Distinct Numbers in Each Subarray - Practice Coding Problems

Distinct Numbers in Each Subarray - Problem

You are given an integer array nums of length n and an integer k. Your task is to find the number of distinct elements in every subarray of size k within nums.

Return an array ans such that ans[i] is the count of distinct elements in nums[i..(i + k - 1)] for each index 0 <= i < n - k.

Example: For nums = [1,2,1,3,4,2,3] and k = 4, the subarrays of size 4 are:

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

So the answer is [3,4,4,3].

Input & Output

Example 1 — Basic Case

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

› Output: [3,4,4,3]

💡 Note: Subarrays: [1,2,1,3]→3 distinct, [2,1,3,4]→4 distinct, [1,3,4,2]→4 distinct, [3,4,2,3]→3 distinct

Example 2 — Minimum Size

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

› Output: [2,2]

💡 Note: Subarrays: [1,2]→2 distinct, [2,3]→2 distinct

Example 3 — All Same Elements

$ Input: nums = [5,5,5,5], k = 3

› Output: [1,1]

💡 Note: Subarrays: [5,5,5]→1 distinct, [5,5,5]→1 distinct

Constraints

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

Visualization

Tap to expand

Understanding the Visualization

Input

Array [1,2,1,3,4,2,3] with k=4

Process

Count distinct elements in each subarray of size k

Output

Array [3,4,4,3] with counts

Key Takeaway

🎯 Key Insight: Use sliding window with frequency tracking to efficiently count distinct elements without recalculation

Asked in

a Amazon 25 M Microsoft 18 G Google 15 f Facebook 12

The sliding window with hash map approach is optimal. Track element frequencies as the window slides - when adding an element, increment its frequency (increase distinct count if frequency was 0). When removing, decrement frequency (decrease distinct count if frequency becomes 0). Time: O(n), Space: O(k)

Common Approaches

Approach	Time	Space	Notes
✓ Sliding Window with Hash Map	O(n)	O(k)	Maintain element frequencies as the window slides
Brute Force	O(n×k)	O(k)	For each subarray, use a set to count distinct elements

Sliding Window with Hash Map — Algorithm Steps

Step 1: Initialize window with first k elements
Step 2: For each slide, remove leftmost element and add new rightmost
Step 3: Track distinct count by monitoring frequency changes

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Set up frequency map for first window

Slide

Add new element, remove old element

Update Count

Adjust distinct count based on frequency changes

Code -

solution.c — C

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

#define MAX_VAL 200001

int* solution(int* nums, int numsSize, int k, int* returnSize) {
    *returnSize = numsSize - k + 1;
    if (*returnSize <= 0) {
        *returnSize = 0;
        return NULL;
    }
    
    int* result = (int*)malloc(*returnSize * sizeof(int));
    int freqMap[MAX_VAL] = {0};
    int distinctCount = 0;
    
    // Initialize first window
    for (int i = 0; i < k; i++) {
        int idx = nums[i] + 100000; // Offset for negative numbers
        if (freqMap[idx] == 0) {
            distinctCount++;
        }
        freqMap[idx]++;
    }
    
    result[0] = distinctCount;
    
    // Slide the window
    for (int i = k; i < numsSize; i++) {
        // Add new element
        int newIdx = nums[i] + 100000;
        if (freqMap[newIdx] == 0) {
            distinctCount++;
        }
        freqMap[newIdx]++;
        
        // Remove old element
        int oldIdx = nums[i - k] + 100000;
        freqMap[oldIdx]--;
        if (freqMap[oldIdx] == 0) {
            distinctCount--;
        }
        
        result[i - k + 1] = distinctCount;
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple array parsing
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        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)

Each element is added and removed exactly once as window slides

✓ Linear Growth

Space Complexity

O(k)

Hash map stores frequencies of at most k distinct elements

✓ Linear Space

23.4K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sliding Window with Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler