Introduction

In this blog post, we will explore how to solve the LeetCode problem "2048" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

An integer x is numerically balanced if for every digit d in the number x, there are exactly d occurrences of that digit in x. Given an integer n, return the smallest numerically balanced number strictly greater than n. Example 1: Input: n = 1 Output: 22 Explanation: 22 is numerically balanced since: - The digit 2 occurs 2 times. It is also the smallest numerically balanced number strictly greater than 1. Example 2: Input: n = 1000 Output: 1333 Explanation: 1333 is numerically balanced since: - The digit 1 occurs 1 time. - The digit 3 occurs 3 times. It is also the smallest numerically balanced number strictly greater than 1000. Note that 1022 cannot be the answer because 0 appeared more than 0 times. Example 3: Input: n = 3000 Output: 3133 Explanation: 3133 is numerically balanced since: - The digit 1 occurs 1 time. - The digit 3 occurs 3 times. It is also the smallest numerically balanced number strictly greater than 3000. Constraints: 0 <= n <= 106

Explanation

• Generate Candidates: Generate numerically balanced numbers in increasing order.
  • Check and Return: Check if each generated number is greater than the input n. If it is, return it.
  • Optimization: Prune the search space by generating numbers of the same length as n's length, and potentially one digit longer.

• Runtime Complexity: O(1) (practically constant, though technically depends on the upper bound of numerically balanced numbers, which is limited due to the digit balance constraint). Storage Complexity: O(1) (constant storage for the generated number).

Code

    def is_balanced(x):
    s = str(x)
    counts = {}
    for digit in s:
        counts[digit] = counts.get(digit, 0) + 1
    for digit in s:
        if int(digit) == 0 and counts['0'] > 0:
            return False
        if str(int(digit)) not in counts or counts[digit] != int(digit):
            return False
    return True

def generate_balanced(length):
    if length > 7:  # largest possible balanced number has 7 digits (e.g. 6210001)
        return

    digits = [0] * length

    def backtrack(index):
        if index == length:
            num = int("".join(map(str, digits)))
            if is_balanced(num):
                yield num
            return

        for digit in range(length + 1):
            digits[index] = digit
            yield from backtrack(index + 1)

    yield from backtrack(0)

def next_balanced(n):
    n_str = str(n)
    length = len(n_str)

    candidates = []

    #Generate balanced numbers of the same length as n
    generated = sorted(num for num in generate_balanced(length) if num > n)
    if generated:
        candidates.extend(generated)

    #Generate balanced numbers with length + 1, which might be needed if all balanced numbers with the same length are smaller or equal to n.
    generated = sorted(generate_balanced(length + 1))
    if generated:
        candidates.extend(generated)


    if candidates:
        return candidates[0]
    else: #no such balanced numbers exists
        return -1


# Test Cases
# print(next_balanced(1))
# print(next_balanced(1000))
# print(next_balanced(3000))
# print(next_balanced(4325))
# print(next_balanced(0))
# print(next_balanced(59))

    return next_balanced(n)