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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi.
A connected trio is a set of three nodes where there is an edge between every pair of them.
The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not.
Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.
Example 1:
Input: n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]] Output: 3 Explanation: There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.
Example 2:
Input: n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]] Output: 0 Explanation: There are exactly three trios: 1) [1,4,3] with degree 0. 2) [2,5,6] with degree 2. 3) [5,6,7] with degree 2.
Constraints:
2 <= n <= 400edges[i].length == 21 <= edges.length <= n * (n-1) / 21 <= ui, vi <= nui != viProblem summary: You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi. A connected trio is a set of three nodes where there is an edge between every pair of them. The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not. Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
6 [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]]
7 [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]]
add-edges-to-make-degrees-of-all-nodes-even)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1761: Minimum Degree of a Connected Trio in a Graph
class Solution {
public int minTrioDegree(int n, int[][] edges) {
boolean[][] g = new boolean[n][n];
int[] deg = new int[n];
for (var e : edges) {
int u = e[0] - 1, v = e[1] - 1;
g[u][v] = true;
g[v][u] = true;
++deg[u];
++deg[v];
}
int ans = 1 << 30;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (g[i][j]) {
for (int k = j + 1; k < n; ++k) {
if (g[i][k] && g[j][k]) {
ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
}
}
}
}
}
return ans == 1 << 30 ? -1 : ans;
}
}
// Accepted solution for LeetCode #1761: Minimum Degree of a Connected Trio in a Graph
func minTrioDegree(n int, edges [][]int) int {
g := make([][]bool, n)
deg := make([]int, n)
for i := range g {
g[i] = make([]bool, n)
}
for _, e := range edges {
u, v := e[0]-1, e[1]-1
g[u][v], g[v][u] = true, true
deg[u]++
deg[v]++
}
ans := 1 << 30
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
if g[i][j] {
for k := j + 1; k < n; k++ {
if g[i][k] && g[j][k] {
ans = min(ans, deg[i]+deg[j]+deg[k]-6)
}
}
}
}
}
if ans == 1<<30 {
return -1
}
return ans
}
# Accepted solution for LeetCode #1761: Minimum Degree of a Connected Trio in a Graph
def min(a: int, b: int) -> int:
return a if a < b else b
class Solution:
def minTrioDegree(self, n: int, edges: List[List[int]]) -> int:
g = [[False] * n for _ in range(n)]
deg = [0] * n
for u, v in edges:
u, v = u - 1, v - 1
g[u][v] = g[v][u] = True
deg[u] += 1
deg[v] += 1
ans = inf
for i in range(n):
for j in range(i + 1, n):
if g[i][j]:
for k in range(j + 1, n):
if g[i][k] and g[j][k]:
ans = min(ans, deg[i] + deg[j] + deg[k] - 6)
return -1 if ans == inf else ans
// Accepted solution for LeetCode #1761: Minimum Degree of a Connected Trio in a Graph
/**
* [1761] Minimum Degree of a Connected Trio in a Graph
*
* You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi.
* A connected trio is a set of three nodes where there is an edge between every pair of them.
* The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not.
* Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.
*
* Example 1:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/01/26/trios1.png" style="width: 388px; height: 164px;" />
* Input: n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]]
* Output: 3
* Explanation: There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.
*
* Example 2:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/01/26/trios2.png" style="width: 388px; height: 164px;" />
* Input: n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]]
* Output: 0
* Explanation: There are exactly three trios:
* 1) [1,4,3] with degree 0.
* 2) [2,5,6] with degree 2.
* 3) [5,6,7] with degree 2.
*
*
* Constraints:
*
* 2 <= n <= 400
* edges[i].length == 2
* 1 <= edges.length <= n * (n-1) / 2
* 1 <= ui, vi <= n
* ui != vi
* There are no repeated edges.
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph/
// discuss: https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
// Credit: https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph/solutions/3209233/just-a-runnable-solution/
pub fn min_trio_degree(n: i32, edges: Vec<Vec<i32>>) -> i32 {
let n = n as usize;
let mut al = vec![vec![]; n + 1];
let mut cnt = vec![0; n + 1];
let mut result = std::i32::MAX;
for e in edges.iter() {
let t1 = e[0].min(e[1]) as usize;
let t2 = e[0].max(e[1]) as usize;
al[t1].push(t2);
cnt[t1] += 1;
cnt[t2] += 1;
}
for t1 in 1..=n {
for &t2 in al[t1].iter() {
for &t3 in al[t1].iter() {
if t2 < t3 && al[t2].contains(&t3) {
result = result.min(cnt[t1] + cnt[t2] + cnt[t3] - 6);
}
}
}
}
if result == std::i32::MAX { -1 } else { result }
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_1761_example_1() {
let n = 6;
let edges = vec![
vec![1, 2],
vec![1, 3],
vec![3, 2],
vec![4, 1],
vec![5, 2],
vec![3, 6],
];
let result = 3;
assert_eq!(Solution::min_trio_degree(n, edges), result);
}
#[test]
fn test_1761_example_2() {
let n = 7;
let edges = vec![
vec![1, 3],
vec![4, 1],
vec![4, 3],
vec![2, 5],
vec![5, 6],
vec![6, 7],
vec![7, 5],
vec![2, 6],
];
let result = 0;
assert_eq!(Solution::min_trio_degree(n, edges), result);
}
}
// Accepted solution for LeetCode #1761: Minimum Degree of a Connected Trio in a Graph
function minTrioDegree(n: number, edges: number[][]): number {
const g = Array.from({ length: n }, () => Array(n).fill(false));
const deg: number[] = Array(n).fill(0);
for (let [u, v] of edges) {
u--;
v--;
g[u][v] = g[v][u] = true;
++deg[u];
++deg[v];
}
let ans = Infinity;
for (let i = 0; i < n; ++i) {
for (let j = i + 1; j < n; ++j) {
if (g[i][j]) {
for (let k = j + 1; k < n; ++k) {
if (g[i][k] && g[j][k]) {
ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
}
}
}
}
}
return ans === Infinity ? -1 : ans;
}
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.