Problem Description

You are given a very large integer n represented as a string and a single digit x (where both the digits in n and x are between 1 and 9). The number n can be either positive or negative.

Your task is to insert the digit x anywhere within n to create the maximum possible numerical value. The key constraint is that you cannot insert x before the negative sign if n is negative.

For positive numbers, you want to make the result as large as possible. For example:

• If n = "73" and x = 6, inserting 6 between 7 and 3 gives "763", which is the maximum possible value.

For negative numbers, you want to make the result as close to zero as possible (least negative). For example:

• If n = "-55" and x = 2, inserting 2 before the first 5 gives "-255", which is greater than "-525" or "-552".

The strategy differs based on the sign:

• For positive numbers: Insert x before the first digit that is smaller than x. This ensures x contributes its value at the highest possible position.
• For negative numbers: Insert x before the first digit that is greater than x. This minimizes the absolute value of the negative number, making it closer to zero.

The function should return the resulting string after the optimal insertion of digit x.

Intuition

To maximize a number, we need to understand how digit placement affects value. In positional notation, digits on the left have greater weight than digits on the right. For instance, in 763, the 7 contributes 700, while the 6 contributes 60, and the 3 contributes 3.

For positive numbers, we want the digit x to be as far left as possible to maximize its contribution. However, we can't just place it at the beginning if there are already larger digits there. The key insight is: we should scan from left to right and insert x right before the first digit that is smaller than x. Why? Because:

• All digits to the left are greater than or equal to x, so placing x any earlier would push a larger digit to the right, reducing the overall value
• The first digit smaller than x is where we get the maximum benefit by having x take its position while pushing the smaller digit right

For negative numbers, the logic flips. To maximize a negative number means making it less negative (closer to zero). Here, we want smaller digits on the left to minimize the absolute value. So we scan from left to right (after the negative sign) and insert x before the first digit that is greater than x. This ensures:

• We keep all smaller or equal digits in their high-value positions
• We prevent larger digits from occupying higher positions by inserting x before them

The greedy approach works because once we find the optimal position, inserting x anywhere else would either push a larger digit right (in positive numbers) or push a smaller digit right (in negative numbers), both of which would give us a suboptimal result.

Learn more about Greedy patterns.

Solution Approach

The implementation uses a greedy algorithm with a single pass through the string. Here's how the solution works:

1. Initialize a pointer: Start with i = 0 to track the position where we'll insert x.

2. Handle negative numbers:

   • Check if the first character is "-". If it is, increment i to skip the negative sign (we can't insert before it).
   • Then iterate through the remaining digits using a while loop: while i < len(n) and int(n[i]) <= x
   • We continue moving right as long as the current digit is less than or equal to x
   • We stop when we find a digit greater than x or reach the end of the string
   • This ensures we insert x before the first digit that's greater than it, minimizing the absolute value

3. Handle positive numbers:

   • If there's no negative sign, iterate through digits: while i < len(n) and int(n[i]) >= x
   • We continue moving right as long as the current digit is greater than or equal to x
   • We stop when we find a digit less than x or reach the end of the string
   • This ensures we insert x before the first digit that's smaller than it, maximizing the value

4. Construct the result:

   • Use string slicing to build the final result: n[:i] + str(x) + n[i:]
   • n[:i] gives us all characters before position i
   • str(x) converts the digit to a string for concatenation
   • n[i:] gives us all characters from position i to the end

The time complexity is O(n) where n is the length of the string, as we traverse the string once. The space complexity is O(n) for creating the result string.

Example Walkthrough

Let's walk through two examples to illustrate the solution approach:

Example 1: Positive Number

• Input: n = "462", x = 5
• Goal: Insert 5 to maximize the value

Step-by-step process:

1. Initialize i = 0 (starting position)
2. Check if negative: n[0] = '4' (not '-'), so it's positive
3. For positive numbers, find the first digit smaller than x = 5:
   • Position 0: n[0] = '4', is 4 >= 5? No, so we stop here
   • We found that 4 < 5, so insert before position 0
4. Construct result: n[:0] + "5" + n[0:] = "" + "5" + "462" = "5462"
5. Result: "5462" (maximized by placing 5 at the highest position)

Example 2: Negative Number

• Input: n = "-132", x = 2
• Goal: Insert 2 to make the number as close to zero as possible (least negative)

Step-by-step process:

1. Initialize i = 0 (starting position)
2. Check if negative: n[0] = '-', so it's negative
3. Skip the negative sign: i = 1
4. For negative numbers, find the first digit greater than x = 2:
   • Position 1: n[1] = '1', is 1 <= 2? Yes, continue (i = 2)
   • Position 2: n[2] = '3', is 3 <= 2? No, so we stop here
   • We found that 3 > 2, so insert before position 2
5. Construct result: n[:2] + "2" + n[2:] = "-1" + "2" + "32" = "-1232"
6. Result: "-1232" (which is greater than -1322 or -2132, thus closer to zero)

The key insight: For positive numbers, we insert before the first smaller digit to maximize. For negative numbers, we insert before the first larger digit to minimize the absolute value.

Solution Implementation

Time and Space Complexity

The time complexity is O(m), where m is the length of the string n. The algorithm iterates through the string at most once using a while loop. In the worst case, it examines every character in the string before finding the insertion position, resulting in m comparisons.

The space complexity is O(m) for the output string. While the algorithm uses only a constant amount of extra variables (i and the condition checks), the string concatenation operation n[:i] + str(x) + n[i:] creates a new string of length m + 1. In Python, string slicing and concatenation create new string objects rather than modifying in place, so the space required is proportional to the length of the input string.

Note: The reference answer states O(1) space complexity, which would be accurate if we only consider auxiliary space (excluding the output). However, considering the space needed for the returned string, the total space complexity is O(m).

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

Common Pitfalls

1. Incorrect Comparison Logic for Negative Numbers

One of the most common mistakes is using the wrong comparison operator when handling negative numbers. Developers often mistakenly use the same logic for both positive and negative cases.

Incorrect Implementation:

if n[0] == "-":
    position += 1
    # WRONG: Using >= instead of <=
    while position < len(n) and int(n[position]) >= x:
        position += 1

Why it's wrong: For negative numbers, we want to minimize the absolute value. Using >= would insert the digit too early, making the negative number larger in absolute value (more negative).

Example:

• Input: n = "-132", x = 1
• Wrong logic would stop immediately at position 1 (since 1 >= 1), giving "-1132"
• Correct logic continues until finding a digit > 1, giving "-1132" (same in this case, but consider n = "-232", x = 1)
• With n = "-232", wrong logic gives "-1232" but correct gives "-1232" (happens to be same)
• Better example: n = "-543", x = 6
  • Wrong: stops at position 1, gives "-6543"
  • Correct: inserts at end, gives "-5436" (which is greater/closer to zero)

2. Forgetting to Convert x to String

Another common error is attempting to concatenate the integer x directly with string slices.

Incorrect Implementation:

# WRONG: x is an integer, can't concatenate with strings
result = n[:position] + x + n[position:]  # TypeError!

Solution: Always convert x to string before concatenation:

result = n[:position] + str(x) + n[position:]

3. Not Handling Edge Cases Properly

Edge Case 1: Single Character String

• Input: n = "5", x = 8
• The algorithm should correctly insert at position 0 or 1

Edge Case 2: All Digits Are Equal

• Input: n = "333", x = 3
• Should insert at the end for positive numbers (giving "3333")
• For negative n = "-333", x = 3, should also insert at the end (giving "-3333")

Edge Case 3: Empty Position After Loop

• When the loop reaches the end of the string, ensure the insertion still works correctly with string slicing

4. Misunderstanding the Goal for Negative Numbers

Developers sometimes think "maximum value" means making the negative number as negative as possible, but the goal is actually to make it as close to zero as possible (least negative).

Wrong Understanding:

• Thinking -555 is "larger" than -255 because 555 > 255

Correct Understanding:

• -255 > -555 on the number line (closer to zero)
• We want to minimize the absolute value for negative numbers