2048. Next Greater Numerically Balanced Number
Problem Description
An integer is considered numerically balanced if for each digit d
that appears in the number, there are exactly d
occurrences of that digit within the number itself. For example, the number 122
is numerically balanced because there is exactly one 1
and two 2
s.
The problem at hand involves finding the smallest numerically balanced number that is strictly greater than a given integer n
. That means if n
is 1, then the next numerically balanced number is 22 because it's the smallest number greater than 1 that satisfies the numerically balanced condition (two instances of the digit 2
).
Flowchart Walkthrough
Let's analyze the problem using the provided algorithm flowchart. Hereās how we can establish the appropriate algorithm:
-
Is it a graph?
- No: The problem doesn't involve nodes and edges typically found in graph-related problems.
-
Need to solve for kth smallest/largest?
- No: This problem is not about finding the kth smallest or largest element.
-
Involves Linked Lists?
- No: There's no mention or use of linked lists in this problem.
-
Does the problem have small constraints?
- Yes: The numerically balanced number within the given constraints suggests we should explore candidate numbers within a manageable range.
-
Brute force / Backtracking?
- Yes: Since brute force or backtracking is a recommended approach for managing smaller constraint problems where we explore various configurations, it fits well for finding such specific numerically balanced numbers.
Conclusion: The flowchart suggests that a Brute force or Backtracking pattern is appropriate for generating and testing numbers to find the next greater numerically balanced number.
Intuition
The intuition behind the solution is to start from the number immediately greater than n
and check each successive number to see if it meets the criteria of being numerically balanced. This is essentially a brute-force approach since it examines each number one by one until it finds the solution.
The function check(num)
assesses whether a number num
is numerically balanced. It creates a list counter
that keeps track of the occurrences of each digit. The number is converted into a string to iterate over its digits, updating counter
accordingly. Following that, the function verifies whether the occurrence of each digit in counter
matches its value, thus confirming the number is numerically balanced or not.
This method is not the most efficient solution, particularly for large n
, because it potentially evaluates many numbers until it finds the next numerically balanced number. However, it's a straightforward and easy-to-understand approach.
Learn more about Math and Backtracking patterns.
Solution Approach
The solution uses a straightforward brute-force approach. Here's a step-by-step guide to how the algorithm is implemented:
-
Validation Function (
check
): A key part of the solution is thecheck
function, which takes an integernum
as input and determines ifnum
is numerically balanced. The function does the following:- Initialize a
counter
list of 10 zeros, with each index representing digits0
to9
. - Convert
num
to a string to iterate over its digits. - Increment the
counter
at the index corresponding to each digitd
by 1 for each occurrence ofd
innum
. - Iterate over the digits of
num
again, and for each digitd
, check ifcounter[d]
is equal to the digitd
itself. This step confirms that the number of occurrences of each digitd
is equal to its value. - If any digit
d
does not meet the balance condition, returnFalse
. - If all digits satisfy the condition, return
True
indicating the number is numerically balanced.
- Initialize a
-
Main Function (
nextBeautifulNumber
): This function looks for the smallest numerically balanced number strictly greater thann
.- Start iterating from
n+1
, since we are looking for a number strictly greater thann
. - Call the
check
function for each numberi
in the iteration. - Stop the iteration and return the current number
i
as soon as a numerically balanced number is found. - The upper limit of
10**7
is an arbitrary large number to ensure that the next numerically balanced number is found within practical computational time.
- Start iterating from
-
Efficiency: While the provided approach is correct, it's important to note that the efficiency isn't optimal for large numbers because it may require checking a large number of candidates. Advanced techniques or optimizations such as precomputing possible balanced numbers or using combinatorics to generate candidates more intelligently could potentially improve efficiency.
The algorithm uses iteration and simple counting, making it easily understandable. The check
function uses a list as a counting mechanism, which is an instance of the array data structure, and the iterations over the sequence of numbers and their digits are basic patterns used in brute-force algorithm implementations.
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Start EvaluatorExample Walkthrough
Let's take an integer n = 300
. According to the problem description, we need to find the smallest numerically balanced number greater than n
.
