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#### 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:

``````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) {
if (y === 1) break
}
if (x === 1) break
}
return Array.from(ans)
}
``````

#### Python Code:

``````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:
if y == 1: break
yj *= y
if x == 1: break
xi *= x
return list(ans)
``````

#### Java Code:

``````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) {
if (y == 1) break;
}
if (x == 1) break;
}
return new ArrayList<Integer>(ans);
}
}
``````

#### C++ Code:

``````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());
}
};
``````