## Algorithm

Problem Name: 970. Powerful Integers

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.

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 <= 106`

## Code Examples

### #1 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def powerfulIntegers(self, x, y, bound):
"""
:type x: int
:type y: int
:type bound: int
:rtype: List[int]
"""
res = set()
i = j = 0
while x ** i <= bound:
while x ** i + y ** j <= bound:
if x ** i + y ** j not in res:
res.add(x ** i + y ** j)
j += 1
if y == 1:
break
j = 0
i += 1
if x == 1:
break
return list(res)
``````
Copy The Code &

Input

cmd
x = 2, y = 3, bound = 10

Output

cmd
[2,3,4,5,7,9,10]

### #2 Code Example with C# Programming

```Code - C# Programming```

``````
using System.Collections.Generic;
using System.Linq;

namespace LeetCode
{
public class _0970_PowerfulIntegers
{
public IList < int> PowerfulIntegers(int x, int y, int bound)
{
var results = new HashSet<int>();
for (int i = 1; i  <  bound; i *= x)
{
for (int j = 1; i + j  < = bound; j *= y)
{
if (y == 1) break;
}
if (x == 1) break;
}

return results.ToArray();
}
}
}
``````
Copy The Code &

Input

cmd
x = 2, y = 3, bound = 10

Output

cmd
[2,3,4,5,7,9,10]