# 970. Powerful Integers

## Description

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

## Solutions

Solution 1: Hash Table + Enumeration

According to the description of the problem, a powerful integer can be represented as $x^i + y^j$, where $i \geq 0$, $j \geq 0$.

The problem requires us to find all powerful integers that do not exceed $bound$. We notice that the value range of $bound$ does not exceed $10^6$, and $2^{20} = 1048576 \gt 10^6$. Therefore, if $x \geq 2$, then $i$ is at most $20$ to make $x^i + y^j \leq bound$ hold. Similarly, if $y \geq 2$, then $j$ is at most $20$.

Therefore, we can use double loop to enumerate all possible $x^i$ and $y^j$, denoted as $a$ and $b$ respectively, and ensure that $a + b \leq bound$, then $a + b$ is a powerful integer. We use a hash table to store all powerful integers that meet the conditions, and finally convert all elements in the hash table into the answer list and return it.

Note that if $x=1$ or $y=1$, then the value of $a$ or $b$ is always equal to $1$, and the corresponding loop only needs to be executed once to exit.

The time complexity is $O(\log^2 bound)$, and the space complexity is $O(\log^2 bound)$.

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

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

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


• func powerfulIntegers(x int, y int, bound int) (ans []int) {
s := map[int]struct{}{}
for a := 1; a <= bound; a *= x {
for b := 1; a+b <= bound; b *= y {
s[a+b] = struct{}{}
if y == 1 {
break
}
}
if x == 1 {
break
}
}
for x := range s {
ans = append(ans, x)
}
return ans
}

• function powerfulIntegers(x: number, y: number, bound: number): number[] {
const ans = new Set<number>();
for (let a = 1; a <= bound; a *= x) {
for (let b = 1; a + b <= bound; b *= y) {
if (y === 1) {
break;
}
}
if (x === 1) {
break;
}
}
return Array.from(ans);
}


• /**
* @param {number} x
* @param {number} y
* @param {number} bound
* @return {number[]}
*/
var powerfulIntegers = function (x, y, bound) {
const ans = new Set();
for (let a = 1; a <= bound; a *= x) {
for (let b = 1; a + b <= bound; b *= y) {
if (y === 1) {
break;
}
}
if (x === 1) {
break;
}
}
return [...ans];
};