grid47 Exploring patterns and algorithms
Oct 16, 2024
4 min read
Solution to LeetCode 217: Contains Duplicate Problem
Given an array of integers, determine if there are any duplicate values present. Return true if any value appears more than once, otherwise return false.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of integers nums.
Example: [10, 20, 30, 10]
Constraints:
• 1 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9
Output: Return true if there are duplicates in the array, otherwise return false.
Example: For input [10, 20, 30, 10], the output is true.
Constraints:
Goal: Identify whether there are any duplicate values in the array.
Steps:
• Use a set or map to track the elements in the array as you iterate through it.
• If you encounter an element that is already in the set or map, return true indicating a duplicate is found.
• If no duplicates are found after processing all elements, return false.
Goal: The solution must handle arrays up to the size of 10^5 efficiently.
Steps:
• The array size will be at most 10^5 elements.
• Element values can range from -10^9 to 10^9.
Assumptions:
• The array will not be empty.
• Input: Example 1
• Explanation: In this example, the value 10 occurs twice, so the answer is true.
• Input: Example 2
• Explanation: In this case, all elements are distinct, so the output is false.
• Input: Example 3
• Explanation: Here, several elements such as 7, 8, and 4 repeat, so the output is true.
Approach: Use a set or a map to efficiently track elements and identify duplicates.
Observations:
• A set or map can help efficiently identify duplicates in O(n) time by keeping track of seen elements.
• By leveraging the properties of a set or map (constant-time lookups), we can check for duplicates while iterating through the array.
Steps:
• Initialize an empty set or map.
• Iterate through the array.
• For each element, check if it exists in the set or map.
• If it does, return true (indicating a duplicate).
• If it does not, add it to the set or map.
• If no duplicates are found, return false.
Empty Inputs:
• The array will always have at least one element.
Large Inputs:
• Ensure the solution works efficiently for arrays with 10^5 elements.
Special Values:
• The array may contain very large or very small integers, but the solution must handle all values within the range.
Defines the 'containsDuplicate' function which takes a vector of integers 'nums' as input and returns a boolean value indicating whether any element in the vector appears more than once.
2 : Map Initialization
map<int, int> ma;
Initializes a map 'ma' to keep track of the count of each integer in the 'nums' vector. The key is the integer, and the value is its count.
3 : Loop Iteration
for(intx: nums)
Iterates over each element 'x' in the 'nums' vector.
4 : Check for Duplicates
if(ma.count(x)) returntrue;
Checks if the current element 'x' is already present in the map 'ma'. If it is, that means a duplicate has been found, and the function returns 'true'.
5 : Map Update
else ma[x] =1;
If the element 'x' is not found in the map, it is added with a count of 1, indicating that this element has been encountered once.
6 : Return Statement
returnfalse;
If no duplicates are found after iterating through all elements in the vector, the function returns 'false'.
Best Case: O(n), where n is the length of the array. This occurs when the first duplicate is encountered early in the iteration.
Average Case: O(n), as we perform one lookup and one insertion for each element.
Worst Case: O(n), when no duplicates are found and we need to process every element.
Description: The solution operates in linear time since the set or map allows constant-time lookups and insertions.
Best Case: O(1), if there are no duplicates and only a single element is stored.
Worst Case: O(n), where n is the number of elements in the array, since we may need to store all elements in the set or map.
Description: The space complexity is determined by the number of unique elements in the array.
