Valid Perfect Square - Practice Coding Problems

Valid Perfect Square - Problem

Imagine you're given a mysterious number and need to determine if it's a perfect square without using any built-in square root functions - like a mathematical detective!

A perfect square is a number that can be expressed as the product of an integer with itself. For example:

16 is a perfect square because 4 × 4 = 16
25 is a perfect square because 5 × 5 = 25
14 is NOT a perfect square because no integer multiplied by itself equals 14

Your task: Given a positive integer num, return true if it's a perfect square, false otherwise.

The Challenge: You cannot use any built-in library functions like sqrt() or Math.sqrt(). You must implement your own logic to solve this mathematical puzzle!

Input & Output

example_1.py — Perfect Square

$ Input: num = 16

› Output: true

💡 Note: 16 is a perfect square because 4 × 4 = 16. The square root of 16 is exactly 4, which is an integer.

example_2.py — Not Perfect Square

$ Input: num = 14

› Output: false

💡 Note: 14 is not a perfect square. The square root of 14 is approximately 3.74, which is not an integer. No integer multiplied by itself equals 14.

example_3.py — Edge Case

$ Input: num = 1

› Output: true

💡 Note: 1 is a perfect square because 1 × 1 = 1. This is the smallest perfect square.

Constraints

1 ≤ num ≤ 2³¹ - 1
Cannot use built-in sqrt functions
Must handle integer overflow properly in calculations

Visualization

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Understanding the Visualization

Set the Playing Field

Establish search boundaries from 1 to the target number

Make Smart Guesses

Use binary search to guess the middle value and check its square

Eliminate Impossibilities

If guess² is too big, eliminate the upper half; if too small, eliminate lower half

Find the Answer

Continue until you find exact match or determine no perfect square exists

Key Takeaway

🎯 Key Insight: Binary search transforms a linear O(√n) problem into a logarithmic O(log n) solution by intelligently eliminating half the search space in each iteration

Asked in

G Google 45 a Amazon 38 ⊞ Microsoft 32 f Meta 28

The binary search approach is optimal for this problem, achieving O(log n) time complexity. Instead of checking every number linearly, we efficiently narrow down the search space by half in each iteration, making it perfect for large inputs while avoiding built-in sqrt functions.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Linear Search)	O(√n)	O(1)	Check every number from 1 to sqrt(num) by testing if i² equals num
Binary Search (Optimal)	O(log n)	O(1)	Use binary search to find the square root efficiently in O(log n) time

Brute Force (Linear Search) — Algorithm Steps

Start with i = 1
Calculate i² and compare with num
If i² == num, return true (found perfect square)
If i² > num, return false (no perfect square exists)
Otherwise, increment i and repeat

Visualization

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Step-by-Step Walkthrough

Start at 1

Begin checking from i = 1, calculate 1² = 1

Check 2

Calculate 2² = 4, compare with target

Check 3

Calculate 3² = 9, compare with target

Continue

Keep incrementing until i² = num or i² > num

Code -

solution.c — C

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

bool isPerfectSquare(int num) {
    if (num < 1) return false;
    
    long long i = 1;
    while (i * i <= num) {
        if (i * i == num) {
            return true;
        }
        i++;
    }
    
    return false;
}

int main() {
    int num;
    scanf("%d", &num);
    printf("%s\n", isPerfectSquare(num) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(√n)

We potentially check every integer from 1 to √n, where n is the input number

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space for the loop variable

✓ Linear Space

34.5K Views

Medium Frequency

~15 min Avg. Time

1.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Linear Search) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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