Maximize shortest path between given vertices by adding a single edge
Last Updated :
12 Jul, 2025
Given an undirected graph of N nodes and M vertices. You are also given a K edges as selected[]. The task to maximize the shortest path length between node 1 to node N by adding single edges between any two vertices from the given selected edges.
Note: You may add an edge between any two selected vertices who have already an edge between them.
Input: N = 5, M = 4, K = 2, selected[] = {2, 4}
Below is the given graph:
Output: 3
Explanation:
Before adding an edge between 2 and 4, the Shortest Path becomes: 1–>2–>3–>4–>5.
After adding an edge between 2 and 4, the Shortest Path becomes 1–>2–>4–>5. Below is the graph after adding edges. denoted by the dashed line.
Input: N = 5 M = 5 K = 3 selected[] = {1, 3, 5}
Below is the given graph:
Output: 3
Explanation:
We can add an edge between 3 and 5 as they have already an edge between them. so, the shortest path becomes 1–>2–>3–>5. Below is the graph after adding edges. denoted by the dashed line.
Approach: The idea is to use Breadth-First Search to find the distance from vertices 1 and N to each selected vertex. For selected vertex i, let xi denote the distance to node 1, and yi denote the distance to node N. Below are the steps:
- Maintain a 2D matrix(say dist[2][]) having 2 rows and N columns.
- In the first row, Maintain the shortest distance between node 1 and other vertices in the graph using BFS Traversal.
- In the second row, Maintain the shortest distance between node N and the other vertices of the graph using BFS Traversal.
- Now, choose two selected vertices a and b from selected[] to minimize the value of min(xa + yb, ya + xb). For this do the following:
- Create a vector of pairs and store the value of (xi - yi) with their respective selected node.
- Sort the above vector of pairs.
- Initialize best to 0 and max to -INF.
- Now traverse the above vector of pairs and for each selected node(say a) update the value of best to maximum of (best, max + dist[1][a]) and update max to maximum of (max, dist[0][a]).
- After the above operations the maximum of (dist[0][N-1] and best + 1) given the minimum shortest path.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
int N, M;
// To store graph as adjacency list
vector<int> edges[200005];
// To store the shortest path
int dist[2][200000];
// Function that performs BFS Traversal
void bfs(int* dist, int s)
{
int q[200000];
// Fill initially each distance as INF
fill(dist, dist + N, INF);
int qh = 0, qt = 0;
q[qh++] = s;
dist[s] = 0;
// Perform BFS
while (qt < qh) {
int x = q[qt++];
// Traverse the current edges
for (int y : edges[x]) {
if (dist[y] == INF) {
// Update the distance
dist[y] = dist[x] + 1;
// Insert in queue
q[qh++] = y;
}
}
}
}
// Function that maximizes the shortest
// path between source and destination
// vertex by adding a single edge between
// given selected nodes
void shortestPathCost(int selected[], int K)
{
vector<pair<int, int> > data;
// To update the shortest distance
// between node 1 to other vertices
bfs(dist[0], 0);
// To update the shortest distance
// between node N to other vertices
bfs(dist[1], N - 1);
for (int i = 0; i < K; i++) {
// Store the values x[i] - y[i]
data.emplace_back(dist[0][selected[i]]
- dist[1][selected[i]],
selected[i]);
}
// Sort all the vectors of pairs
sort(data.begin(), data.end());
int best = 0;
int MAX = -INF;
// Traverse data[]
for (auto it : data) {
int a = it.second;
best = max(best,
