1985. Find the Kth Largest Integer in the Array
Problem Description
You are given an array of strings called nums
where each string is a non-negative integer without leading zeros. Alongside this array, you're given an integer k
. The task is to determine the kth
largest integer in the array, when the integers are considered in their numeric value rather than their string order, and return it as a string.
The problem stipulates that duplicate numbers should count as separate entities in the ordering. For example, if the given array contains multiple instances of the same number, each instance should be considered separately in the ranking of largest numbers. Essentially, every string is a distinct entry when determining the kth
largest number, even if its numeric value is identical to that of another string.
It's important to remember that integers represented as longer strings are intrinsically larger than shorter strings when comparing numeric values, regardless of the characters within those strings. For instance, "100" is larger than "99" simply because it represents a larger integer, even though "9" comes after "1" in lexicographic order.
Intuition
The intuition behind the solution is based on custom sorting. Since we're dealing with strings representing numbers, we can't directly sort them as strings because the lexicographic sorting would yield incorrect results for numeric comparisons. For example, "30" would come before "12" in lexicographic order, even though it's numerically larger.
Therefore, we employ a custom comparison function that first compares the lengths of the strings. If two strings have different lengths, the longer one represents a larger number and should come first in a descending sort. When the string lengths are equal, we compare them lexicographically, as numeric equality in string form means the comparative order of digits will determine the larger number.
The cmp_to_key
method from the functools
module in Python is used to convert this custom comparator into a sorting key. With this method, we use Python's built-in sort
function which sorts based on the comparisons defined in our custom function, ensuring that the largest numbers (highest numeric values) come first in the array.
Once the array is sorted in descending numerical order, retrieving the kth
largest number is straightforward: we access the element at the position k - 1
since arrays are zero-indexed in Python. The resulting element, which is still in string form, is our desired solution and is returned directly.
Learn more about Divide and Conquer, Sorting and Heap (Priority Queue) patterns.
Solution Approach
The solution provided uses Python's built-in sorting algorithm with a custom comparison function to address the need for a numerical rather than lexicographical order of the strings in the nums
array.
The solution is outlined as follows:
-
Define a custom comparator
cmp
that takes two stringsa
andb
:- The comparator first checks the lengths of the two strings. If they have different lengths, the function returns the difference between the lengths of
b
anda
to ensure that the longer string is considered larger (since we want a descending sort). - If both strings are of the same length, lexicographical comparison is used to determine which one is larger. It returns
1
ifb
is greater thana
, and-1
ifb
is less than or equal toa
.
- The comparator first checks the lengths of the two strings. If they have different lengths, the function returns the difference between the lengths of
-
Use the
cmp_to_key
function from thefunctools
module to convert the custom comparatorcmp
to a key function. This key function is then used with thesort
method onnums
. Thesort
method will internally use the key function to perform comparisons between elements during the sort process. -
With the array now sorted in descending numerical order by the values represented by the strings, the solution directly accesses the
k-1
indexed element to account for zero-based indexing.
Here is how the custom comparison and sorting work out in code:
def cmp(a, b):
if len(a) != len(b):
return len(b) - len(a)
return 1 if b > a else -1
nums.sort(key=cmp_to_key(cmp))
The cmp
function is critical in achieving the desired order without converting strings to integers, enabling efficient sorting even for very large numbers.
After correctly sorting nums
, the kth
largest number is simply retrieved using:
return nums[k - 1]
By ensuring all steps adhere to the problem's requirements, the solution successfully retrieves the kth
largest integer in its string representation from nums
array.
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Start EvaluatorExample Walkthrough
Let's walk through a small example to illustrate the given solution approach.
Imagine we are given an array of strings nums = ["3", "123", "34", "30", "5"]
representing integers, and we want to find the 2nd
largest integer among these. We will follow the steps outlined in the solution approach.
-
First, we need to construct the custom comparator. The
cmp
function defined will compare the strings based on their lengths to decide which string represents a larger integer, and if the lengths are equal, it will compare them lexicographically. -
We utilize the
cmp_to_key
method from thefunctools
module to use this comparator in the sorting process. With the sorting key ready, we will apply thesort
method tonums
. -
After sorting, the array of strings should reflect the correct numerical order, descending from the largest number. The
nums
array would look like this after sorting:["123", "34", "30", "5", "3"]
. -
Finally, to get the
2nd
largest number, we select the element at indexk - 1
, which is1
in our zero-indexed array (sincek = 2
). The element at index1
is"34"
, which is the2nd
largest number in our array.
Here is how the example would be implemented in Python based on our solution approach:
from functools import cmp_to_key
# Define the comparator function
def cmp(a, b):
if len(a) != len(b):
return len(b) - len(a) # Strings with longer length are numerically bigger
return 1 if b > a else -1 # For equal lengths, compare lexicographically
# Example array and k value
nums = ["3", "123", "34", "30", "5"]
k = 2
# Sort the strings using the custom comparator converted to a key
nums.sort(key=cmp_to_key(cmp))
# Getting the kth largest number in its string representation
kth_largest = nums[k - 1]
print(kth_largest) # Output should be "34"
The output of the program is "34"
, which is what we expected for the 2nd
largest number. The custom sorting ensures we are directly comparing the numbers not just as strings but as numerical values they represent, adhering to the desired outcome of the given task.
