Solution: Powerful Integers

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Leetcode Problem #970 (Medium): Powerful Integers

Description:

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Given three integers x, y, and bound, return a list of all the powerful integers that have a value less than or equal to bound.

An integer is powerful if it can be represented as xi + yj for some integers i >= 0 and j >= 0.

You may return the answer in any order. In your answer, each value should occur at most once.

Examples:

Example 1:
Input:	x = 2, y = 3, bound = 10
Output:	[2,3,4,5,7,9,10]
Explanation:	2 = 20 + 30 3 = 21 + 30 4 = 20 + 31 5 = 21 + 31 7 = 22 + 31 9 = 23 + 30 10 = 20 + 32

Example 2:
Input:	x = 3, y = 5, bound = 15
Output:	[2,4,6,8,10,14]

Constraints:

• 1 <= x, y <= 100
• 0 <= bound <= 10^6

Idea:

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This problem is a pretty straightforward one. Since we need to return all powerful integers and not just a count of them, there aren't many shortcuts we can take; we'll have to actually come up with the solution iteratively with nested loops.

First, we can use a set structure (ans) to prevent duplicate answers. Then we can have our nested loops increment the power of the x and y values while adding the appropriate results to our set.

One somewhat tricky situation occurs when one or more of the values is a 1, as that power will continue to be 1 forever, regardless of the exponent. To deal with that, we can force each nested loop to break after the first iteration if its original value was a 1.

Once we've iterated over all possible combinations, we can convert ans to an array and return it.

Implementation:

There are only minor differences in the code of each language.

Javascript Code:

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var powerfulIntegers = function(x, y, bound) { let ans = new Set() for (let xi = 1; xi < bound; xi *= x) { for (let yj = 1; xi + yj <= bound; yj *= y) { ans.add(xi + yj) if (y === 1) break } if (x === 1) break } return Array.from(ans) } 

Python Code:

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class Solution: def powerfulIntegers(self, x: int, y: int, bound: int) -> List[int]: ans, xi = set(), 1 while xi < bound: yj = 1 while xi + yj <= bound: ans.add(xi + yj) if y == 1: break yj *= y if x == 1: break xi *= x return list(ans) 

Java Code:

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class Solution { public List<Integer> powerfulIntegers(int x, int y, int bound) { Set<Integer> ans = new HashSet<>(); for (int xi = 1; xi < bound; xi *= x) { for (int yj = 1; xi + yj <= bound; yj *= y) { ans.add(xi + yj); if (y == 1) break; } if (x == 1) break; } return new ArrayList<Integer>(ans); } } 

C++ Code:

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class Solution { public: vector<int> powerfulIntegers(int x, int y, int bound) { unordered_set<int> ans; for (int xi = 1; xi < bound; xi *= x) { for (int yj = 1; xi + yj <= bound; yj *= y) { ans.insert(xi + yj); if (y == 1) break; } if (x == 1) break; } return vector<int>(ans.begin(), ans.end()); } };