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
Example 2:
Input: nums = [3,1,3,4,2] Output: 3
Constraints:
1 <= n <= 105
nums.length == n + 1
1 <= nums[i] <= n
- All the integers in
nums
appear only once except for precisely one integer which appears two or more times.
Find the Duplicate Number Solutions
✅Time: O(n)
✅Space: O(1)
C++
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int slow = nums[nums[0]];
int fast = nums[nums[nums[0]]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
slow = nums[0];
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
};
Java
class Solution {
public int findDuplicate(int[] nums) {
int slow = nums[nums[0]];
int fast = nums[nums[nums[0]]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
slow = nums[0];
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
}
Python
class Solution:
def findDuplicate(self, nums: List[int]) -> int:
slow = nums[nums[0]]
fast = nums[nums[nums[0]]]
while slow != fast:
slow = nums[slow]
fast = nums[nums[fast]]
slow = nums[0]
while slow != fast:
slow = nums[slow]
fast = nums[fast]
return slow
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