LeetCode #1893 — EASY

Check if All the Integers in a Range Are Covered

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 2D integer array ranges and two integers left and right. Each ranges[i] = [start_i, end_i] represents an inclusive interval between start_i and end_i.

Return true if each integer in the inclusive range [left, right] is covered by at least one interval in ranges. Return false otherwise.

An integer x is covered by an interval ranges[i] = [start_i, end_i] if start_i <= x <= end_i.

Example 1:

Input: ranges = [[1,2],[3,4],[5,6]], left = 2, right = 5
Output: true
Explanation: Every integer between 2 and 5 is covered:
- 2 is covered by the first range.
- 3 and 4 are covered by the second range.
- 5 is covered by the third range.

Example 2:

Input: ranges = [[1,10],[10,20]], left = 21, right = 21
Output: false
Explanation: 21 is not covered by any range.

Constraints:

1 <= ranges.length <= 50
1 <= start_i <= end_i <= 50
1 <= left <= right <= 50

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array ranges and two integers left and right. Each ranges[i] = [starti, endi] represents an inclusive interval between starti and endi. Return true if each integer in the inclusive range [left, right] is covered by at least one interval in ranges. Return false otherwise. An integer x is covered by an interval ranges[i] = [starti, endi] if starti <= x <= endi.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[1,2],[3,4],[5,6]]
2
5

Example 2

[[1,10],[10,20]]
21
21

Core Insight

What unlocks the optimal approach

Iterate over every integer point in the range [left, right].
For each of these points check if it is included in one of the ranges.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1893: Check if All the Integers in a Range Are Covered
class Solution {
    public boolean isCovered(int[][] ranges, int left, int right) {
        int[] diff = new int[52];
        for (int[] range : ranges) {
            int l = range[0], r = range[1];
            ++diff[l];
            --diff[r + 1];
        }
        int s = 0;
        for (int i = 0; i < diff.length; ++i) {
            s += diff[i];
            if (s <= 0 && left <= i && i <= right) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1893: Check if All the Integers in a Range Are Covered
func isCovered(ranges [][]int, left int, right int) bool {
	diff := [52]int{}
	for _, e := range ranges {
		l, r := e[0], e[1]
		diff[l]++
		diff[r+1]--
	}
	s := 0
	for i, x := range diff {
		s += x
		if s <= 0 && left <= i && i <= right {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #1893: Check if All the Integers in a Range Are Covered
class Solution:
    def isCovered(self, ranges: List[List[int]], left: int, right: int) -> bool:
        diff = [0] * 52
        for l, r in ranges:
            diff[l] += 1
            diff[r + 1] -= 1
        s = 0
        for i, x in enumerate(diff):
            s += x
            if s <= 0 and left <= i <= right:
                return False
        return True

// Accepted solution for LeetCode #1893: Check if All the Integers in a Range Are Covered
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn is_covered(ranges: Vec<Vec<i32>>, left: i32, right: i32) -> bool {
        let mut a = HashSet::new();
        let mut b = HashSet::new();
        for i in left..=right {
            b.insert(i);
        }
        for r in ranges {
            let start = r[0];
            let end = r[1];
            for i in start..=end {
                if b.contains(&i) {
                    a.insert(i);
                }
            }
        }
        a == b
    }
}

#[test]
fn test() {
    let ranges = vec_vec_i32![[1, 2], [3, 4], [5, 6]];
    let left = 2;
    let right = 5;
    let res = true;
    assert_eq!(Solution::is_covered(ranges, left, right), res);
    let ranges = vec_vec_i32![[1, 10], [10, 20]];
    let left = 21;
    let right = 21;
    let res = false;
    assert_eq!(Solution::is_covered(ranges, left, right), res);
}

// Accepted solution for LeetCode #1893: Check if All the Integers in a Range Are Covered
function isCovered(ranges: number[][], left: number, right: number): boolean {
    const diff: number[] = Array(52).fill(0);
    for (const [l, r] of ranges) {
        ++diff[l];
        --diff[r + 1];
    }
    let s = 0;
    for (let i = 0; i < diff.length; ++i) {
        s += diff[i];
        if (s <= 0 && left <= i && i <= right) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + M)

Space

O(M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.