MAX + dist[1][a]);
// Maximize x[a] - y[b]
MAX= max(MAX, dist[0][a]);
}
// Print minimum cost
printf("%d\n", min(dist[0][N - 1], best + 1));
}
// Driver Code
int main()
{
// Given nodes and edges
N = 5, M = 4;
int K = 2;
int selected[] = { 1, 3 };
// Sort the selected nodes
sort(selected, selected + K);
// Given edges
edges[0].push_back(1);
edges[1].push_back(0);
edges[1].push_back(2);
edges[2].push_back(1);
edges[2].push_back(3);
edges[3].push_back(2);
edges[3].push_back(4);
edges[4].push_back(3);
// Function Call
shortestPathCost(selected, K);
return 0;
}
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG{
static int INF = (int)1e9 + 7;
static int N, M;
// To store graph as adjacency list
static ArrayList<ArrayList<Integer>> edges;
// To store the shortest path
static int[][] dist = new int[2][200000];
// Function that performs BFS Traversal
static void bfs(int[] dist, int s)
{
int[] q = new int[200000];
// Fill initially each distance as INF
Arrays.fill(dist, INF);
int qh = 0, qt = 0;
q[qh++] = s;
dist[s] = 0;
// Perform BFS
while (qt < qh)
{
int x = q[qt++];
// Traverse the current edges
for(Integer y : edges.get(x))
{
if (dist[y] == INF)
{
// Update the distance
dist[y] = dist[x] + 1;
// Insert in queue
q[qh++] = y;
}
}
}
}
// Function that maximizes the shortest
// path between source and destination
// vertex by adding a single edge between
// given selected nodes
static void shortestPathCost(int selected[], int K)
{
ArrayList<int[]> data = new ArrayList<>();
// To update the shortest distance
// between node 1 to other vertices
bfs(dist[0], 0);
// To update the shortest distance
// between node N to other vertices
bfs(dist[1], N - 1);
for(int i = 0; i < K; i++)
{
// Store the values x[i] - y[i]
data.add(new int[]{dist[0][selected[i]] -
dist[1][selected[i]],
selected[i]});
}
// Sort all the vectors of pairs
Collections.sort(data, (a, b) -> a[0] - b[0]);
int best = 0;
int MAX = -INF;
// Traverse data[]
for(int[] it : data)
{
int a = it[1];
best = Math.max(best,
MAX + dist[1][a]);
// Maximize x[a] - y[b]
MAX = Math.max(MAX, dist[0][a]);
}
// Print minimum cost
System.out.println(Math.min(dist[0][N - 1],
best + 1));
}
// Driver code
public static void main (String[] args)
{
// Given nodes and edges
N = 5; M = 4;
int K = 2;
int selected[] = { 1, 3 };
// Sort the selected nodes
Arrays.sort(selected);
edges = new ArrayList<>();
for(int i = 0; i < 200005; i++)
edges.add(new ArrayList<Integer>());
// Given edges
edges.get(0).add(1);
edges.get(1).add(0);
edges.get(1).add(2);
edges.get(2).add(1);
edges.get(2).add(3);
edges.get(3).add(2);
edges.get(3).add(4);
edges.get(4).add(3);
// Function Call
shortestPathCost(selected, K);
}
}
// This code is contributed by offbeat
# Python3 program for the above approach
# Function that performs BFS Traversal
def bfs(x, s):
global edges, dist
q = [0 for i in range(200000)]
# Fill initially each distance as INF
# fill(dist, dist + N, INF)
qh, qt = 0, 0
q[qh] = s
qh += 1
dist[x][s] = 0
# Perform BFS
while (qt < qh):
xx = q[qt]
qt += 1
# Traverse the current edges
for y in edges[xx]:
if (dist[x][y] == 10**18):
# Update the distance
dist[x][y] = dist[x][xx] + 1
# Insert in queue
q[qh] = y
qh += 1
# Function that maximizes the shortest
# path between source and destination
# vertex by adding a single edge between
# given selected nodes
def shortestPathCost(selected, K):
global dist, edges
data = []
# To update the shortest distance
# between node 1 to other vertices
bfs(0, 0)