Starting from 301
(the next integer after n
), we apply check(num)
to each subsequent integer:
-
Check
301
: Is there one3
, zero0
's, and one1
? No, because there has to be zero0
's (which is correct) but also three3
's and one1
, so301
is not numerically balanced. -
Check
302
,303
, ..., up until312
: None of these numbers will be numerically balanced because there will never be three3
s, or two2
s until at least322
(the instance where the digit matches its count). -
Check
312
: It is not numerically balanced either because there's only one1
and one3
, but there should be three3
s and one1
. -
Skip forward to
322
: This is the first number where the digit2
appears twice, satisfying its requirement. But we need three occurrences of3
and none of the numbers313
to322
meet this criteria.
Proceeding with this method, the first number to satisfy the checking criteria after 300
is 333
. In 333
, there are exactly three 3
s, which meets the requirement of being numerically balanced.
This example shows that the check
function will iterate over the numbers 301
, 302
, 303
, ..., 332
, 333
, and when it reaches 333
, it will return True
, indicating that this is the smallest numerically balanced number greater than 300
. The number 333
will then be returned by the nextBeautifulNumber
function.
Thus, applying the brute-force approach outlined in the solution, we conclude that the next numerically balanced number greater than 300
is 333
. While this method is straightforward, it is also clear that for a large number the number of iterations can become significant, which highlights the inefficiency for larger inputs.
Solution Implementation
1class Solution:
2 def nextBeautifulNumber(self, n: int) -> int:
3 # Define a function to check if a number is 'beautiful'.
4 # A number is beautiful if the frequency of each digit is equal to the digit itself.
5 def is_beautiful(num):
6 # Initialize a counter for each digit from 0 to 9.
7 frequency_counter = [0] * 10
8
9 # Increment the counter for each digit found in the number.
10 for digit_char in str(num):
11 digit = int(digit_char)
12 frequency_counter[digit] += 1
13
14 # Check if the frequency of each digit is equal to the digit itself.
15 for digit_char in str(num):
16 digit = int(digit_char)
17 if frequency_counter[digit] != digit:
18 return False
19 return True
20
21 # Iterate through numbers greater than n to find the next beautiful number.
22 for i in range(n + 1, 10**7):
23 if is_beautiful(i):
24 # If the number is beautiful, return it.
25 return i
26
27 # If no beautiful number is found (which is highly improbable), return -1.
28 # This is a safeguard, and in practice, this return statement should not be reached.
29 return -1
30
1class Solution {
2
3 // Finds the next beautiful number that is greater than a given number `n`
4 public int nextBeautifulNumber(int n) {
5 // Start from the next integer value and check up to an upper limit
6 for (int i = n + 1; i < 10000000; ++i) {
7 // Utilize the check method to determine if the number is beautiful
8 if (isBeautiful(i)) {
9 // Return the first beautiful number found
10 return i;
11 }
12 }
13 // Return -1 if no beautiful number is found (should not occur with the given constraints)
14 return -1;
15 }
16
17 // Helper method to check if a number is beautiful
18 private boolean isBeautiful(int num) {
19 // Initialize a counter to store the frequency of each digit
20 int[] digitCounter = new int[10];
21
22 // Convert the number to a character array
23 char[] chars = String.valueOf(num).toCharArray();
24
25 // Increment the count for each digit found
26 for (char c : chars) {
27 digitCounter[c - '0']++;
28 }
29
30 // Check if each digit appears exactly as many times as the digit itself
31 for (char c : chars) {
32 if (digitCounter[c - '0'] != c - '0') {
33 // If not, the number is not beautiful
34 return false;
35 }
36 }
37
38 // If all digits match the criteria, the number is beautiful
39 return true;
40 }
41}
42
1#include <vector>
2#include <string>
3using namespace std;
4
5class Solution {
6public:
7 // Function to find the next beautiful number greater than a given number n.
8 int nextBeautifulNumber(int n) {
9 // Loop starts from the next number after n and stops when it finds a beautiful number.
