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Program to check number is perfect square or not without sqrt function in Python
Suppose we have a number n, we have to check whether n is a perfect square number or not. A perfect square number k can be represented as k = a * a for some integer a. We have to solve this without using built-in square root function.
So, if the input is like n = 121, then the output will be True because 121 = 11*11.
To solve this, we will follow these steps −
if n is same as 0 or n is same as 1, then
return True
start := 2
stop := floor of n / 2
while start <= stop, do
temp := a list of all numbers from start to stop
k := middle element of temp
k_squared := k * k
if k_squared is same as n, then
return True
if k_squared > n, then
start := temp[0]
stop := k - 1
otherwise,
start := k + 1
stop := last element of temp
return False
Example
Let us see the following implementation to get better understanding
def solve(n): if n == 0 or n == 1: return True start = 2 stop = n // 2 while start <= stop: temp = range(start, stop + 1) k = temp[len(temp) // 2] k_squared = k * k if k_squared == n: return True if k_squared > n: start = temp[0] stop = k - 1 else: start = k + 1 stop = temp[-1] return False n = 121 print(solve(n))
Input
121
Output
True
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