Leetcode•Sep 19, 2025
Redundant Connection II
Hazrat Ali
Leetcode
The given input is a directed graph that started as a rooted tree with n nodes (with distinct values from 1 to n), with one additional directed edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed.
The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [ui, vi] that represents a directed edge connecting nodes ui and vi, where ui is a parent of child vi.
Return an edge that can be removed so that the resulting graph is a rooted tree of n nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.
Example 1:
Input: edges = [[1,2],[1,3],[2,3]] Output: [2,3]
Example 2:
Input: edges = [[1,2],[2,3],[3,4],[4,1],[1,5]] Output: [4,1]
Solution
/**
* @param {number[][]} edges
* @return {number[]}
*/
const findRedundantDirectedConnection = edges => {
let canA = [-1, -1];
let canB = [-1, -1];
let parent = Array(2001).fill(0);
for (let i = 0; i < edges.length; i++) {
const edge = edges[i];
const u = edge[0];
const v = edge[1];
if (parent[v] === 0) {
parent[v] = u;
} else {
canB = [u, v];
canA = [parent[v], v];
edges[i][1] = 0;
}
}
parent = Array(2001).fill(-1);
const find = i => {
if (parent[i] === -1) {
return i;
}
return find(parent[i]);
};
for (let i = 0; i < edges.length; i++) {
const edge = edges[i];
if (edge[1] === 0) {
continue;
}
const x = find(edge[0]);
const y = find(edge[1]);
if (x === y) {
if (canA[0] === -1) {
return edge;
}
return canA;
}
parent[x] = y;
}
return canB;
};