Count Distinct in Windows - Problem

Given an array of integers and a window size K, count the number of distinct elements in every contiguous window of size K.

For example, if the array is [1, 2, 1, 3, 4, 2, 3] and K = 4, then the windows are:

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

Return an array containing the count of distinct elements for each window.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,1,3,4,2,3], k = 4

› Output: [3,4,4,3]

💡 Note: Window [1,2,1,3] has distinct elements {1,2,3} = 3 count. Window [2,1,3,4] has {1,2,3,4} = 4 count. Window [1,3,4,2] has {1,2,3,4} = 4 count. Window [3,4,2,3] has {2,3,4} = 3 count.

Example 2 — All Same Elements

$ Input: nums = [1,1,1,1,1], k = 3

› Output: [1,1,1]

💡 Note: Every window of size 3 contains only the element 1, so distinct count is always 1.

Example 3 — All Different Elements

$ Input: nums = [1,2,3,4], k = 2

› Output: [2,2,2]

💡 Note: Window [1,2] has 2 distinct, [2,3] has 2 distinct, [3,4] has 2 distinct elements.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ k ≤ nums.length
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

G Google 42 a Amazon 38 M Microsoft 29 f Meta 24

The key insight is to use a sliding window with frequency tracking instead of recalculating each window from scratch. The optimal sliding window approach maintains a HashMap to track element frequencies and adjusts the distinct count as elements enter and leave the window. Time: O(n), Space: O(k).

Common Approaches

✓ Brute Force - Check Each Window Separately

⏱️ Time: O(n*k) Space: O(k)

Examine every possible window of size K and count distinct elements using a set. For each window, create a new set and add all elements to count unique values.

Sliding Window with HashMap

⏱️ Time: O(n) Space: O(k)

Maintain a sliding window of size K and use a frequency map to track element counts. When sliding the window, remove the leftmost element and add the rightmost element, updating counts efficiently.

Brute Force - Check Each Window Separately — Algorithm Steps

Step 1: Iterate through all possible window positions (0 to n-k)
Step 2: For each window, create a new set and add all K elements
Step 3: Count the size of the set as distinct elements
Step 4: Store the count in result array

Visualization

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Step-by-Step Walkthrough

Window 1

Create new set, add [1,2,1,3], count = 3

Window 2

Create new set, add [2,1,3,4], count = 4

Continue

Repeat for all remaining windows

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* nums, int numsSize, int k, int* returnSize) {
    if (!nums || k <= 0 || k > numsSize) {
        *returnSize = 0;
        return NULL;
    }
    
    *returnSize = numsSize - k + 1;
    int* result = (int*)malloc(*returnSize * sizeof(int));
    
    // Check each window of size k
    for (int i = 0; i <= numsSize - k; i++) {
        int distinct[1001] = {0}; // Assuming nums[i] in range [-500, 500]
        int count = 0;
        
        // Add all elements in current window
        for (int j = i; j < i + k; j++) {
            int idx = nums[j] + 500; // Offset for negative numbers
            if (distinct[idx] == 0) {
                distinct[idx] = 1;
                count++;
            }
        }
        result[i] = count;
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL && numsSize < 1000) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(nums, numsSize, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n*k)

For each of (n-k+1) windows, we process k elements

✓ Linear Growth

Space Complexity

O(k)

Set stores at most k distinct elements per window

✓ Linear Space

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Count Distinct in Windows - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Window Separately — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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