Valid Perfect Square

Valid Perfect Square - Problem

Given a positive integer num, return true if num is a perfect square or false otherwise.

A perfect square is an integer that is the square of an integer. In other words, it is the product of some integer with itself.

You must not use any built-in library function, such as sqrt.

Input & Output

Example 1 — Perfect Square

$ Input: num = 16

› Output: true

💡 Note: 16 is a perfect square because 4 * 4 = 16

Example 2 — Not Perfect Square

$ Input: num = 14

› Output: false

💡 Note: 14 is not a perfect square. √14 ≈ 3.74, which is not an integer

Example 3 — Edge Case

$ Input: num = 1

› Output: true

💡 Note: 1 is a perfect square because 1 * 1 = 1

Constraints

1 ≤ num ≤ 2³¹ - 1

Visualization

Tap to expand

Asked in

f Facebook 15 in LinkedIn 12

The key insight is to find the integer square root efficiently without using built-in sqrt function. Best approach is binary search with Time: O(log n), Space: O(1), offering the optimal balance of speed and simplicity.

Common Approaches

✓ Binary Search

⏱️ Time: O(log n) Space: O(1)

Instead of checking every number linearly, use binary search on the range [1, num] to quickly narrow down to the exact square root.

Brute Force

⏱️ Time: O(√n) Space: O(1)

Iterate through all possible square roots from 1 up to num, calculating each square until we find a match or exceed num.

Newton's Method

⏱️ Time: O(log log n) Space: O(1)

Apply Newton's method for finding square roots, which converges quickly to the exact value. Start with an initial guess and refine it iteratively.

Binary Search — Algorithm Steps

Set left=1, right=num
Calculate mid and mid²
If mid²=num, return true
If mid²<num, search right half; else search left half
Continue until found or exhausted

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Set left=1, right=num/2

Calculate mid, compare mid² with num

Adjust

Move left or right pointer based on comparison

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool solution(int num) {
    if (num == 1) {
        return true;
    }
    
    long long left = 1, right = num / 2;
    
    while (left <= right) {
        long long mid = (left + right) / 2;
        long long square = mid * mid;
        
        if (square == num) {
            return true;
        } else if (square < num) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return false;
}

int main() {
    int num;
    scanf("%d", &num);
    
    bool result = solution(num);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Binary search divides the search space in half each iteration

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for left, right, and mid pointers

✓ Linear Space

125.0K Views

Medium Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler