Solution Explanation for Remove 9

The problem asks to find the nth number in a sequence where numbers containing the digit '9' are removed. A brute-force approach would be inefficient for large values of n. Instead, we can use a mathematical approach to efficiently determine the nth number.

Core Idea:

The key observation is that the sequence without '9's can be mapped to a base-9 system. Each digit in the base-9 representation corresponds to a digit in the modified sequence (0 maps to 0, 1 maps to 1, ..., 8 maps to 8). Converting n to base 9 effectively gives us the digits of the result in the modified sequence. However, we must adjust since base 9 starts from 0, while we need 1-based indexing.

Algorithm:

1. Convert n to base 9: Convert the input n to its base 9 representation. We subtract 1 from n before conversion because our sequence is 1-indexed, while base-9 representation is 0-indexed.

2. Construct the result: Iterate through the base 9 digits. Each digit represents a digit in the final answer. However, since we skipped numbers with '9', we replace any instance of 0, 1, 2, ..., 7 with the same digit. This mapping directly from base 9 to the sequence of numbers without 9s.

3. Return the result: The constructed number is the nth number in the modified sequence.

Time Complexity Analysis:

The dominant operation is converting n to base 9. The number of digits in the base 9 representation of n is proportional to log9(n). Each digit conversion takes constant time. Therefore, the overall time complexity is O(log9n), which is effectively O(log n).

Space Complexity Analysis:

The space complexity is O(log n) because we need to store the base 9 digits, the number of which is proportional to log9(n).

Code Implementation (Python):

def newInteger(n: int) -> int:
    """
    Finds the nth integer in the sequence without numbers containing '9'.
 
    Args:
        n: The index of the desired integer (1-indexed).
 
    Returns:
        The nth integer in the modified sequence.
    """
    n -= 1  # Adjust for 0-based indexing in base 9 conversion
    base9_digits = []
    while n > 0:
        digit = n % 9
        base9_digits.insert(0, digit)  # Prepend for correct order
        n //= 9
    
    result = 0
    for digit in base9_digits:
        result = result * 10 + digit
    return result
 
 

Code Implementation (Java):

class Solution {
    public int newInteger(int n) {
        n--; // Adjust for 0-based indexing
        List<Integer> base9Digits = new ArrayList<>();
        while (n > 0) {
            int digit = n % 9;
            base9Digits.add(0, digit); // Prepend for correct order
            n /= 9;
        }
 
        int result = 0;
        for (int digit : base9Digits) {
            result = result * 10 + digit;
        }
        return result;
    }
}

Code Implementation (C++):

class Solution {
public:
    int newInteger(int n) {
        n--; // Adjust for 0-based indexing
        vector<int> base9Digits;
        while (n > 0) {
            int digit = n % 9;
            base9Digits.insert(base9Digits.begin(), digit); // Prepend for correct order
            n /= 9;
        }
 
        int result = 0;
        for (int digit : base9Digits) {
            result = result * 10 + digit;
        }
        return result;
    }
};

Code Implementation (Go):

func newInteger(n int) int {
	n-- // Adjust for 0-based indexing
	base9Digits := []int{}
	for n > 0 {
		digit := n % 9
		base9Digits = append([]int{digit}, base9Digits...) // Prepend for correct order
		n /= 9
	}
 
	result := 0
	for _, digit := range base9Digits {
		result = result*10 + digit
	}
	return result
}

These implementations all follow the same algorithm, achieving the optimal time and space complexity. Remember to choose the implementation that best suits your preferred programming language.