Problem Description

You are given an array of strings nums where each string represents an integer without leading zeros. Your task is to find and return the kth largest integer in the array as a string.

The key points to understand:

1. Input format: The numbers are given as strings, not integers. This is important because the numbers might be very large and exceed the normal integer range.

2. Finding the kth largest: You need to identify which number would be at position k if all numbers were sorted in descending order (largest to smallest).

3. Handling duplicates: Duplicate numbers are counted separately. For example, if you have ["1", "2", "2"] and need to find the 2nd largest number:

   • The 1st largest is "2" (the first occurrence)
   • The 2nd largest is "2" (the second occurrence)
   • The 3rd largest is "1"

4. Return format: The result should be returned as a string, maintaining the same format as the input.

For example, if nums = ["3", "6", "7", "10"] and k = 4, the numbers in descending order would be ["10", "7", "6", "3"], so the 4th largest number would be "3".

The solution uses the nlargest function with a custom key that converts strings to integers for proper numerical comparison, then returns the kth element from the resulting list of the k largest numbers.

Intuition

The main challenge here is that we're dealing with numbers stored as strings. If we try to sort strings directly using lexicographic comparison, we'd get incorrect results. For instance, "9" would come after "10" in string comparison because "9" > "1" when comparing character by character.

To solve this, we need to compare the actual numerical values, not the string representations. The natural approach is to convert each string to an integer during comparison. This ensures that "10" is correctly identified as larger than "9".

Since we need the kth largest element, we have two main strategies:

1. Sort everything: Convert and sort all numbers in descending order, then pick the element at index k-1. This works but might be inefficient if k is small and the array is large.

2. Partial selection: We only need to find the top k elements. Using nlargest with k elements is more efficient because it maintains a heap of size k internally, avoiding the need to sort the entire array.

The solution uses nlargest(k, nums, key=lambda x: int(x)) which:

• Finds the k largest elements from nums
• Uses int(x) as the comparison key to ensure numerical comparison
• Returns these k elements in descending order
• We then take the last element [k-1] from this result, which is the kth largest

This approach elegantly handles the string-to-integer conversion issue while being efficient for finding the kth largest element without sorting the entire array when k is small relative to the array size.

Learn more about Divide and Conquer, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The solution can be implemented using two different approaches as mentioned in the reference:

Approach 1: Using nlargest (Shown in the code)

The implementation uses Python's heapq.nlargest function with a custom key:

return nlargest(k, nums, key=lambda x: int(x))[k - 1]

Breaking down this implementation:

1. nlargest(k, nums, key=lambda x: int(x)): This function finds the k largest elements from the nums array
   • The key=lambda x: int(x) parameter tells the function to convert each string to an integer for comparison
   • This returns a list of k strings, ordered from largest to smallest
2. [k - 1]: Since we want the kth largest element and the returned list is 0-indexed, we access the element at index k-1

Time Complexity: O(n log k) where n is the length of the array. The nlargest function maintains a min-heap of size k.

Space Complexity: O(k) for storing the heap of k largest elements.

Approach 2: Sorting (Alternative mentioned in reference)

An alternative approach would be to sort the entire array:

nums.sort(key=lambda x: int(x), reverse=True)
return nums[k - 1]

This approach:

1. Sorts all strings in descending order based on their integer values
2. Returns the element at index k-1

Time Complexity: O(n log n) for sorting the entire array.

Space Complexity: O(1) if sorting in-place, or O(n) if creating a new sorted array.

The nlargest approach is generally more efficient when k is small relative to n, while sorting might be simpler to understand and implement. Both approaches handle the critical aspect of converting strings to integers for proper numerical comparison.

Example Walkthrough

Let's walk through an example with nums = ["3", "6", "7", "10", "2", "6"] and k = 3.

