Valid Compressed String

A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string  T is valid for the plaintext string S.

Examples:

Input: S = "GEEKSFORGEEKS", T = "G7G3S"
Output: 1
Explanation: We can clearly see that T is a valid compressed string for S.

Input: S = "DFS", T = "D1D"
Output: 0
Explanation: T is not a valid compressed string.

Approach: To solve the problem follow the below idea:

Idea is to traverse through the string T using variable i and take a variable j and initialize it to 0, if we find an integer in string T, increase the j pointer by that integer and then compare S[j] and T[i]. If at any point, it doesn't match, the compressed String isn't valid.

Below are the steps for the above approach:

• Initialize 3 variables flag = 1, j = 0, and n = 0.
• Run a loop from i = 0 to i < length of t.
• Check if t[i] is a digit, retrieve the digit, and decrement j by 1.
• Else update j += n and check if t[i] != s[j], mark the flag = 0, and break the loop. Reset n to 0 and increment j by 1.
• When the loop ends, Increment j by n.
• Check if j != s.length(), and mark the flag = 0.
• Check if flag == 1, return 1, else return 0.

Below is the implementation of the above approach:

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

int checkCompressed(string s, string t)
{
    int n = 0;
    int flag = 1;
    int j = 0;
    for (int i = 0; i < t.length(); i++) {
        if (t[i] >= '0' && t[i] <= '9') {
            n *= 10;
            if (n > 100000)
                return 0;
            n += t[i] - '0';
            j--;
        }
        else {
            j += n;
            if (t[i] != s[j]) {
                flag = 0;
                break;
            }
            n = 0;
        }
        j++;
    }
    j += n;
    if (j != s.length())
        flag = 0;

    if (flag)
        return 1;
    else
        return 0;
}

// Drivers code
int main()
{
    string S = "GEEKSFORGEEKS", T = "G7G3S";
    int ans = checkCompressed(S, T);
    cout << ans;
}

// Java code for the above approach:

import java.io.*;

class GFG {

    // Function to check the compressed string.
    static int checkCompressed(String s, String t)
    {
        int n = 0;
        boolean flag = true;
        int j = 0;
        for (int i = 0; i < t.length(); i++) {
            if (t.charAt(i) >= '0' && t.charAt(i) <= '9') {
                n *= 10;
                if (n > 100000)
                    return 0;
                n += (int)t.charAt(i) - '0';
                j--;
            }
            else {
                j += n;
                if (t.charAt(i) != s.charAt(j)) {
                    flag = false;
                    break;
                }
                n = 0;
            }
            j++;
        }
        j += n;
        if (j != s.length())
            flag = false;

        if (flag)
            return 1;
        else
            return 0;
    }

    public static void main(String[] args)
    {
        String S = "GEEKSFORGEEKS", T = "G7G3S";
        int ans = checkCompressed(S, T);
        System.out.println(ans);
    }
}

// This code is contributed by karthik.

# Python3 code for the above approach:
def checkCompressed(s, t):
    n = 0
    flag = 1
    j = 0
    for i in range(len(t)):
        if t[i].isdigit():
            n *= 10
            if n > 100000:
                return 0
            n += int(t[i])
            j -= 1
        else:
            j += n
            if t[i] != s[j]:
                flag = 0
                break
            n = 0
        j += 1
    j += n
    if j != len(s):
        flag = 0

    if flag:
        return 1
    else:
        return 0

# Driver code
S = "GEEKSFORGEEKS"
T = "G7G3S"
ans = checkCompressed(S, T)
print(ans)

# This code is contributed by Prajwal Kandekar

// JavaScript code for the above approach

// Function to check the compressed string.
function checkCompressed(s, t) {
    let n = 0;
    let flag = true;
    let j = 0;
    for (let i = 0; i < t.length; i++) {
        if (t[i] >= '0' && t[i] <= '9') {
            n *= 10;
            if (n > 100000)
                return 0;
            n += parseInt(t[i]);
            j--;
        }
        else {
            j += n;
            if (t[i] != s[j]) {
                flag = false;
                break;
            }
            n = 0;
        }
        j++;
    }
    j += n;
    if (j !== s.length)
        flag = false;

    if (flag)
        return 1;
    else
        return 0;
}

// Driver code
let S = "GEEKSFORGEEKS";
let T = "G7G3S";

// Function call
let ans = checkCompressed(S, T);
console.log(ans);

using System;

public class GFG{

   // Function to check the compressed string.
static int checkCompressed(string s, string t) {
    int n = 0;
    bool flag = true;
    int j = 0;
    
    for (int i = 0; i < t.Length; i++) {
        if (t[i] >= '0' && t[i] <= '9') {
            n *= 10;
            if (n > 100000)
                return 0;
            n += (int)t[i] - '0';
            j--;
        }
        else {
            j += n;
            if (t[i] != s[j]) {
                flag = false;
                break;
            }
            n = 0;
        }
        j++;
    }
    j += n;
    if (j != s.Length)
        flag = false;
    
    if (flag)
        return 1;
    else
        return 0;
}

  // Driver code
public static void Main(string[] args) {
    string S = "GEEKSFORGEEKS", T = "G7G3S";
    int ans = checkCompressed(S, T);
    Console.WriteLine(ans);
}
    }

1

Time Complexity: O(T)  //In the above-given approach, there is one loop for iterating over string which takes O(T) time in worst case. Therefore, the time complexity for this approach will be O(T).
Auxiliary Space: O(1) //since no extra array or data structure is used so the space taken by the algorithm is constant