
Problem Statement
Lucky numbers are a special set of numbers in mathematics defined by the presence of specific digits. In this problem's context, only the digits 4
and 7
are considered lucky. Consequently, a number is termed as lucky if it consists exclusively of these digits – either 4
or 7
or both. The task here is to determine the kth
lucky number where the definition of lucky has been specifically restricted to numbers containing only the digits 4
and 7
. Given an integer k
, your challenge is to find and return this kth
lucky number, with the output represented as a string.
Examples
Example 1
Input:
k = 4
Output:
"47"
Explanation:
The first lucky number is 4, the second one is 7, the third one is 44 and the fourth one is 47.
Example 2
Input:
k = 10
Output:
"477"
Explanation:
Here are lucky numbers sorted in increasing order: 4, 7, 44, 47, 74, 77, 444, 447, 474, 477. So the 10th lucky number is 477.
Example 3
Input:
k = 1000
Output:
"777747447"
Explanation:
It can be shown that the 1000th lucky number is 777747447.
Constraints
1 <= k <= 109
Approach and Intuition
Understanding the sequence and generation of lucky numbers is imperative to tackling this problem. Here's how we can think about approaching this:
Lucky Number Generation:
- Lucky numbers are essentially binary-like with only two digits,
4
and7
. Starting from the smallest (smallest in terms of value), the sequence begins with4
, then7
. - After single digits, we expand by considering all possible combinations of
4
and7
with two digits:44
,47
,74
,77
, and so forth. - The lengthier the string, the permutations rise exponentially as every additional digit spot can be filled with either
4
or7
.
- Lucky numbers are essentially binary-like with only two digits,
Ordered Sequence Extraction:
- As we are interested in the
kth
lucky number, knowing the position in an ascending sequence is essential. As such, generating the lucky numbers in order might not be computationally feasible, especially for largek
, due to the exponential increase in possibilities.
- As we are interested in the
Kth Element Extraction Technique:
- Instead of generation and counting, leveraging a mathematical or algorithmic way of directly pinpointing the
kth
sequence element based on its properties (similar to binary conversion but limited to4
and7
) might be computationally efficient. - This can include breaking down
k
in a manner that reflects which 'binary' position (4
or7
) applies at various place values, treating4
as0
(binary) and7
as1
.
- Instead of generation and counting, leveraging a mathematical or algorithmic way of directly pinpointing the
By understanding and structuring the process of lucky number generation and sequence determination, we can navigate towards finding the kth
element effectively, considering the constraints provided. The aforementioned examples also illustrate how this ordering and accumulation of lucky-number sequences unfold, allowing us to grasp the run-through of gradually complex sequences as k
increases.
Solutions
- C++
- Java
- Python
class Solution {
public:
string getLuckyNumber(int n) {
n += 1;
int numDigits = log2(n);
string luckyNum(numDigits, ' ');
while (n > 1) {
numDigits--;
luckyNum[numDigits] = (n % 2) ? '7' : '4';
n /= 2;
}
return luckyNum;
}
};
This solution in C++ focuses on finding the k-th lucky number, where a lucky number is composed of digits 4 and 7 only. Here's the breakdown of how this solution works:
- Increment the input
n
by 1 to adjust for zero-based indexing, as counting starts from 1 but array indexing in programming languages typically starts from 0. - Calculate the number of digits required to represent
n
in binary form usinglog2(n)
. This value determines the number of characters in the resulting stringluckyNum
. - Initialize
luckyNum
with the calculated number of spaces. This string will store the resulting lucky number. - Use a
while
loop to convertn
into a binary-like form where instead of binary digits (0 and 1), you use '4' and '7'. This conversion works as follows:- Continuously decrement
numDigits
to fill characters in reverse (from least significant digit to the most). - Determine the character to place ('7' for odd numbers and '4' for even numbers) by checking if
n % 2
results in 1 (odd) or not (even). - Update
n
by dividing it by 2 iteratively to move to the next binary digit.
- Continuously decrement
- Return
luckyNum
once all positions are filled, representing the k-th lucky number where the computation is a creative use of binary representation with custom substitutions.
public class Solution {
public String findLuckyNumber(int index) {
// Adjust index for 1-based calculation
index = index + 1;
// Determine binary representation and remove leading '1'
String binaryStr = Integer.toBinaryString(index).substring(1);
char[] luckyNumber = binaryStr.toCharArray();
// Transform binary '0' to '4' and '1' to '7'
for (int pos = 0; pos < luckyNumber.length; pos++) {
luckyNumber[pos] = (luckyNumber[pos] == '1') ? '7' : '4';
}
return new String(luckyNumber);
}
}
The Java program defines a method findLuckyNumber
designed to find the K-th lucky number by converting its index in a unique sequence. The lucky numbers specifically use the numerals 4 and 7, which are popular in certain cultures.
- The program begins by adjusting the
index
to accommodate a 1-based system as Java inherently uses a 0-based index. - To generate the lucky number, the method computes the binary representation of the adjusted index and then modifies it by discarding the initial character.
- The resulting binary string is manipulated by replacing each '0' with '4' and each '1' with '7'.
- This replacement is done through a loop that iterates over the characters of the binary string, making the appropriate substitutions.
The method finally returns the constructed string, which represents the K-th lucky number derived from its binary form. Make sure to pass the appropriate index to retrieve the correct lucky number sequence.
class Solution:
def findLuckyNumber(self, index: int) -> str:
# Update index to consider 1-based index system
index += 1
# Convert index to binary, skip '0b1' prefix
lucky_number = bin(index)[3:]
# Transform '0' to '4' and '1' to '7'
lucky_number = lucky_number.replace("0", "4").replace("1", "7")
return lucky_number
The Python solution focuses on finding the K-th "lucky number," which transforms a 1-based index into a string representation of binary digits, then replaces those digits with specific numbers, '4' or '7'. Here's how you can understand the approach taken in the provided solution:
First, adjust the given K-th index to accommodate a 1-based indexing system by incrementing the index by 1.
Convert the revised index into its binary form. Python's binary conversion outputs a string that starts with '0b', indicating that the number is binary.
Remove the initial '0b1' from the string to reshape the binary representation properly.
Replace each binary '0' digit with '4' and each '1' digit with '7' to construct the K-th lucky number in its final string form.
This method ensures that you convert an integral index into a special transformation based on a binary representation, transitioning each bit into either '4' or '7' rather than the typical '0' or '1'.
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