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feat: add solutions to lc problem: No.0970
No.0970.Powerful Integers
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‎solution/0900-0999/0970.Powerful Integers/README.md

Lines changed: 101 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -48,6 +48,18 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表 + 枚举**
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根据题目描述，一个强整数可以表示成 $x^i + y^j$，其中 $i \geq 0$, $j \geq 0$。
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题目需要我们找出所有不超过 $bound$ 的强整数，我们注意到 $bound$ 的取值范围不超过 $10^6$，而 $2^{20} = 1048576 \gt 10^6$。因此，如果 $x \geq 2$，那么 $i$ 最大不超过 $20$，才有可能使得 $x^i + y^j \leq bound$ 成立。同理，如果 $y \geq 2$，那么 $j$ 最大不超过 $20$。
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因此我们可以使用双重循环，枚举所有可能的 $x^i$ 和 $y^j$，分别记为 $a$ 和 $b$，并保证 $a + b \leq bound$，此时 $a + b$ 即为一个强整数。我们使用哈希表存储所有满足条件的强整数，最后将哈希表中的所有元素转换成答案列表返回即可。
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> 注意，如果 $x=1$ 或者 $y=1$，那么 $a$ 或者 $b$ 的值恒等于 $1$，对应的循环只需要执行一次即可退出。
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时间复杂度 $O(\log^2 bound)$，空间复杂度 $O(\log^2 bound)$。
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<!-- tabs:start -->
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### **Python3**
@@ -57,20 +69,19 @@
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```python
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class Solution:
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def powerfulIntegers(self, x: int, y: int, bound: int) -> List[int]:
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s = set()
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i = 1
62-
while i < bound:
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j = 1
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while j < bound:
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if i + j <= bound:
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s.add(i + j)
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ans = set()
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a = 1
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while a <= bound:
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b = 1
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while a + b <= bound:
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ans.add(a + b)
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b *= y
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if y == 1:
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break
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j *= y
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if x == 1:
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break
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i *= x
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return list(s)
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a *= x
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return list(ans)
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```
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### **Java**
@@ -80,12 +91,33 @@ class Solution:
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```java
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class Solution {
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public List<Integer> powerfulIntegers(int x, int y, int bound) {
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Set<Integer> s = new HashSet<>();
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for (int i = 1; i < bound; i *= x) {
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for (int j = 1; j < bound; j *= y) {
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if (i + j <= bound) {
87-
s.add(i + j);
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Set<Integer> ans = new HashSet<>();
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for (int a = 1; a <= bound; a *= x) {
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for (int b = 1; a + b <= bound; b *= y) {
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ans.add(a + b);
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if (y == 1) {
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break;
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}
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}
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if (x == 1) {
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break;
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}
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}
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return new ArrayList<>(ans);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> powerfulIntegers(int x, int y, int bound) {
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unordered_set<int> ans;
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for (int a = 1; a <= bound; a *= x) {
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for (int b = 1; a + b <= bound; b *= y) {
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ans.insert(a + b);
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if (y == 1) {
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break;
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}
@@ -94,8 +126,51 @@ class Solution {
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break;
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}
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}
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return new ArrayList<>(s);
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return vector<int>(ans.begin(), ans.end());
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}
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};
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```
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### **Go**
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```go
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func powerfulIntegers(x int, y int, bound int) (ans []int) {
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s := map[int]struct{}{}
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for a := 1; a <= bound; a *= x {
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for b := 1; a+b <= bound; b *= y {
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s[a+b] = struct{}{}
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if y == 1 {
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break
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}
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}
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if x == 1 {
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break
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}
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}
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for x := range s {
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ans = append(ans, x)
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function powerfulIntegers(x: number, y: number, bound: number): number[] {
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const ans = new Set<number>();
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for (let a = 1; a <= bound; a *= x) {
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for (let b = 1; a + b <= bound; b *= y) {
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ans.add(a + b);
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if (y === 1) {
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break;
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}
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}
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if (x === 1) {
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break;
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}
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}
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return Array.from(ans);
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}
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```
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@@ -109,17 +184,19 @@ class Solution {
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* @return {number[]}
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*/
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var powerfulIntegers = function (x, y, bound) {
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let res = new Set();
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for (let i = 1; i < bound; i *= x) {
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for (let j = 1; j < bound; j *= y) {
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if (i + j <= bound) {
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res.add(i + j);
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const ans = new Set();
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for (let a = 1; a <= bound; a *= x) {
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for (let b = 1; a + b <= bound; b *= y) {
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ans.add(a + b);
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if (y === 1) {
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break;
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}
118-
if (y == 1) break;
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}
120-
if (x == 1) break;
195+
if (x === 1) {
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break;
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}
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}
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return [...res];
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return [...ans];
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};
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```
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‎solution/0900-0999/0970.Powerful Integers/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -43,40 +43,72 @@
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## Solutions
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**Solution 1: Hash Table + Enumeration**
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According to the description of the problem, a powerful integer can be represented as $x^i + y^j$, where $i \geq 0$, $j \geq 0$.
