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Contains Duplicate (easy)
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Problem Statement
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.
Examples
Example 1:
Input: nums= [1, 2, 3, 4]
Output: false
Explanation: There are no duplicates in the given array.
Example 2:
Input: nums= [1, 2, 3, 1]
Output: true
Explanation: '1' is repeating.
Example 3:
Input: nums= [3, 2, 6, -1, 2, 1]
Output: true
Explanation: '2' is repeating.
Constraints:
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Try it yourself
Try solving this question here:
For a detailed solution, see the next lesson.
Mohamed Zehraoui
· 22 days ago
type Solution struct{} // containsDuplicate checks for duplicates in a slice of integers func (s *Solution) containsDuplicate(nums []int) bool { seen := make(map[int]bool) for _, num := range nums { if seen[num] { return true } seen[num] = true } return false }
Shubham Pokale
· a month ago
A one liner solution to this would be like :
return len(set(nums)) != len(nums)
Shraddha
· 2 years ago
package main
import "fmt"
// containsDuplicate checks for duplicates in a slice of integers
func containsDuplicate(nums []int) bool {
if len(nums) == 0 || nums == nil {
return false
}
for i := 0; i< len(nums) - 1; i++ {
for j := i+1 ; j < len(nums) ; j++ {
if nums[i] == nums[j] {
return true
}
}
}
// ToDo: Write Your Code Here.
return false
}
func main() {
arr := [][]int{
{1,2,3,4},
{1,2,3,1},
}
for _, input := range arr {
isDuplicate := containsDuplicate(input)
fmt.Println(isDuplicate)
}
}
pratiksh.patel91
· 3 years ago
Computing plan
Brute Force: Compare each element with every other element in the array. If any two elements are equal, return true. Time complexity: O(n^2), space complexity: O(1).
Sorting: Sort the array and check if adjacent elements are equal. Time complexity: O(n log n) due to sorting, space complexity: O(1).
Hash Set: Use a HashSet to store unique elements. If adding an element fails (because it’s already in the set), return true. Otherwise, return false. Time complexity: O(n) for the loop, space complexity: O(n) for the HashSet.
Gokul Rama
· 3 years ago
public boolean containsDuplicate(int[] nums) { if (nums == null || nums.length == 0) { return false; } // O(N*logN) - sorting Arrays.sort(nums); // O(N) - iteration for (int i=1; i<nums.length; i++) { if (nums[i-1] == nums[i]) { return true; } } return false; }
Shane
· 3 years ago
rust support soon?
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Problem Statement
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