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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 <= 1000 <= 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 $O(\log^2 bound)$.
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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) { ans.add(a + b); 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: ans.add(a + b) 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) { ans.add(a + b); 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) { ans.add(a + b); if (y === 1) { break; } } if (x === 1) { break; } } return [...ans]; };