Check If All 1's Are at Least Length K Places Away

Check If All 1's Are at Least Length K Places Away - Problem

Array Easy

Given a binary array nums and an integer k, return true if all 1's are at least k places away from each other, otherwise return false.

In other words, check if the distance between any two consecutive 1's in the array is at least k.

Input & Output

Example 1 — Valid Spacing

$ Input: nums = [1,0,0,0,1,0,0,1], k = 2

› Output: true

💡 Note: All 1's are at least 2 places away: positions [0,4,7], distances are 4-0=4≥2+1 and 7-4=3≥2+1

Example 2 — Invalid Spacing

$ Input: nums = [1,0,0,1,0,1], k = 2

› Output: false

💡 Note: 1's at positions [0,3,5], distance 5-3=2≤k, violates the k=2 minimum distance requirement

Example 3 — Single Element

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

› Output: true

💡 Note: Only one 1 exists, so no distance constraints to check

Constraints

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

Visualization

Tap to expand

Understanding the Visualization

Input

Binary array with 1's that need k-distance validation

Process

Check distances between consecutive 1's

Output

Return true if all distances > k, false otherwise

Key Takeaway

🎯 Key Insight: Only need to track the previous 1's position and check distance on-the-fly

Asked in

f Facebook 15 G Google 12 M Microsoft 8

The key insight is that we only need to track the distance from the previous 1 to the current 1. The optimal approach uses a single pass to check if consecutive 1's are more than k positions apart. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Pairs	O(n)	O(n)	Find all 1's positions, then check every pair of consecutive 1's
Single Pass - Track Last Position	O(n)	O(1)	Walk through array once, remembering the last 1's position

Brute Force - Check All Pairs — Algorithm Steps

Step 1: Collect all indices where nums[i] == 1
Step 2: For each consecutive pair of indices, check if distance ≥ k

Visualization

Tap to expand

Step-by-Step Walkthrough

Find Positions

Scan array and collect indices of all 1's

Check Distances

Compare each consecutive pair of positions

Result

Return false if any distance ≤ k, otherwise true

Code -

solution.c — C

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

bool solution(int* nums, int numsSize, int k) {
    // Find all positions of 1's
    int* onesPositions = (int*)malloc(numsSize * sizeof(int));
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 1) {
            onesPositions[count++] = i;
        }
    }
    
    // Check distances between consecutive 1's
    for (int i = 1; i < count; i++) {
        if (onesPositions[i] - onesPositions[i-1] <= k) {
            free(onesPositions);
            return false;
        }
    }
    
    free(onesPositions);
    return true;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    bool result = solution(nums, numsSize, k);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

One pass to find positions, one pass through positions (at most n/2)

✓ Linear Growth

Space Complexity

O(n)

Array to store all 1's positions (worst case n/2 positions)

⚡ Linearithmic Space

23.4K Views

Medium Frequency

~15 min Avg. Time

892 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

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler