Problem Description

You are given a string s that contains only digit characters (0-9). Your task is to find a valid pair of digits in the string.

A valid pair is defined as two adjacent digits in the string that satisfy both of these conditions:

1. The two digits must be different from each other (not equal)
2. Each digit in the pair must appear in the entire string s exactly as many times as its numeric value

For example:

• The digit 3 would need to appear exactly 3 times in the string
• The digit 5 would need to appear exactly 5 times in the string
• The digit 0 would need to appear exactly 0 times (meaning it shouldn't exist in the string at all)

You need to traverse the string from left to right and return the first valid pair you find. The pair should be returned as a string containing the two adjacent digits.

If no valid pair exists in the entire string, return an empty string "".

The solution approach counts the frequency of each digit (0-9) in the string using an array cnt. Then it checks each pair of adjacent digits using pairwise to see if they form a valid pair. A pair (x, y) is valid when x ≠ y and cnt[x] == x and cnt[y] == y. The first such valid pair found is returned as a string f"{x}{y}", otherwise an empty string is returned.

Intuition

The key insight is that we need to check two conditions for each adjacent pair: the digits must be different, and each digit must appear in the string exactly as many times as its numeric value.

To efficiently check the second condition, we can preprocess the string by counting how many times each digit appears. This way, when we examine any adjacent pair, we can immediately know if each digit satisfies the frequency requirement without having to count again.

Why count first? If we tried to count the frequency of digits for each adjacent pair separately, we would be doing redundant work - counting the same digits over and over again. Since the frequency of each digit is fixed for the entire string, we can count once at the beginning and reuse this information.

The natural approach then becomes:

1. First, count the frequency of all digits (0-9) in the string - this takes one pass through the string
2. Then, iterate through all adjacent pairs from left to right - another pass through the string
3. For each pair (x, y), check if x ≠ y (different digits) and if cnt[x] == x and cnt[y] == y (frequency matches numeric value)
4. Return the first pair that satisfies both conditions

This approach is efficient because we only need two passes through the string - one for counting and one for checking pairs. The checking step is O(1) for each pair since we're just looking up precomputed values in our frequency array.

Solution Approach

The implementation follows a straightforward counting and validation approach:

Step 1: Count digit frequencies

cnt = [0] * 10
for x in map(int, s):
    cnt[x] += 1

We create an array cnt of size 10 (for digits 0-9) initialized with zeros. Then we iterate through each character in the string s, convert it to an integer using map(int, s), and increment the corresponding counter. After this step, cnt[i] contains the total number of times digit i appears in the string.

Step 2: Check adjacent pairs

for x, y in pairwise(map(int, s)):
    if x != y and cnt[x] == x and cnt[y] == y:
        return f"{x}{y}"

We use the pairwise function to iterate through consecutive pairs of digits. The pairwise function takes an iterable and returns consecutive overlapping pairs. For example, if s = "1234", then pairwise(map(int, s)) yields (1,2), (2,3), (3,4).

For each pair (x, y), we check three conditions:

• x != y: The two digits must be different
• cnt[x] == x: The first digit appears exactly as many times as its numeric value
• cnt[y] == y: The second digit appears exactly as many times as its numeric value

If all three conditions are satisfied, we immediately return the pair as a string using f-string formatting: f"{x}{y}".

Step 3: Return empty string if no valid pair found

return ""

If we complete the iteration without finding any valid pair, we return an empty string.

The time complexity is O(n) where n is the length of the string, as we make two passes through the string. The space complexity is O(1) since the cnt array has a fixed size of 10 regardless of input size.

Example Walkthrough

Let's walk through the solution with the string s = "21233".

Step 1: Count digit frequencies

First, we count how many times each digit appears in the string:

• Digit '2' appears 2 times
• Digit '1' appears 1 time
• Digit '3' appears 2 times

Our frequency array becomes: cnt = [0, 1, 2, 2, 0, 0, 0, 0, 0, 0]

• cnt[1] = 1 (digit 1 appears once)
• cnt[2] = 2 (digit 2 appears twice)
• cnt[3] = 2 (digit 3 appears twice)

Step 2: Check adjacent pairs from left to right

Now we examine each adjacent pair:

1. Pair "21" (digits 2 and 1):
   • Are they different? Yes, 2 ≠ 1 ✓
   • Does digit 2 appear exactly 2 times? Yes, cnt[2] = 2 ✓
   • Does digit 1 appear exactly 1 time? Yes, cnt[1] = 1 ✓
   • All conditions satisfied! Return "21"

We found a valid pair immediately, so we return "21" without checking the remaining pairs.

For completeness, let's see what would happen if we continued:

• Pair "12": 1 ≠ 2, cnt[1] = 1 (valid), but cnt[2] = 2 (valid) → Would be valid
• Pair "23": 2 ≠ 3, cnt[2] = 2 (valid), but cnt[3] = 2 (not equal to 3) → Invalid
• Pair "33": 3 = 3 → Invalid (same digits)

The algorithm correctly returns "21" as the first valid pair found when traversing from left to right.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. This is because:

• The first loop iterates through all characters in s to count digit frequencies: O(n)
• The second loop uses pairwise to iterate through consecutive pairs of digits in s, which results in n-1 iterations: O(n)
• Inside the second loop, all operations (comparisons and dictionary lookups) are O(1)
• Therefore, the overall time complexity is O(n) + O(n) = O(n)

The space complexity is O(|Σ|), where Σ is the character set of the string s. In this problem, Σ = {0, 1, 2, ..., 9}, so |Σ| = 10. This is because:

• The cnt array has a fixed size of 10 to store the frequency of each digit: O(10) = O(1)
• The pairwise iterator and map functions use O(1) additional space as they process elements one at a time
• Since the size of the character set is constant (10 digits), the space complexity is O(|Σ|) or equivalently O(1) for this specific problem

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

Common Pitfalls

Pitfall 1: Misunderstanding the digit '0' requirement

A common mistake is not properly handling the digit '0'. According to the problem, if a digit's value is 0, it must appear exactly 0 times in the string. This means the digit '0' cannot exist in the string at all for it to be part of a valid pair.

Incorrect assumption: Some might think that having zero '0's in the string automatically satisfies the condition for '0'.

The issue: If '0' appears in a pair position (like "10" or "01"), the validation digit_count[0] == 0 would return True if there are no '0's in the string. But this is contradictory - we're checking a '0' that exists in the pair while requiring it doesn't exist at all.

Solution: The current implementation actually handles this correctly! When we encounter a '0' in a pair during iteration, digit_count[0] will be at least 1 (since we counted it), so the condition digit_count[0] == 0 will be False, correctly rejecting any pair containing '0'.

Pitfall 2: Early termination without checking all pairs

Some might be tempted to optimize by stopping early once they find digits that individually satisfy their count requirements.

Example of incorrect optimization:

# WRONG: Don't do this
valid_digits = set()
for i in range(10):
    if digit_count[i] == i:
        valid_digits.add(i)

# Then trying to find pairs only from valid_digits

The issue: This approach misses the adjacency requirement. Even if digits '2' and '3' both appear the correct number of times, they might never be adjacent in the string.

Solution: Stick to the original approach of checking each adjacent pair in order. The validation must happen on actual adjacent pairs, not just on digits that meet the frequency requirement.

Pitfall 3: Using the wrong iteration method for pairs

Attempting to manually iterate through pairs using indices can lead to off-by-one errors.

Error-prone approach:

# Risky: Manual index management
for i in range(len(s) - 1):
    x = int(s[i])
    y = int(s[i + 1])
    # ... validation logic

The issue: While this works, it's more error-prone. Forgetting the - 1 in the range would cause an index out of bounds error.

Solution: Use pairwise from itertools as shown in the original solution. It's cleaner, more readable, and less prone to boundary errors. If pairwise is not available (Python < 3.10), you can use zip(s, s[1:]) as an alternative:

# Alternative for older Python versions
for x, y in zip(map(int, s), map(int, s[1:])):
    if x != y and digit_count[x] == x and digit_count[y] == y:
        return f"{x}{y}"