How do you solve Integers With Multiple Sum of Two Cubes on LeetCode?

Integers With Multiple Sum of Two Cubes (LeetCode #3890) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Good integers can be expressed as sums of cubes in multiple ways.

What is the time complexity of Integers With Multiple Sum of Two Cubes?

The optimal solution for Integers With Multiple Sum of Two Cubes runs in O(n) time and uses O(n) space. The complexity is reduced by limiting the range of a and b to the cube root of n, leading to a more efficient solution.

What is the brute force solution for Integers With Multiple Sum of Two Cubes?

The brute force approach for Integers With Multiple Sum of Two Cubes is Brute Force, with O(n²) time complexity and O(n) space complexity. We can generate all pairs of integers (a, b) such that a <= b and compute their cubes' sum. By counting how many times each sum appears, we can identify good integers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Integers With Multiple Sum of Two Cubes?

Integers With Multiple Sum of Two Cubes uses the following concepts: Hash Table, Sorting, Counting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Integers With Multiple Sum of Two Cubes?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Integers With Multiple Sum of Two Cubes?

Not considering the distinctness of pairs (a, b). Forgetting to limit the range of a and b based on n.

#3890

Integers With Multiple Sum of Two Cubes

Medium

Hash TableSortingCountingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all pairs, we can compute sums directly and count occurrences using a hash map, which reduces unnecessary computations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a hash map to count sums of cubes.
2Step 2: Loop through possible values of a and b, calculating their cubes and storing sums in the map.
3Step 3: Collect keys from the map that have counts greater than 1 and return them sorted.

solution.py11 lines

1def findGoodIntegers(n):
2    from collections import defaultdict
3    count = defaultdict(int)
4    limit = int(n**(1/3))
5    for a in range(1, limit + 1):
6        for b in range(a, limit + 1):
7            s = a**3 + b**3
8            if s > n:
9                break
10            count[s] += 1
11    return sorted(k for k, v in count.items() if v > 1)

ℹ

Complexity note: The complexity is reduced by limiting the range of a and b to the cube root of n, leading to a more efficient solution.

1Good integers can be expressed as sums of cubes in multiple ways.
2Using a hash map allows efficient counting of occurrences.

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