Check if All the Integers in a Range Are Covered

Check if All the Integers in a Range Are Covered - Problem

Imagine you're a security guard monitoring a building with multiple surveillance cameras. Each camera covers a specific range of floors, and you need to ensure that every floor in a critical section is under surveillance.

You are given a 2D integer array ranges where each ranges[i] = [start_i, end_i] represents an inclusive interval of floors covered by camera i. You also have two integers left and right representing the critical section that needs complete coverage.

Your task: Return true if every single floor in the range [left, right] is covered by at least one camera. Return false if any floor is left unmonitored.

A floor x is covered by camera i if start_i ≤ x ≤ end_i.

Input & Output

example_1.py — Basic Coverage

$ Input: ranges = [[1,2],[3,4],[5,6]], left = 2, right = 5

› Output: true

💡 Note: Every integer from 2 to 5 is covered: 2∈[1,2], 3∈[3,4], 4∈[3,4], 5∈[5,6]

example_2.py — Gap in Coverage

$ Input: ranges = [[1,10],[10,20]], left = 21, right = 21

› Output: false

💡 Note: The number 21 is not covered by any range. All ranges end at 20 or earlier.

example_3.py — Single Point Range

$ Input: ranges = [[1,50]], left = 25, right = 25

› Output: true

💡 Note: The single number 25 is covered by the range [1,50] since 1 ≤ 25 ≤ 50

Constraints

1 ≤ ranges.length ≤ 50
1 ≤ start_i ≤ end_i ≤ 50
1 ≤ left ≤ right ≤ 50
All ranges and target values are small integers

Visualization

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Understanding the Visualization

Deploy Cameras

Each camera covers a range of floors [start, end]

Mark Coverage

Create a coverage map showing all monitored floors

Check VIP Section

Verify every floor in [left, right] is monitored

Security Report

Return true if fully covered, false if any gaps exist

Key Takeaway

🎯 Key Insight: Instead of checking each floor against all cameras repeatedly, create a master coverage map first, then verify the VIP section in one efficient pass!

Asked in

a Amazon 42 G Google 38 ⊞ Microsoft 25 f Meta 18

The optimal solution uses a hash set to mark all covered numbers first, then checks the target range in a single pass. This achieves O(total_coverage + range_size) time complexity with clear separation of concerns: mark all covered positions, then verify the target range.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Each Number)	O((right-left+1) * ranges.length)	O(1)	For each number in [left, right], check if any range covers it
Sorted Intervals with Merging	O(n log n + n)	O(n)	Sort intervals and merge overlapping ones, then check coverage
Hash Set Marking (Optimal)	O(sum of all range sizes + (right-left+1))	O(total numbers covered by all ranges)	Mark all covered numbers in a set, then check the target range

Brute Force (Check Each Number) — Algorithm Steps

Loop through each integer from left to right
For each integer, check all ranges to see if any range covers it
If any integer is not covered by any range, return false
If all integers are covered, return true

Visualization

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Step-by-Step Walkthrough

Check number 2

Test if 2 is covered by any range

Check number 3

Test if 3 is covered by any range

Check number 4

Test if 4 is covered by any range

Check number 5

Test if 5 is covered by any range

Code -

solution.c — C

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

bool isCovered(int ranges[][2], int rangesSize, int left, int right) {
    for (int num = left; num <= right; num++) {
        bool covered = false;
        for (int i = 0; i < rangesSize; i++) {
            if (ranges[i][0] <= num && num <= ranges[i][1]) {
                covered = true;
                break;
            }
        }
        if (!covered) {
            return false;
        }
    }
    return true;
}

int main() {
    int ranges[][2] = {{1,2},{3,4},{5,6}};
    int left = 2, right = 5;
    printf("%s\n", isCovered(ranges, 3, left, right) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((right-left+1) * ranges.length)

For each number in the target range, we check all ranges

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no additional data structures

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Check Each Number) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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