Solution Implementation
1from functools import cmp_to_key
2
3class Solution:
4 def kthLargestNumber(self, nums: List[str], k: int) -> str:
5 # Define a comparison function to use in sorting.
6 # This function will compare two numbers based on their length first,
7 # and then by their value if the lengths are equal.
8 def compare_numbers(num1: str, num2: str) -> int:
9 if len(num1) != len(num2):
10 # Sort primarily by length of the string representation of the numbers.
11 return len(num2) - len(num1)
12 else:
13 # If lengths are equal, sort by the string representation itself.
14 # -1 if num1 should come before num2, 1 if num2 should come before num1.
15 return -1 if num2 > num1 else 1
16
17 # Sort the list of numbers using our custom comparison function.
18 nums.sort(key=cmp_to_key(compare_numbers))
19
20 # Return the k-th largest number as a string.
21 # Since the list is sorted from largest to smallest,
22 # the k-th largest number is at index k-1.
23 return nums[k - 1]
24
1import java.util.Arrays; // Import necessary class for sorting
2
3class Solution {
4
5 // Method to find the kth largest number in the form of a string from a given array of strings
6 public String kthLargestNumber(String[] nums, int k) {
7 // Sort the array using a custom comparator
8 Arrays.sort(nums, (firstNumber, secondNumber) -> {
9 // If the lengths of numbers are equal, compare them lexicographically in descending order
10 if(firstNumber.length() == secondNumber.length()) {
11 return secondNumber.compareTo(firstNumber);
12 } else {
13 // Sort based on length of strings to handle large numbers accurately
14 // Longer numbers should come first since they are larger
15 return secondNumber.length() - firstNumber.length();
16 }
17 });
18
19 // Return the kth element in the sorted array which is the kth largest number
20 // (k - 1) is used because array indices start from 0
21 return nums[k - 1];
22 }
23}
24
1#include <string> // Include the string library
2#include <vector> // Include the vector library
3#include <algorithm> // Include the algorithms library for the sort function
4
5// Define the Solution class with the public member function 'kthLargestNumber'
6class Solution {
7public:
8 // Define the 'kthLargestNumber' function that returns the k-th largest number from a vector of strings
9 string kthLargestNumber(vector<string>& nums, int k) {
10 // Define a custom comparator lambda function to sort the numbers based on length and value
11 auto comparator = [](const string& a, const string& b) {
12 // Compare sizes first. If sizes are equal, compare strings lexicographically
13 return a.size() == b.size() ? a > b : a.size() > b.size();
14 };
15
16 // Sort the vector of strings using the custom comparator
17 sort(nums.begin(), nums.end(), comparator);
18
19 // Return the k-th largest number, which is at index k-1 after sorting
20 return nums[k - 1];
21 }
22};
23
1// Import necessary functionalities from util libraries (not applicable in TypeScript as it's often executed in a browser or Node.js environment)
2// However, TypeScript has built-in sort functionality for arrays
3
4// Define the 'kthLargestNumber' function that returns the k-th largest number from an array of strings
5function kthLargestNumber(nums: string[], k: number): string {
6 // Define a custom comparator function to sort the numbers based on length and value
7 const comparator = (a: string, b: string): number => {
8 // Compare sizes first. If sizes are equal, compare strings lexicographically
9 if (a.length === b.length) {
10 return b.localeCompare(a);
11 }
12 return b.length - a.length;
13 };
14
15 // Sort the array of strings using the custom comparator
16 nums.sort(comparator);
17
18 // Return the k-th largest number, which is at index k-1 after sorting
19 return nums[k - 1];
20}
21
22// The TypeScript function can be used as follows:
23// let result = kthLargestNumber(["3", "6", "2", "10"], 2); // result would be "6"
24
Time and Space Complexity
Time Complexity
The given code defines a custom comparator and sorts a list of strings representing numbers based on their numeric values. The time complexity of this algorithm is dominated by the sorting step.
The sort()
function in Python typically has a time complexity of O(n log n)
where n
is the number of elements in the list. However, because a custom comparator is used, there is an additional overhead of comparing each pair of strings. The comparison involves checking the length of the strings and possibly comparing the strings themselves.
The worst-case time complexity for comparing two strings is O(m)
where m
is the length of the longer string among the two being compared.
Thus, considering both the sorting of n
elements and the custom comparison, the total time complexity is O(n * m * log n)
, where n
is the length of the input list nums
and m
is the maximum length of a string within nums
.
Space Complexity
The space complexity is mainly due to the space required for sorting the strings. Python's sort function can be O(n)
in the worst case for space, since it may require allocating space for the entire list. Moreover, since we are dealing with strings, the space complexity does not only depend on the number of elements n
, but also on the total space required to hold all of the strings.
If we let k
be the total amount of space required to store all strings in the list (sum of the lengths of all strings), the overall space complexity will be O(k)
.
The other space overhead in the solution is the space required for the function cmp_to_key(cmp)
. This is a function object that consumes constant space, so it does not significantly impact the overall space complexity, which remains O(k)
.
Learn more about how to find time and space complexity quickly using problem constraints.
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