Find the Duplicate Number

Introduction

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

Note:

• You must not modify the array (assume the array is read only).
• You must use only constant, O(1) extra space.
• Your runtime complexity should be less than O(n2).
• There is only one duplicate number in the array, but it could be repeated more than once.

Solution

Simple Solution:

func findDuplicate(nums []int) int {

    n := len(nums)
    seen := make([]bool, n+1)

    for _, n := range nums {

        if ok := seen[n]; ok {
          return n
        }

        seen[n] = true
    }

    return -1
}

Cycle Detection Solution:

func findDuplicate(nums []int) int {

    n := len(nums)

    if n <= 1 {
        return -1
    }

    for i, j := nums[0], nums[nums[0]];; i, j = nums[i], nums[nums[j]] {
        if i == j {
            meet := i
            if meet == 0 {
               return 0
            }
           for i, j := nums[meet], nums[0];; i, j = nums[i], nums[j] {
               if i == j {
                   return i
               }
           }             
        }
    }

    return -1
}

Explanation

Simple solution based on fact that numbers are in range [1..n], that gives ability to build lookup array seen. The disadvantage of this approach is memory consumption.

Another way is to solve the problem by transforming this task to Linked List In Cycle, where i is the node reference, and the nums[i] is the next node index.