 1
for i in range(1, len(S)):
if S[i] != S[i - 1]:
V.append((S[i - 1], count))
count = 0
count += 1
V.append((S[-1], count))
# Function to Check if a string can be
# converted into another by performing
# the given operation any number of times
def isPossible(A, B):
# Declaring vectors V1 and V2 to store
# Run Length Encoding of Strings A and B
V1 = []
V2 = []
runLengthEncoding(A, V1)
runLengthEncoding(B, V2)
# Checking for 1st condition
if len(V1) != len(V2):
return False
for i in range(len(V1)):
# Checking for second condition
if V1[i][0] != V2[i][0]:
return False
# Checking for third condition
if not (V1[i][1] == V2[i][1] or (V1[i][1] < V2[i][1] and V1[i][1] >= 2)):
return False
# If all three conditions are satisfied, return true
return True
# Driver Code
if __name__ == '__main__':
A = "abbaac"
B = "abbbbaaac"
answer = isPossible(A, B)
print("YES") if answer else print("NO")
# This code is contributed by Tapesh(tapeshdua420)
// C# implementation
using System;
using System.Collections.Generic;
// represent pair
public class Pair<T, U> {
public Pair() {
}
public Pair(T first, U second) {
this.First = first;
this.Second = second;
}
public T First { get; set; }
public U Second { get; set; }
};
public class GFG{
// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
public static void runLengthEncoding(string S, List<Pair<char, int>> V)
{
// Pair<String, int> pair = new Pair<String, int>("test", 2);
int count = 1;
for (int i = 1; i < S.Length; i++) {
if (S[i] != S[i - 1]) {
Pair<char, int> pair = new Pair<char, int>(S[i - 1], count);
V.Add(pair);
count = 0;
}
count++;
}
Pair<char, int> pair2 = new Pair<char, int>(S[S.Length -1], count);
V.Add(pair2);
}
// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
public static bool isPossible(string A, string B)
{
// Declaring vectors V1 and V2 to store
// Run Length Encoding of Strings A and B
//vector<pair<char, int> > V1, V2;
var V1 = new List<Pair<char, int>>();
var V2 = new List<Pair<char, int>>();
runLengthEncoding(A, V1);
runLengthEncoding(B, V2);
// Checking for 1st condition
if (V1.Count != V2.Count) {
return false;
}
for (int i = 0; i < V1.Count; i++) {
// Checking for second condition
if (V1[i].First != V2[i].First) {
return false;
}
// Checking for third condition
if (!(V1[i].Second == V2[i].Second || (V1[i].Second < V2[i].Second && V1[i].Second >= 2))) {
return false;
}
}
// If all three conditions are
// satisfied, return true
return true;
}
static public void Main (){
// Code
string A = "abbaac";
string B = "abbbbaaac";
// Function Call
bool answer = isPossible(A, B);
if (answer == true) {
Console.WriteLine("YES");
}
else {
Console.WriteLine("NO");
}
}
}
// this code is contributed by ksam24000
// Javascript code for the above approach
// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
function runLengthEncoding(S, V)
{
let count = 1;
for (let i = 1; i < S.length; i++) {
if (S[i] != S[i - 1]) {
V.push({ "first":S[i - 1],
"second":count });
count = 0;
}
count++;
}
V.push({ "first":S[S.length-1],
"second":count });
}
// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
function isPossible( A, B)
{
// Declaring vectors V1 and V2 to store
// Run Length Encoding of Strings A and B
let V1 = [];
let V2 = [];
runLengthEncoding(A, V1);
runLengthEncoding(B, V2);
// Checking for 1st condition
if (V1.length != V2.length) {
return false;
}
for (let i = 0; i < V1.length; i++) {
// Checking for second condition
if (V1[i].first != V2[i].first) {
return false;
}
// Checking for third condition
if (!(V1[i].second == V2[i].second || (V1[i].second < V2[i].second && V1[i].second >= 2))) {
return false;
}
}
// If all three conditions are
// satisfied, return true
return true;
}
// Driver Code
let A = "abbaac";
let B = "abbbbaaac";
// Function Call
let answer = isPossible(A, B);
if (answer == 1) {
console.log("YES");
}
else {
console.log("NO");
}
// This code is contributed by akashish__
Time Complexity: O(M + N), where M and N are lengths of string A and B respectively
Auxiliary Space: O(N)
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