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Solution: Powerful Integers

seanpgallivan profile image seanpgallivan ・3 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #970 (Medium): Powerful Integers


Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

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:


(Jump to: Problem Description || Solution Idea)

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)
}
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Python Code:


(Jump to: Problem Description || Solution Idea)

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)
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Java Code:


(Jump to: Problem Description || Solution Idea)

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);
    }
}
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C++ Code:


(Jump to: Problem Description || Solution Idea)

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());
    }
};
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