Day 7 - Count distinct elements in every window.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	two-pointer-algorithm sliding-window Hash

🚀 Day 7. Count Distinct Elements in Every Window 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given an integer array arr[] and a number k, find the count of distinct elements in every window of size k in the array.

🔍 Example Walkthrough:

Input:
arr[] = [1, 2, 1, 3, 4, 2, 3], k = 4
Output:
[3, 4, 4, 3]

Explanation:

• Window 1: [1, 2, 1, 3] → 3 distinct elements (1, 2, 3)
• Window 2: [2, 1, 3, 4] → 4 distinct elements (1, 2, 3, 4)
• Window 3: [1, 3, 4, 2] → 4 distinct elements (1, 2, 3, 4)
• Window 4: [3, 4, 2, 3] → 3 distinct elements (2, 3, 4)

Input:
arr[] = [4, 1, 1], k = 2
Output:
[2, 1]

Explanation:

• Window 1: [4, 1] → 2 distinct elements (4, 1)
• Window 2: [1, 1] → 1 distinct element (1)

Input:
arr[] = [1, 1, 1, 1, 1], k = 3
Output:
[1, 1, 1]

Explanation:

• Each window has only 1 distinct element (1) as all values in arr are identical.

Constraints

• $( 1 \leq k \leq \text{arr.size()} \leq 10^5 ) $
• $( 1 \leq \text{arr}[i] \leq 10^5 ) $

🎯 My Approach:

1. Sliding Window with Frequency Map:

   • Use an unordered map (or dictionary) to keep track of the frequency of elements in the current window.
   • Iterate over the array while maintaining the size of the window as k.

2. Steps:

   • Add the current element to the frequency map and increment its count.
   • If the current index exceeds k - 1, count the distinct elements by checking the size of the map.
   • Remove the element that is sliding out of the window by decrementing its count. If its count becomes 0, remove it from the map.

3. Edge Cases:

   • If $( k = 1 )$, each element is its own window, so the count of distinct elements is $( n )$.
   • If $( n \leq k )$, handle the window size appropriately without exceeding bounds.

🕒 Time and Auxiliary Space Complexity

Expected Time Complexity: $( O(n) )$, where $( n )$ is the size of the array. Each element is added to and removed from the map at most once.

Expected Auxiliary Space Complexity: $( O(n) )$, as the size of the frequency map is proportional to the size of the window.

📝 Solution Code

Code (C++)

class Solution {
public:
    vector<int> countDistinct(vector<int> &arr, int k) {
        unordered_map<int, int> freq;
        vector<int> res;
        for (int i = 0; i < arr.size(); i++) {
            freq[arr[i]]++;
            if (i >= k - 1) {
                res.push_back(freq.size());
                if (--freq[arr[i - k + 1]] == 0) freq.erase(arr[i - k + 1]);
            }
        }
        return res;
    }
};

Code (Java)

class Solution {
    ArrayList<Integer> countDistinct(int arr[], int k) {
        HashMap<Integer, Integer> freq = new HashMap<>();
        ArrayList<Integer> res = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1);
            if (i >= k - 1) {
                res.add(freq.size());
                freq.put(arr[i - k + 1], freq.get(arr[i - k + 1]) - 1);
                if (freq.get(arr[i - k + 1]) == 0) freq.remove(arr[i - k + 1]);
            }
        }
        return res;
    }
}

Code (Python)

class Solution:
    def countDistinct(self, arr, k):
        freq = {}
        res = []
        for i in range(len(arr)):
            freq[arr[i]] = freq.get(arr[i], 0) + 1
            if i >= k - 1:
                res.append(len(freq))
                freq[arr[i - k + 1]] -= 1
                if freq[arr[i - k + 1]] == 0:
                    del freq[arr[i - k + 1]]
        return res

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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