A naive approach for this problem is to iterate through each character in the ransom note. For each character, check if it exists in the magazine or not. If it does, we remove that character from the magazine to ensure that each character in the magazine is only used once. If, at any point, we find that a character is not present in the magazine, we return FALSE, indicating that it is not possible to construct a ransom note from the given magazine. If we successfully iterate through all characters in the ransom note, we return TRUE.
We indeed got our desired output. However, this approach is computationally expensive as we have to check the presence of every character of the ransom note in the magazine, giving us the time complexity of
A naive approach for this problem is to iterate through each character in the ransom note. For each character, check if it exists in the magazine or not. If it does, we remove that character from the magazine to ensure that each character in the magazine is only used once. If, at any point, we find that a character is not present in the magazine, we return FALSE, indicating that it is not possible to construct a ransom note from the given magazine. If we successfully iterate through all characters in the ransom note, we return TRUE.
We indeed got our desired output. However, this approach is computationally expensive as we have to check the presence of every character of the ransom note in the magazine, giving us the time complexity of