Description: Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

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Example 1:

Input : nums = [1,2,3,1] Output: true

Example 2:

Input : nums = [1,2,3,4] Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2] Output: true


  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109


Brute Force:

  • Compare every element of the array with other elements. Return true if two elements are same else return false.
  • Time: O(n^2) -> for every elements the whole array is iterated making it n*n
  • Space: O(1) -> no additional memory
for i in range(len(nums)-1):
    for j in range(i+1, len(nums)):
        if nums[i] == nums[j]:
            return True
return False

Revised Solution:

  • Sort the array. Return true if two adjacent elements are same, else return false.
  • Time: O(n.log(n)) -> single loop of the array + time complexity of the sorting algo
  • Space: O(1) -> No additional memory
for i in range(len(sorted_nums)-1):
    for j in range(i+1, len(sorted_nums)):
        if sorted_nums[i] == sorted_nums[j]:
            return True
return False

Optimal Solution:

  • As you are looping through the array add it to a hash set.
  • Time: O(n) -> single loop of the array
  • Space: O(n) -> space occupied by hastset

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