# To update the shortest distance
# between node N to other vertices
bfs(1, N - 1)
for i in range(K):
# Store the values x[i] - y[i]
data.append([dist[0][selected[i]]- dist[1][selected[i]], selected[i]])
# Sort all the vectors of pairs
data = sorted(data)
best = 0
MAX = -10**18
# Traverse data[]
for it in data:
a = it[1]
best = max(best,MAX + dist[1][a])
# Maximize x[a] - y[b]
MAX= max(MAX, dist[0][a])
# Print minimum cost
print(min(dist[0][N - 1], best + 1))
# Driver Code
if __name__ == '__main__':
# Given nodes and edges
edges = [[] for i in range(5)]
dist = [[10**18 for i in range(1000005)] for i in range(2)]
N,M = 5, 4
K = 2
selected = [1, 3]
# Sort the selected nodes
selected = sorted(selected)
# Given edges
edges[0].append(1)
edges[1].append(0)
edges[1].append(2)
edges[2].append(1)
edges[2].append(3)
edges[3].append(2)
edges[3].append(4)
edges[4].append(3)
# Function Call
shortestPathCost(selected, K)
# This code is contributed by mohit kumar 29
using System;
using System.Linq;
class Program
{
static int N, M, K;
static int[][] edges, dist;
static int[] selected;
//Function that performs BFS Traversal
static void BFS(int x, int s)
{
int[] q = new int[200000];
//
//Fill initially each distance as INF
//fill(dist, dist + N, INF)
int qh = 0, qt = 0;
q[qh] = s;
qh++;
dist[x][s] = 0;
//Perform BFS
while (qt < qh)
{
int xx = q[qt];
qt++;
for (int i = 0; i < edges[xx].Length; i++)
{
int y = edges[xx][i];
if (dist[x][y] == int.MaxValue)
{
dist[x][y] = dist[x][xx] + 1;
q[qh] = y;
qh++;
}
}
}
}
// Function that maximizes the shortest
// path between source and destination
// vertex by adding a single edge between
// given selected nodes
static void ShortestPathCost(int[] selected, int K)
{
BFS(0, 0);
BFS(1, N - 1);
Tuple<int, int>[] data = new Tuple<int, int>[K];
for (int i = 0; i < K; i++)
{
data[i] = Tuple.Create(dist[0][selected[i]] - dist[1][selected[i]], selected[i]);
}
Array.Sort(data, (a, b) => a.Item1.CompareTo(b.Item1));
int best = 0, MAX = int.MinValue;
foreach (Tuple<int, int> it in data)
{
int a = it.Item2;
best = Math.Max(best, MAX + dist[1][a]);
MAX = Math.Max(MAX, dist[0][a]);
}
Console.WriteLine(Math.Min(dist[0][N - 1], best + 1));
}
//Driver code
static void Main(string[] args)
{
N = 5;
M = 4;
K = 2;
selected = new int[] { 1, 3 };
Array.Sort(selected);
edges = new int[5][];
edges[0] = new int[] { 1 };
edges[1] = new int[] { 0, 2 };
edges[2] = new int[] { 1, 3 };
edges[3] = new int[] { 2, 4 };
edges[4] = new int[] { 3 };
dist = new int[2][];
dist[0] = Enumerable.Repeat(int.MaxValue, 1000005).ToArray();
dist[1] = Enumerable.Repeat(int.MaxValue, 1000005).ToArray();
ShortestPathCost(selected, K);
}
}
<script>
// Javascript program for the above approach
let INF = 1e9 + 7;
let N, M;
// To store graph as adjacency list
let edges=[];
// To store the shortest path
let dist=new Array(2);
for(let i=0;i<2;i++)
{
dist[i]=new Array(200000);
for(let j=0;j<200000;j++)
{
dist[i][j]=INF;
}
}
// Function that performs BFS Traversal
function bfs(dist,s)
{
let q = new Array(200000);
// Fill initially each distance as INF
let qh = 0, qt = 0;
q[qh++] = s;
dist[s] = 0;
// Perform BFS
while (qt < qh)
{
let x = q[qt++];
// Traverse the current edges
for(let y=0;y< edges[x].length;y++)
{
if (dist[edges[x][y]] == INF)
{
// Update the distance
dist[edges[x][y]] = dist[x] + 1;
// Insert in queue
q[qh++] = edges[x][y];
}
}
}
}
// Function that maximizes the shortest
// path between source and destination
// vertex by adding a single edge between
