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.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Triangle
+-------------+------+ | Column Name | Type | +-------------+------+ | x | int | | y | int | | z | int | +-------------+------+ In SQL, (x, y, z) is the primary key column for this table. Each row of this table contains the lengths of three line segments.
Report for every three line segments whether they can form a triangle.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Triangle table: +----+----+----+ | x | y | z | +----+----+----+ | 13 | 15 | 30 | | 10 | 20 | 15 | +----+----+----+ Output: +----+----+----+----------+ | x | y | z | triangle | +----+----+----+----------+ | 13 | 15 | 30 | No | | 10 | 20 | 15 | Yes | +----+----+----+----------+
Problem summary: Table: Triangle +-------------+------+ | Column Name | Type | +-------------+------+ | x | int | | y | int | | z | int | +-------------+------+ In SQL, (x, y, z) is the primary key column for this table. Each row of this table contains the lengths of three line segments. Report for every three line segments whether they can form a triangle. Return the result table in any order. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Triangle":["x","y","z"]},"rows":{"Triangle":[[13,15,30],[10,20,15]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #610: Triangle Judgement
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #610: Triangle Judgement
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #610: Triangle Judgement
// # Write your MySQL query statement below
// SELECT
// *,
// IF(x + y > z AND x + z > y AND y + z > x, 'Yes', 'No') AS triangle
// FROM Triangle;
// "#
// }
// Accepted solution for LeetCode #610: Triangle Judgement
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #610: Triangle Judgement
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #610: Triangle Judgement
// # Write your MySQL query statement below
// SELECT
// *,
// IF(x + y > z AND x + z > y AND y + z > x, 'Yes', 'No') AS triangle
// FROM Triangle;
// "#
// }
# Accepted solution for LeetCode #610: Triangle Judgement
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #610: Triangle Judgement
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #610: Triangle Judgement
# # Write your MySQL query statement below
# SELECT
# *,
# IF(x + y > z AND x + z > y AND y + z > x, 'Yes', 'No') AS triangle
# FROM Triangle;
# "#
# }
// Accepted solution for LeetCode #610: Triangle Judgement
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #610: Triangle Judgement
# Write your MySQL query statement below
SELECT
*,
IF(x + y > z AND x + z > y AND y + z > x, 'Yes', 'No') AS triangle
FROM Triangle;
"#
}
// Accepted solution for LeetCode #610: Triangle Judgement
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #610: Triangle Judgement
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #610: Triangle Judgement
// # Write your MySQL query statement below
// SELECT
// *,
// IF(x + y > z AND x + z > y AND y + z > x, 'Yes', 'No') AS triangle
// FROM Triangle;
// "#
// }
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.