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Merge pull request #3918 from kosuri-indu/err/pr-score-descreased
Concern Regarding Score Reduction and Disappearance of Pull Requests
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‎dsa-solutions/gfg-solutions/Easy problems/square-root.md

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---
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id: square-root
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title: Square Root
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sidebar_label: Square-Root
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tags:
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- Math
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- Binary Search
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description: "This document provides solutions to the problem of finding the Square Root of an integer."
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---
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## Problem
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Given an integer `x`, find the square root of `x`. If `x` is not a perfect square, then return the floor value of √x.
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### Examples
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**Example 1:**
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```
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Input: x = 5
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Output: 2
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Explanation: Since 5 is not a perfect square, the floor of the square root of 5 is 2.
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```
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**Example 2:**
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```
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Input: x = 4
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Output: 2
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Explanation: Since 4 is a perfect square, its square root is 2.
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```
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### Your Task
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You don't need to read input or print anything. The task is to complete the function `floorSqrt()` which takes `x` as the input parameter and returns its square root. Note: Try solving the question without using the sqrt function. The value of `x` ≥ 0.
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**Expected Time Complexity:** $O(log N)$
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**Expected Auxiliary Space:** $O(1)$
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**Constraints**
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- `1 ≤ x ≤ 10^7`
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## Solution
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### Intuition & Approach
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To find the square root of a number without using the built-in `sqrt` function, we can use binary search. This approach leverages the fact that the square root of `x` must lie between `0` and `x`. By repeatedly narrowing down the range using binary search, we can efficiently find the floor value of the square root.
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### Implementation
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<Tabs>
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<TabItem value="python" label="Python">
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```python
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class Solution:
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def floorSqrt(self, x: int) -> int:
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if x == 0 or x == 1:
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return x
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start, end = 1, x
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ans = 0
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while start <= end:
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mid = (start + end) // 2
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if mid * mid == x:
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return mid
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if mid * mid < x:
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start = mid + 1
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ans = mid
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else:
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end = mid - 1
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return ans
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```
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</TabItem>
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<TabItem value="java" label="Java">
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```java
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class Solution {
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long floorSqrt(long x) {
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if (x == 0 || x == 1) {
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return x;
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}
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long start = 1, end = x, ans = 0;
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while (start <= end) {
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long mid = (start + end) / 2;
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if (mid * mid == x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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<TabItem value="cpp" label="C++">
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```cpp
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class Solution {
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public:
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long long int floorSqrt(long long int x) {
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if (x == 0 || x == 1)
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return x;
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long long int start = 1, end = x, ans = 0;
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while (start <= end) {
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long long int mid = (start + end) / 2;
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if (mid * mid == x)
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return mid;
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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};
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```
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</TabItem>
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<TabItem value="javascript" label="JavaScript">
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```javascript
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class Solution {
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floorSqrt(x) {
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if (x === 0 || x === 1) {
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return x;
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}
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let start = 1,
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end = x,
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ans = 0;
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while (start <= end) {
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let mid = Math.floor((start + end) / 2);
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if (mid * mid === x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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<TabItem value="typescript" label="TypeScript">
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```typescript
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class Solution {
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floorSqrt(x: number): number {
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if (x === 0 || x === 1) {
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return x;
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}
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let start = 1,
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end = x,
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ans = 0;
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while (start <= end) {
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let mid = Math.floor((start + end) / 2);
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if (mid * mid === x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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</Tabs>
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## Complexity Analysis
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The provided solutions efficiently find the floor value of the square root of a given integer `x` using binary search. This approach ensures a time complexity of $ O(log N) and an auxiliary space complexity of $O(1)$. The algorithms are designed to handle large values of `x` up to 10^7 efficiently without relying on built-in square root functions.
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**Time Complexity:** $O(log N)$
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**Auxiliary Space:** $O(1)$
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---
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id: square-root
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title: Square Root
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sidebar_label: Square-Root
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tags:
6+
- Math
7+
- Binary Search
8+
description: "This document provides solutions to the problem of finding the Square Root of an integer."