Step 1: Understanding the problem

• We have numbers as strings: ["3", "6", "7", "10", "2", "6"]
• We need to find the 3rd largest number
• If we sorted these numerically in descending order: ["10", "7", "6", "6", "3", "2"]
• The 3rd largest would be "6" (first occurrence)

Step 2: Apply the solution using nlargest

When we call nlargest(3, nums, key=lambda x: int(x)):

1. The function creates a min-heap to track the 3 largest elements

2. It processes each string, converting to integer for comparison:

   • Process "3" → int value 3, heap: [3]
   • Process "6" → int value 6, heap: [3, 6]
   • Process "7" → int value 7, heap: [3, 6, 7]
   • Process "10" → int value 10, larger than min (3), so replace: heap: [6, 7, 10]
   • Process "2" → int value 2, smaller than min (6), so skip
   • Process "6" → int value 6, equal to min, both kept: heap: [6, 7, 10]

3. The function returns the 3 largest elements in descending order: ["10", "7", "6"]

4. We access index [k-1] = [2] to get "6"

Step 3: Verify with sorting approach

If we used the sorting approach:

1. Sort with key=lambda x: int(x) and reverse=True
2. Result: ["10", "7", "6", "6", "3", "2"]
3. Access index [k-1] = [2] to get "6"

Both approaches correctly return "6" as the 3rd largest number.

Key insight: The string "10" is correctly identified as larger than "7" because we compare int("10") = 10 vs int("7") = 7, not the strings themselves (which would incorrectly order "7" > "10" lexicographically).

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log k) where n is the length of the nums array and k is the parameter k. The nlargest function from Python's heapq module maintains a min-heap of size k and processes each element, taking O(log k) time for heap operations per element. Since we need to convert each string to an integer using the key function, this adds O(m) time per element where m is the average length of the strings, making the total time complexity O(n × (m + log k)). However, if k approaches n, the time complexity can degrade to O(n × log n) as the heap size grows.

The space complexity is O(k) for storing the heap of k largest elements, plus O(m) for temporary integer conversions where m is the maximum length of a string number. The overall space complexity is O(k + m).

Note that the reference answer mentions O(n) time complexity which would apply if a linear-time selection algorithm like quickselect were used instead of the heap-based approach. The O(log n) space complexity in the reference could refer to the recursion stack depth if quickselect were implemented recursively, while O(1) would apply to an iterative implementation with in-place partitioning.

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

Common Pitfalls

1. String Comparison Instead of Numerical Comparison

The most critical pitfall is comparing strings lexicographically instead of numerically. Without proper conversion, string comparison would give incorrect results.

Example of the pitfall:

# WRONG: Direct string comparison
nums = ["3", "30", "34", "5", "9"]
result = nlargest(k, nums)  # Missing key function
# This would compare "9" > "5" > "34" > "30" > "3" lexicographically
# But numerically we want 34 > 30 > 9 > 5 > 3

Solution: Always use key=lambda x: int(x) to ensure numerical comparison:

result = nlargest(k, nums, key=lambda x: int(x))

2. Integer Overflow for Very Large Numbers

While Python handles arbitrary precision integers, in other languages or if you try to optimize by using fixed-size integers, you might face overflow issues.

Example of the pitfall:

nums = ["99999999999999999999999999999", "100000000000000000000000000000"]
# In languages with fixed integer sizes, these would overflow

Solution:

• In Python: The current solution handles this correctly as Python's int() supports arbitrary precision
• For other languages: Implement custom string comparison that handles length first, then digit-by-digit comparison

3. Handling Leading Zeros (Even Though Problem States No Leading Zeros)

If the input accidentally contains leading zeros, direct integer conversion handles it correctly, but you should be aware of this edge case.

Example of the pitfall:

nums = ["003", "3", "0003"]  # All represent the same number
k = 2
# These would all convert to 3, making all three equal

Solution: The current implementation actually handles this correctly since int("003") equals int("3"). However, if you need to preserve the original string format, you might need additional logic.

4. Off-by-One Error in Index Access

A common mistake is forgetting that nlargest returns a 0-indexed list while k is 1-indexed.

Example of the pitfall:

# WRONG: Using k directly as index
return nlargest(k, nums, key=lambda x: int(x))[k]  # IndexError when k equals length

Solution: Always use [k-1] to access the kth largest element:

return nlargest(k, nums, key=lambda x: int(x))[k - 1]

5. Not Handling Edge Cases

Failing to consider edge cases like single element arrays or when k equals the array length.

Example of the pitfall:

nums = ["1"]
k = 1
# Should return "1", but implementation might fail if not careful

Solution: The current implementation handles these cases correctly, but always test with:

• Single element arrays
• k = 1 (finding the largest)
• k = len(nums) (finding the smallest)
• Arrays with duplicate values