Problem Description

The problem asks to determine if all numbers in a given sentence are strictly increasing. A sentence is defined as a list of tokens, which are either a positive number or a word, separated by single spaces. There are no leading zeros in numbers and no leading or trailing spaces in the sentence. The numbers must increase from left to right, meaning that each number should be smaller than the number that follows it. The task is to return true if the condition of strictly increasing numbers is met, otherwise return false.

Intuition

The intuition behind solving this problem is to iterate over the tokens in the sentence and keep track of the last number encountered. We split the sentence into tokens based on spaces, and then scan each token. If a token is a number (we can check this by looking at the first character of the token), we then compare it with the previous number we have seen (if any). To verify that the numbers are strictly increasing, we need to ensure that the current number is greater than the last number seen. If at any point we encounter a number that is not greater than the previous one, we return false as the sequence is no longer strictly increasing. If we successfully go through all the numbers without violating the condition, we return true, which means all numbers are strictly increasing.

Solution Approach

The implementation of the solution uses a simple iteration approach. The only data structure used here is an integer variable pre to store the previously seen number, which initially is set to 0, being smaller than any positive number as defined in the problem. We employ Python's built-in str.split() method to break the sentence s into tokens based on space as a delimiter.

The solution involves the following steps:

1. We start by splitting the sentence s into tokens by using the split method. Each token is either a word or a number.

2. We iterate over each token t in the sentence:

   • We check if the token is a number by looking at the first character of the token with t[0].isdigit().
   • If it is a number, we convert the token to an integer cur with int(t).
   • We then check if cur is less than or equal to the previous number pre. If so, we immediately return false since the numbers must be strictly increasing.
   • If the condition is not met, we update pre to be the current number cur.

3. If we finish iterating through all the tokens without returning false, we return true as no condition was violated and therefore the numbers in the sentence are strictly increasing.

Here is the essence of the code, showing the straightforward loop and checks:

1pre = 0
2for t in s.split():
3    if t[0].isdigit():
4        cur = int(t)
5        if cur <= pre:
6            return False
7        pre = cur
8return True

The code is efficient, running in O(n) time complexity – where n is the number of characters in the string – because it checks each token exactly once. Additionally, the space complexity is also O(n) due to the space taken by the split tokens. By using tuple unpacking with the walrus operator (:=), the code is concise and avoids a nested if statement.

Example Walkthrough

Let's illustrate the solution approach using a simple example:

Given the sentence "cat 1 dog 2 fish 3 cow 5 lion 8"

1. We start by splitting the sentence into tokens: ["cat", "1", "dog", "2", "fish", "3", "cow", "5", "lion", "8"].

2. Now, we initialize pre = 0, and iterate over the tokens:

   • The first token is "cat", which starts with a letter, so we ignore it.
   • The second token is "1", which starts with a digit. We convert "1" to integer 1 and compare it with pre (0). Since 1 > 0, the condition is satisfied. We set pre = 1.
   • The next token "dog" is ignored since it doesn't start with a digit.
   • The following token is "2", which is again a number. We convert "2" to integer 2 and compare it with pre (1). Since 2 > 1, the condition is satisfied. We set pre = 2.
   • This process continues for each number token.

3. Each number is greater than the last, so the condition of strictly increasing numbers is satisfied throughout.

4. After checking all the tokens, since we've found no violation of the condition, the function will return True.

In this example, the sentence follows the strictly increasing numerical order despite being interspersed with words. Thus, the output is True.

Solution Implementation

Time and Space Complexity

Time Complexity:

The time complexity of the code is O(n), where n is the length of the string s. This is because the split method runs in O(n), splitting the string into words based on spaces, and the for loop iterates over each word once. The loop itself contains a constant-time check (if t[0].isdigit()) and a constant-time integer conversion (int(t)), so overall, each iteration of the loop adds a constant amount of work. Consequently, the total time taken is proportional to the number of words, which is at most proportional to the length of the input string s.

Space Complexity:

The space complexity of the code is O(m), where m is the number of words in the string s. This is because the split method creates a list of all the words in the string, which requires space proportionate to the number of words. The for loop does not allocate any additional data structures that grow with the size of the input; it only uses a couple of variables to hold the current and previous numerical values. Note that in the worst case, every character could be a word (if they are all digits separated by spaces), making m roughly n/2, but this does affect the big-O notation and the space complexity remains linear with respect to the length of the input string.

Learn more about how to find time and space complexity quickly using problem constraints.