How do you solve Kth Smallest Number in Multiplication Table on LeetCode?

Kth Smallest Number in Multiplication Table (LeetCode #668) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m log(m * n)) time complexity and O(1) space complexity. Key insight: The multiplication table is structured such that smaller numbers appear more frequently in the earlier rows and columns.

What is the brute force solution for Kth Smallest Number in Multiplication Table?

The brute force approach for Kth Smallest Number in Multiplication Table is Brute Force, with O(n² log(n²)) time complexity and O(n²) space complexity. The brute-force approach involves generating all the elements of the multiplication table and then sorting them to find the k-th smallest element. This is straightforward but inefficient for larger tables. The optimal Optimal Solution approach improves this to O(m log(m * n)).

What algorithm or data structure is used to solve Kth Smallest Number in Multiplication Table?

Kth Smallest Number in Multiplication Table uses the following concepts: Math, Binary Search. The recommended patterns to study are: Binary Search, Counting Sort.

What are the interview tips for Kth Smallest Number in Multiplication Table?

Always discuss your thought process and approach before jumping into coding. Consider edge cases, such as the smallest and largest values for m, n, and k. Practice explaining your solution as if teaching someone else; this helps solidify your understanding.

What are common mistakes when solving Kth Smallest Number in Multiplication Table?

Students often forget that the multiplication table is sorted in a specific way, leading to incorrect counting of elements. Confusing the indices when returning the k-th smallest element; remember it's 0-indexed in programming.

#668

Kth Smallest Number in Multiplication Table

Hard

MathBinary SearchBinary SearchCounting Sort

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² log(n²))	O(m log(m * n))
Space	O(n²)	O(1)

💡

Intuition

Time O(m log(m * n))Space O(1)

Instead of generating the entire multiplication table, we can use binary search to efficiently find the k-th smallest number by counting how many numbers are less than or equal to a mid value.

⚙️

Algorithm

5 steps

1Step 1: Set low = 1 and high = m * n (the largest possible value in the table).
2Step 2: While low < high, calculate mid = (low + high) // 2.
3Step 3: Count how many numbers in the multiplication table are less than or equal to mid using the formula: count += min(mid // i, n) for i from 1 to m.
4Step 4: If count < k, set low = mid + 1; otherwise, set high = mid.
5Step 5: Return low as the k-th smallest number.

solution.py10 lines

1def kthSmallest(m, n, k):
2    low, high = 1, m * n
3    while low < high:
4        mid = (low + high) // 2
5        count = sum(min(mid // i, n) for i in range(1, m + 1))
6        if count < k:
7            low = mid + 1
8        else:
9            high = mid
10    return low

ℹ

Complexity note: The time complexity is O(m log(m * n)) because we perform binary search on the range of possible values and count the elements in O(m) time. The space complexity is O(1) since we only use a few variables.

1The multiplication table is structured such that smaller numbers appear more frequently in the earlier rows and columns.
2Using binary search allows us to efficiently narrow down the possible values without generating the entire table.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.