10 for (int i = n + 1; i < 10000000; ++i) {
11 // If i is a beautiful number, return it.
12 if (isBeautiful(i)) {
13 return i;
14 }
15 }
16 // Return -1 if no beautiful number is found (this won't happen given the problem constraints).
17 return -1;
18 }
19
20private:
21 // Helper function to check if a number is beautiful.
22 // A beautiful number has each digit appearing exactly d times where d is the actual digit.
23 bool isBeautiful(int num) {
24 // Convert the number to a string.
25 string numStr = to_string(num);
26
27 // Initialize a counter for each digit (0-9).
28 vector<int> digitCounter(10, 0);
29
30 // Increment the counter for each digit found in the number.
31 for (char digit : numStr) {
32 ++digitCounter[digit - '0'];
33 }
34
35 // Check if the count of each digit matches its value.
36 for (char digit : numStr) {
37 if (digitCounter[digit - '0'] != digit - '0') {
38 // If any digit count doesn't match its value, the number isn't beautiful.
39 return false;
40 }
41 }
42
43 // All digit counts match their values, so the number is beautiful.
44 return true;
45 }
46};
47
1function nextBeautifulNumber(n: number): number {
2 // Start checking numbers greater than the given number
3 for (let answer = n + 1; ; answer++) {
4 // Check if the current number is "beautiful"
5 if (isValid(answer)) {
6 // If it's a beautiful number, then return it as the answer
7 return answer;
8 }
9 }
10}
11
12function isValid(n: number): boolean {
13 // Initialize digit occurrence record with zeros
14 let digitFrequency = new Array(10).fill(0);
15
16 // While there are still digits to process
17 while (n > 0) {
18 // Get the rightmost digit of the number
19 const currentDigit = n % 10;
20 // Increment the frequency of the currentDigit
21 digitFrequency[currentDigit]++;
22 // Remove the rightmost digit from the number
23 n = Math.floor(n / 10);
24 }
25
26 // Check all digits from 0 to 9
27 for (let i = 0; i < 10; i++) {
28 // If a digit occurs and its frequency is not equal to the digit itself, it's not valid
29 if (digitFrequency[i] && digitFrequency[i] != i) return false;
30 }
31
32 // If all conditions are met, return true
33 return true;
34}
35
Time and Space Complexity
Time Complexity
The function nextBeautifulNumber
increments a number, starting from n + 1
, and checks each number to see if it is a "beautiful" number as defined by the function check
. The check
function verifies if the number of times each digit appears is equal to the digit itself.
The time complexity of the check function is as follows:
- Converting a number to its string representation is
O(log(num))
, because the length of the string is proportional to the number of digits in the number. - Counting the digits by using an array of size 10 (for each possible digit) is done in
O(log(num))
, wherelog(num)
is the number of digits innum
. - Again, it iterates over each digit to verify if the number is beautiful, which is also
O(log(num))
.
Combining these steps, the function check
alone is O(log(num))
. However, we need to consider that it is called for each number starting from n + 1
.
The loop can be considered to go on until the next beautiful number is found, which is the worst-case scenario since we do not know the distribution of beautiful numbers. If we say finding the next beautiful number from n
has a maximum range of r
, the total time complexity is O(r * log(r))
.
The exact value of r
is not straightforward to determine as it depends on the pattern distribution of beautiful numbers, but given the constraint that the for loop can go up to 10^7
, in the worst case scenario r
would be close to 10^7
, making it O(7 * 10^7)
, which simplifies to O(10^7)
.
Space Complexity
The space complexity is primarily impacted by:
- The counter array which holds 10 integers, representing each digit, so it remains constant
O(1)
. - The string representation of the number being checked inside the loop, which has a space complexity of
O(log(num))
for each numbernum
.
Since the counter array is static and doesn't grow with the input, and the string representation of the current number is only stored temporarily for each check, the space complexity can be simplified to O(1)
because the logarithmic space required for the string representation does not affect the overall space complexity which is dominated by the fixed size of the counter array.
Learn more about how to find time and space complexity quickly using problem constraints.
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