9+
---
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11+
## Problem
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13+
Given an integer `x`, find the square root of `x`. If `x` is not a perfect square, then return the floor value of √x.
14+
15+
### Examples
16+
17+
**Example 1:**
18+
19+
```
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Input: x = 5
21+
Output: 2
22+
Explanation: Since 5 is not a perfect square, the floor of the square root of 5 is 2.
23+
```
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**Example 2:**
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27+
```
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Input: x = 4
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Output: 2
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Explanation: Since 4 is a perfect square, its square root is 2.
31+
```
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### Your Task
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35+
You don't need to read input or print anything. The task is to complete the function `floorSqrt()` which takes `x` as the input parameter and returns its square root. Note: Try solving the question without using the sqrt function. The value of `x` ≥ 0.
36+
37+
**Expected Time Complexity:** $O(log N)$
38+
**Expected Auxiliary Space:** $O(1)$
39+
40+
**Constraints**
41+
42+
- `1 ≤ x ≤ 10^7`
43+
44+
## Solution
45+
46+
### Intuition & Approach
47+
48+
To find the square root of a number without using the built-in `sqrt` function, we can use binary search. This approach leverages the fact that the square root of `x` must lie between `0` and `x`. By repeatedly narrowing down the range using binary search, we can efficiently find the floor value of the square root.
49+
50+
### Implementation
51+
52+
<Tabs>
53+
<TabItem value="python" label="Python">
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```python
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class Solution:
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def floorSqrt(self, x: int) -> int:
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if x == 0 or x == 1:
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return x
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start, end = 1, x
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ans = 0
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while start <= end:
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mid = (start + end) // 2
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if mid * mid == x:
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return mid
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if mid * mid < x:
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start = mid + 1
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ans = mid
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else:
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end = mid - 1
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return ans
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```
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</TabItem>
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<TabItem value="java" label="Java">
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```java
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class Solution {
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long floorSqrt(long x) {
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if (x == 0 || x == 1) {
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return x;
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}
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long start = 1, end = x, ans = 0;
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while (start <= end) {
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long mid = (start + end) / 2;
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if (mid * mid == x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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<TabItem value="cpp" label="C++">
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```cpp
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class Solution {
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public:
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long long int floorSqrt(long long int x) {
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if (x == 0 || x == 1)
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return x;
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long long int start = 1, end = x, ans = 0;
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while (start <= end) {
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long long int mid = (start + end) / 2;
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if (mid * mid == x)
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return mid;
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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};
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```
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</TabItem>
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<TabItem value="javascript" label="JavaScript">
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```javascript
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class Solution {
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floorSqrt(x) {
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if (x === 0 || x === 1) {
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return x;
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}
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let start = 1,
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end = x,
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ans = 0;
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while (start <= end) {
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let mid = Math.floor((start + end) / 2);
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if (mid * mid === x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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<TabItem value="typescript" label="TypeScript">
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```typescript
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class Solution {
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floorSqrt(x: number): number {
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if (x === 0 || x === 1) {
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return x;
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}
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let start = 1,
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end = x,
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ans = 0;
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while (start <= end) {
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let mid = Math.floor((start + end) / 2);
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if (mid * mid === x) {
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return mid;
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}
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if (mid * mid < x) {
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start = mid + 1;
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ans = mid;
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} else {
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end = mid - 1;
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}
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}
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return ans;
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}
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}
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```
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</TabItem>
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</Tabs>
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## Complexity Analysis
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The provided solutions efficiently find the floor value of the square root of a given integer `x` using binary search. This approach ensures a time complexity of $ O(log N) and an auxiliary space complexity of $O(1)$. The algorithms are designed to handle large values of `x` up to 10^7 efficiently without relying on built-in square root functions.
191+
192+
**Time Complexity:** $O(log N)$
193+
**Auxiliary Space:** $O(1)$

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