README_EN.md

970. Powerful Integers

中文文档

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

Python3

class Solution:
    def powerfulIntegers(self, x: int, y: int, bound: int) -> List[int]:
        s = set()
        i = 1
        while i < bound:
            j = 1
            while j < bound:
                if i + j <= bound:
                    s.add(i + j)
                if y == 1:
                    break
                j *= y
            if x == 1:
                break
            i *= x
        return list(s)

Java

class Solution {
    public List<Integer> powerfulIntegers(int x, int y, int bound) {
        Set<Integer> s = new HashSet<>();
        for (int i = 1; i < bound; i *= x) {
            for (int j = 1; j < bound; j *= y) {
                if (i + j <= bound) {
                    s.add(i + j);
                }
                if (y == 1) {
                    break;
                }
            }
            if (x == 1) {
                break;
            }
        }
        return new ArrayList<>(s);
    }
}

JavaScript

/**
 * @param {number} x
 * @param {number} y
 * @param {number} bound
 * @return {number[]}
 */
var powerfulIntegers = function(x, y, bound) {
    let res = new Set();
    for (let i = 1; i < bound; i *= x) {
        for (let j = 1; j < bound; j *= y) {
            if ((i + j) <= bound) {
                res.add(i + j);
            }
            if (y == 1) break;
        }
        if (x == 1) break;
    }
    return [...res];
};

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