From b21c4ec67fcba793eec9f05b05f8a944f7112bec Mon Sep 17 00:00:00 2001
From: Mohammad Shakib <50780268+Mo-Shakib@users.noreply.github.com>
Date: Thu, 31 Oct 2024 02:54:26 +0600
Subject: [PATCH] Accepted

---
 easy/231-power-of-two.py | 75 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 75 insertions(+)
 create mode 100644 easy/231-power-of-two.py

diff --git a/easy/231-power-of-two.py b/easy/231-power-of-two.py
new file mode 100644
index 0000000..4f3dc82
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+++ b/easy/231-power-of-two.py
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+#
+# @lc app=leetcode id=231 lang=python3
+#
+# [231] Power of Two
+#
+# https://leetcode.com/problems/power-of-two/description/
+#
+# algorithms
+# Easy (48.04%)
+# Likes:    6958
+# Dislikes: 447
+# Total Accepted:    1.5M
+# Total Submissions: 3.1M
+# Testcase Example:  '1'
+#
+# Given an integer n, return true if it is a power of two. Otherwise, return
+# false.
+# 
+# An integer n is a power of two, if there exists an integer x such that n ==
+# 2^x.
+# 
+# 
+# Example 1:
+# 
+# 
+# Input: n = 1
+# Output: true
+# Explanation: 2^0 = 1
+# 
+# 
+# Example 2:
+# 
+# 
+# Input: n = 16
+# Output: true
+# Explanation: 2^4 = 16
+# 
+# 
+# Example 3:
+# 
+# 
+# Input: n = 3
+# Output: false
+# 
+# 
+# 
+# Constraints:
+# 
+# 
+# -2^31 <= n <= 2^31 - 1
+# 
+# 
+# 
+# Follow up: Could you solve it without loops/recursion?
+#
+
+# @lc code=start
+class Solution:
+    def isPowerOfTwo(self, n: int) -> bool:
+        if (n==1):
+            return True
+        
+        n_bin = bin(n)
+        n_bin = n_bin[2:]
+        
+        if (n_bin[0]!= '1'):
+            return False
+        
+        elif (n_bin.count('1')>1):
+            return False
+        
+        return True
+                
+# @lc code=end
+