Find-the-Duplicate-Number

Author: Shikha-code36

README.md

@@ -36,4 +36,5 @@ Check the notes for the explaination - [Notes](https://stingy-shallot-4ea.notion
 - [x] [Two Pointers](Two-Pointers)
     - [x] [Squares of a Sorted Array](Two-Pointers/977-Squares-of-a-Sorted-Array.py)
     - [x] [Backspace String Compare](Two-Pointers/844-Backspace-String-Compare.py)
+    - [x] [Find the Duplicate Number](Two-Pointers/287-Find-the-Duplicate-Number.py)
 

Two-Pointers/287-Find-the-Duplicate-Number.py

@@ -0,0 +1,79 @@
+"Leetcode- https://leetcode.com/problems/find-the-duplicate-number/ "
+'''
+Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.
+
+There is only one repeated number in nums, return this repeated number.
+
+You must solve the problem without modifying the array nums and uses only constant extra space.
+
+Example 1:
+
+Input: nums = [1,3,4,2,2]
+Output: 2
+'''
+
+# Solution-1
+def findDuplicate(self, nums):
+    nums.sort()
+
+    for i in range(len(nums)):
+        if nums[i] == nums[i-1]:
+            return nums[i]
+
+# T:O(nlogn)
+# S:O(n)
+
+# Solution-2- using set
+def findDuplicate(self, nums):
+    seen = set()
+    for num in nums:
+        if num in seen:
+            return num
+        seen.add(num)
+
+# T:O(n)
+# S:O(n)
+
+# Solution-3- Binary Search
+def findDuplicate(self, nums):
+    # 'low' and 'high' represent the range of values of the target
+    low = 1
+    high = len(nums) - 1
+
+    while low <= high:
+        cur = (low + high) // 2
+        count = 0
+
+        # Count how many numbers are less than or equal to 'cur'
+        count = sum(num <= cur for num in nums)
+        if count > cur:
+            duplicate = cur
+            high = cur - 1
+        else:
+            low = cur + 1
+
+    return duplicate
+
+# T:O(nlogn)
+# S:O(1)
+
+# Solution-4- Two Pointers Floyd's Tortoise and Hare (Cycle Detection)
+def findDuplicate(self, nums):
+     # Find the intersection point of the two runners.
+    to = hr = nums[0]
+    while True:
+        to = nums[to]
+        hr = nums[nums[hr]]
+        if to == hr:
+            break
+        
+    # Find the "entrance" to the cycle.
+    to = nums[0]
+    while to != hr:
+        to = nums[to]
+        hr = nums[hr]
+        
+    return hr
+    
+# T:O(n)
+# S:O(1)