LeetCode 0970. Powerful Integers Solution in Java, C++, Python & More | Explanation + Code

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 xⁱ + y^j, where i ≥ 0, j ≥ 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 10⁶, and 2²⁰ = 1048576 \gt 10⁶. Therefore, if x ≥ 2, then i is at most 20 to make xⁱ + y^j ≤ bound hold. Similarly, if y ≥ 2, then j is at most 20.

Therefore, we can use double loop to enumerate all possible xⁱ and y^j, denoted as a and b respectively, and ensure that a + b ≤ 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).

PythonJavaC++GoTypeScriptJavaScript

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)(code-box)

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); } }(code-box)

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()); } };(code-box)

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 }(code-box)

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); }(code-box)

/** * @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]; };(code-box)