How do you solve Powerful Integers on LeetCode?

Powerful Integers (LeetCode #970) 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: Using a set ensures that we only store unique powerful integers.

What is the brute force solution for Powerful Integers?

The brute force approach for Powerful Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of x raised to the power of i and y raised to the power of j, checking if their sum is within the given bound. This method is straightforward but inefficient as it checks all combinations exhaustively. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Powerful Integers?

Powerful Integers uses the following concepts: Hash Table, Math, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Powerful Integers?

Always consider edge cases, such as when x or y equals 1. Discuss your thought process clearly; explain why you choose certain approaches. Optimize your solution step by step, explaining the trade-offs involved.

What are common mistakes when solving Powerful Integers?

Not using a set to store results, leading to duplicates. Forgetting to handle the case when x or y is 1, which can cause infinite loops.

#970

Powerful Integers

Medium

Hash TableMathEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a more efficient approach by limiting the number of iterations based on the values of x and y. By using a set to store results and breaking out of loops when necessary, we reduce unnecessary calculations.

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Algorithm

5 steps

1Step 1: Initialize an empty set to store unique powerful integers.
2Step 2: Use a loop for i starting from 0, calculating x^i until it exceeds the bound.
3Step 3: For each x^i, use a loop for j starting from 0, calculating y^j until x^i + y^j exceeds the bound.
4Step 4: Add the powerful integer to the set if valid.
5Step 5: Return the set as a list.

solution.py12 lines

1def powerfulIntegers(x, y, bound):
2    powerful_set = set()
3    i = 0
4    while x**i <= bound:
5        j = 0
6        while x**i + y**j <= bound:
7            powerful_set.add(x**i + y**j)
8            j += 1
9            if y == 1: break
10        i += 1
11        if x == 1: break
12    return list(powerful_set)

ℹ

Complexity note: The optimal solution reduces the number of iterations significantly by breaking out of loops when necessary, leading to a linear relationship with respect to the number of unique powerful integers generated.

1Using a set ensures that we only store unique powerful integers.
2Breaking out of loops when a condition is met prevents unnecessary calculations.

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