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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 $20$.
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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.
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> 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.
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The time complexity is $O(\log^2 bound)$, and the space complexity is $O(\log^2 bound)$.
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<!-- tabs:start -->
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### **Python3**
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```python
5163
class Solution:
5264
def powerfulIntegers(self, x: int, y: int, bound: int) -> List[int]:
53-
s = set()
54-
i = 1
55-
while i < bound:
56-
j = 1
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while j < bound:
58-
if i + j <= bound:
59-
s.add(i + j)
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ans = set()
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a = 1
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while a <= bound:
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b = 1
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while a + b <= bound:
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ans.add(a + b)
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b *= y
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if y == 1:
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break
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j *= y
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if x == 1:
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break
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i *= x
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return list(s)
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a *= x
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return list(ans)
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```
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### **Java**
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```java
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class Solution {
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public List<Integer> powerfulIntegers(int x, int y, int bound) {
74-
Set<Integer> s = new HashSet<>();
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for (int i = 1; i < bound; i *= x) {
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for (int j = 1; j < bound; j *= y) {
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if (i + j <= bound) {
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s.add(i + j);
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Set<Integer> ans = new HashSet<>();
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for (int a = 1; a <= bound; a *= x) {
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for (int b = 1; a + b <= bound; b *= y) {
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ans.add(a + b);
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if (y == 1) {
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break;
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}
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}
93+
if (x == 1) {
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break;
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}
96+
}
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return new ArrayList<>(ans);
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}
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}
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```
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### **C++**
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```cpp
105+
class Solution {
106+
public:
107+
vector<int> powerfulIntegers(int x, int y, int bound) {
108+
unordered_set<int> ans;
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for (int a = 1; a <= bound; a *= x) {
110+
for (int b = 1; a + b <= bound; b *= y) {
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ans.insert(a + b);
80112
if (y == 1) {
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break;
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}
@@ -85,8 +117,51 @@ class Solution {
85117
break;
86118
}
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}
88-
return new ArrayList<>(s);
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return vector<int>(ans.begin(), ans.end());
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}
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};
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```
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### **Go**
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```go
128+
func powerfulIntegers(x int, y int, bound int) (ans []int) {
129+
s := map[int]struct{}{}
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for a := 1; a <= bound; a *= x {
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for b := 1; a+b <= bound; b *= y {
132+
s[a+b] = struct{}{}
133+
if y == 1 {
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break
135+
}
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}
137+
if x == 1 {
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break
139+
}
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}
141+
for x := range s {
142+
ans = append(ans, x)
143+
}
144+
return ans
145+
}
146+
```
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### **TypeScript**
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```ts
151+
function powerfulIntegers(x: number, y: number, bound: number): number[] {
152+
const ans = new Set<number>();
153+
for (let a = 1; a <= bound; a *= x) {
154+
for (let b = 1; a + b <= bound; b *= y) {
155+
ans.add(a + b);
156+
if (y === 1) {
157+
break;
158+
}
159+
}
160+
if (x === 1) {
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break;
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}
89163
}
164+
return Array.from(ans);
90165
}
91166
```
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100175
* @return {number[]}
101176
*/
102177
var powerfulIntegers = function (x, y, bound) {
103-
let res = new Set();
104-
for (let i = 1; i < bound; i *= x) {
105-
for (let j = 1; j < bound; j *= y) {
106-
if (i + j <= bound) {
107-
res.add(i + j);
178+
const ans = new Set();
179+
for (let a = 1; a <= bound; a *= x) {
180+
for (let b = 1; a + b <= bound; b *= y) {
181+
ans.add(a + b);
182+
if (y === 1) {
183+
break;
108184
}
109-
if (y == 1) break;
110185
}
111-
if (x == 1) break;
186+
if (x === 1) {
187+
break;
188+
}
112189
}
113-
return [...res];
190+
return [...ans];
114191
};
115192
```
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‎solution/0900-0999/0970.Powerful Integers/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
class Solution {
2+
public:
3+
vector<int> powerfulIntegers(int x, int y, int bound) {
4+
unordered_set<int> ans;
5+
for (int a = 1; a <= bound; a *= x) {
6+
for (int b = 1; a + b <= bound; b *= y) {
7+
ans.insert(a + b);
8+
if (y == 1) {
9+
break;
10+
}
11+
}
12+
if (x == 1) {
13+
break;
14+
}
15+
}
16+
return vector<int>(ans.begin(), ans.end());
17+
}
18+
};

‎solution/0900-0999/0970.Powerful Integers/Solution.go

Lines changed: 18 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
func powerfulIntegers(x int, y int, bound int) (ans []int) {
2+
s := map[int]struct{}{}
3+
for a := 1; a <= bound; a *= x {
4+
for b := 1; a+b <= bound; b *= y {
5+
s[a+b] = struct{}{}
6+
if y == 1 {
7+
break
8+
}
9+
}
10+
if x == 1 {
11+
break
12+
}
13+
}
14+
for x := range s {
15+
ans = append(ans, x)
16+
}
17+
return ans
18+
}

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