Find All K-Distant Indices in an Array - Practice Coding Problems

Find All K-Distant Indices in an Array - Problem

You are given a 0-indexed integer array nums and two integers key and k.

A k-distant index is an index i of nums for which there exists at least one index j such that |i - j| <= k and nums[j] == key.

Return a list of all k-distant indices sorted in increasing order.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,4,9,1,4,9], key = 4, k = 1

› Output: [0,1,2,3,4,5]

💡 Note: Key 4 appears at indices 1 and 4. For k=1: index 1 covers [0,1,2], index 4 covers [3,4,5]. Together they cover all indices.

Example 2 — Sparse Keys

$ Input: nums = [2,2,2,2,2], key = 2, k = 2

› Output: [0,1,2,3,4]

💡 Note: Key 2 appears at every index. With k=2, every position is within distance 2 of multiple key occurrences.

Example 3 — No Key Found

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

› Output: []

💡 Note: Key 4 does not appear in the array, so no indices are k-distant.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000
1 ≤ key ≤ 1000
0 ≤ k ≤ nums.length

Visualization

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Understanding the Visualization

Input

Array with key occurrences and distance parameter k

Coverage

Each key creates coverage zone of radius k

Output

All indices within any coverage zone

Key Takeaway

🎯 Key Insight: Each key occurrence creates a coverage zone, and we need all indices within any zone

Asked in

M Microsoft 15 G Google 12

The key insight is that each occurrence of the key value creates a coverage zone of radius k. The optimal approach uses range merging to efficiently handle overlapping coverage areas. Time: O(n), Space: O(m) where m is the number of key occurrences.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n²)	O(1)	For each index, check if any key occurrence is within distance k
Two-Pass Optimization	O(n × m)	O(m)	First find all key positions, then mark their coverage ranges
Range Merging	O(n)	O(m)	Calculate coverage ranges for each key and merge overlapping ranges

Brute Force — Algorithm Steps

For each index i from 0 to n-1
Check all indices j to see if |i-j| <= k and nums[j] == key
If found, add i to result list

Visualization

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Step-by-Step Walkthrough

Check Position

For each index i, search entire array for key within distance k

Find Key

If any key occurrence is within |i-j| <= k, mark position as k-distant

Collect Results

Add all k-distant indices to result array

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int key, int k, int* returnSize) {
    int* result = (int*)malloc(numsSize * sizeof(int));
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int found = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == key && abs(i - j) <= k) {
                found = 1;
                break;
            }
        }
        if (found) {
            result[count++] = i;
        }
    }
    
    *returnSize = count;
    return result;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int key, k;
    scanf("%d", &key);
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(nums, numsSize, key, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n indices, we check up to n positions for key occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for iteration, result list not counted as extra space

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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