Program to find duplicate element from n+1 numbers ranging from 1 to n in Python

Suppose we have a list of numbers called nums of length n + 1. These numbers are picked from range 1, 2, ..., n. As we know, using the pigeonhole principle, there must be a duplicate. We have to find that and return it.

So, if the input is like [2, 1, 4, 3, 3], then the output will be 3.

Approach

To solve this, we will follow these steps −

• l := size of nums
• temp := l*(l-1) /2
• temp_sum := sum of all elements in nums
• return (temp_sum - temp)

The logic is based on the mathematical formula. For numbers 1 to n, the sum is n*(n+1)/2. Since we have n+1 numbers where one is duplicate, the extra value is our duplicate number.

Example

Let us see the following implementation to get better understanding −

class Solution:
    def solve(self, nums):
        l = len(nums)
        temp = l*(l-1)/2
        temp_sum = sum(nums)
        return int(temp_sum-temp)

ob = Solution()
print(ob.solve([2, 1, 4, 3, 3]))

The output of the above code is −

3

Alternative Method Using Set

We can also solve this problem using a set to track seen numbers ?

def find_duplicate(nums):
    seen = set()
    for num in nums:
        if num in seen:
            return num
        seen.add(num)

numbers = [2, 1, 4, 3, 3]
print(find_duplicate(numbers))

3

Comparison

Conclusion

The mathematical approach using sum formula is more memory-efficient with O(1) space complexity. The set-based approach is more intuitive but uses O(n) extra space to track seen numbers.