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Copy file name to clipboardexpand all lines: solution/0300-0399/0367.Valid Perfect Square/README_EN.md
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<!-- solution:start -->
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### Solution 1: Binary search
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### Solution 1: Binary Search
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We can use binary search to solve this problem. Define the left boundary $l = 1$ and the right boundary $r = num$ of the binary search, then find the smallest integer $x$ that satisfies $x^2 \geq num$ in the range $[l, r]$. Finally, if $x^2 = num$, then $num$ is a perfect square.
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The time complexity is $O(\log n)$, where $n$ is the given number. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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classSolution:
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defisPerfectSquare(self, num: int) -> bool:
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left, right =1, num
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while left < right:
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mid = (left + right) >>1
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if mid * mid >= num:
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right = mid
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else:
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left = mid +1
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return left * left == num
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l = bisect_left(range(1, num +1), num, key=lambdax: x * x) +1
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return l * l == num
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```
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#### Java
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```java
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classSolution {
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publicbooleanisPerfectSquare(intnum) {
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long left=1, right= num;
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while (left<right) {
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long mid = (left+right) >>>1;
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if (mid * mid >= num) {
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right= mid;
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int l=1, r= num;
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while (l<r) {
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int mid = (l+r) >>>1;
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if (1L*mid * mid >= num) {
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r= mid;
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} else {
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left= mid +1;
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l= mid +1;
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}
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}
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returnleft*left== num;
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returnl*l== num;
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}
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}
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```
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classSolution {
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public:
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bool isPerfectSquare(int num) {
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long left = 1, right = num;
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while (left < right) {
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long mid = left + right >> 1;
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if (mid * mid >= num)
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right = mid;
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else
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left = mid + 1;
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int l = 1, r = num;
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while (l < r) {
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int mid = l + (r - l) / 2;
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if (1LL * mid * mid >= num) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return left * left == num;
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return 1LL * l * l == num;
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}
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};
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```
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```go
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func isPerfectSquare(num int) bool {
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left, right := 1, num
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for left < right {
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mid := (left + right) >> 1
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if mid*mid >= num {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left*left == num
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l := sort.Search(num, func(i int) bool { return i*i >= num })
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return l*l == num
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}
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```
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#### TypeScript
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```ts
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function isPerfectSquare(num:number):boolean {
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let left =1;
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let right =num>>1;
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while (left<right) {
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const mid = (left+right) >>>1;
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if (mid*mid<num) {
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left=mid+1;
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let [l, r] = [1, num];
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while (l<r) {
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const mid = (l+r) >>1;
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if (mid>=num/mid) {
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r=mid;
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} else {
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right=mid;
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l=mid+1;
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}
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}
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returnleft*left===num;
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returnl*l===num;
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}
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```
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#### Rust
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```rust
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usestd::cmp::Ordering;
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implSolution {
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pubfnis_perfect_square(num:i32) ->bool {
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letnum:i64=numasi64;
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letmutleft=1;
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letmutright=num>>1;
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whileleft<right {
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letmid=left+ (right-left) /2;
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match (mid*mid).cmp(&num) {
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Ordering::Less=> {
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left=mid+1;
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}
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Ordering::Greater=> {
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right=mid-1;
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}
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Ordering::Equal=> {
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returntrue;
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}
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letmutl=1;
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letmutr=numasi64;
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whilel<r {
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letmid= (l+r) /2;
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ifmid*mid>= (numasi64) {
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r=mid;
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} else {
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l=mid+1;
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}
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}
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left*left==num
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l*l==(numasi64)
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}
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}
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```
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<!-- solution:start -->
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### Solution 2: Math trick
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### Solution 2: Mathematics
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This is a math problem：
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Since $1 + 3 + 5 + \cdots + (2n - 1) = n^2$, we can gradually subtract $1, 3, 5, \cdots$ from $num$. If $num$ finally equals $0$, then $num$ is a perfect square.
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```bash
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1 = 1
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4 = 1 + 3
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9 = 1 + 3 + 5
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16 = 1 + 3 + 5 + 7
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25 = 1 + 3 + 5 + 7 + 9
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36 = 1 + 3 + 5 + 7 + 9 + 11
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....
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so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = n²
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```
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The time complexity is $O(\sqrt n)$, and the space complexity is $O(1)$.
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