Valid Compressed String

Problem Statement - link #

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

Your Task:

You don’t need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.

Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)

Examples #

Example 1:

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

Example 2:

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

Constraints #

• 1 ≤ |S| ≤ 10^6
• 1 ≤ |T| ≤ 10^6

Solutions #

class Solution{
  public:
    int checkCompressed(string s, string t) {
        int i=0,j=0;
        while(i < s.size() and j < t.size()){
            if(s[i] == t[j]) i++,j++;
            else if(isdigit(t[j])){
                int num = 0;
                while(j < t.size() and isdigit(t[j])){
                    num *= 10; num += (t[j++]-'0');
                }
                i += num;
            }
            else return 0;
        }
        return i == s.size() and j == t.size();
    }
};