// given selected nodes
function shortestPathCost(selected,K)
{
let data = [];
// To update the shortest distance
// between node 1 to other vertices
bfs(dist[0], 0);
// To update the shortest distance
// between node N to other vertices
bfs(dist[1], N - 1);
for(let i = 0; i < K; i++)
{
// Store the values x[i] - y[i]
data.push([dist[0][selected[i]] -
dist[1][selected[i]],
selected[i]]);
}
// Sort all the vectors of pairs
data.sort(function(a, b){return a[0] - b[0];});
let best = 0;
let MAX = -INF;
// Traverse data[]
for(let it=0;it< data.length;it++)
{
let a = data[it][1];
best = Math.max(best,
MAX + dist[1][a]);
// Maximize x[a] - y[b]
MAX = Math.max(MAX, dist[0][a]);
}
// Print minimum cost
document.write(Math.min(dist[0][N - 1],
best + 1));
}
// Driver code
// Given nodes and edges
N = 5; M = 4;
let K = 2;
let selected = [ 1, 3 ];
// Sort the selected nodes
(selected).sort(function(a,b){return a-b;});
edges = [];
for(let i = 0; i < 200005; i++)
edges.push([]);
// Given edges
edges[0].push(1);
edges[1].push(0);
edges[1].push(2);
edges[2].push(1);
edges[2].push(3);
edges[3].push(2);
edges[3].push(4);
edges[4].push(3);
// Function Call
shortestPathCost(selected, K);
// This code is contributed by patel2127
</script>
Time Complexity: O(N*log N + M)
Auxiliary Space: O(N)
Similar Reads
Minimum number of edges between two vertices of a graph using DFS Given an undirected graph G(V, E) with N vertices and M edges. We need to find the minimum number of edges between a given pair of vertices (u, v). We have already discussed this problem using the BFS approach, here we will use the DFS approach. Examples: Input: For the following given graph, find t
10 min read
Minimum number of edges between two vertices of a graph using DFS Given an undirected graph G(V, E) with N vertices and M edges. We need to find the minimum number of edges between a given pair of vertices (u, v). We have already discussed this problem using the BFS approach, here we will use the DFS approach. Examples: Input: For the following given graph, find t
10 min read
Shortest paths from all vertices to a destination Given a Weighted Directed Graph and a destination vertex in the graph, find the shortest distance from all vertex to the destination vertex.Input : 1 Output : 0 6 10 7 5Distance of 0 from 0: 0 Distance of 0 from 1: 1+5 = 6 (1->4->0) Distance of 0 from 2: 10 (2->0) Distance of 0 from 3: 1+1+
11 min read
Shortest paths from all vertices to a destination Given a Weighted Directed Graph and a destination vertex in the graph, find the shortest distance from all vertex to the destination vertex.Input : 1 Output : 0 6 10 7 5Distance of 0 from 0: 0 Distance of 0 from 1: 1+5 = 6 (1->4->0) Distance of 0 from 2: 10 (2->0) Distance of 0 from 3: 1+1+
11 min read
D'Esopo-Pape Algorithm : Single Source Shortest Path Given an undirected weighted graph with v vertices labeled from 0 to v - 1, and a list of edges represented as a 2D array edges[][], where each entry is of the form [a, b, w] indicating an undirected edge between vertex a and vertex b with weight w. You are also given a source vertex src.The task is
11 min read
D'Esopo-Pape Algorithm : Single Source Shortest Path Given an undirected weighted graph with v vertices labeled from 0 to v - 1, and a list of edges represented as a 2D array edges[][], where each entry is of the form [a, b, w] indicating an undirected edge between vertex a and vertex b with weight w. You are also given a source vertex src.The task is
11 min read