Suppose we have a list of elements called nums of size n + 1, they are selected from range 1, 2, ..., n. As we know, by pigeonhole principle, there must be a duplicate. We have to find the duplicate. Here our target is to find the task in O(n) time and constant space.
So, if the input is like nums = [2,1,4,3,5,4], then the output will be 4
To solve this, we will follow these steps −
q := sum of all elements present in nums
n := size of nums
v := floor of ((n - 1)*(n)/2)
return q - v
Example
Let us see the following implementation to get better understanding
def solve(nums): q = sum(nums) n = len(nums) v = (n - 1) * (n) // 2 return q - v nums = [2,1,4,3,5,4] print(solve(nums))
Input
[2,1,4,3,5,4]
Output
4