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Table of contents
Problem statement
Time Complexity 
Space Complexity 
Frequently Asked Questions
Key Takeaways
Last Updated: Mar 27, 2024

Maximize the shortest path between given vertices by adding a single edge

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Problem statement

In the problem, you are given an undirected graph with ‘V’ vertices numbered from 1 to ‘V’ and ‘E’ edges between these ‘V’ vertices also you are given with an array named as operation of ‘K’ size which store ‘K’ number of vertices and you are allowed to join an edge between any two vertices given in the array and by making these connections. Your task is to maximize the shortest path between the first and the last vertex.


Note: You may add an edge between any two selected vertices that have already an edge between them to maximize the shortest path.






An unweighted graph is given with 5 vertices named as 1, 2, 3, 4, 5, and 4 edges between (1-2),(2-3),(3-4), and (4-5). The value of K here is 3 and the operation array is given as  {1, 2, 4}. We have to maximize the shortest path by adding a single edge between the combinations of all the possible vertices in the operation array.     

So let’s explore all the possible combinations of vertices in the operation array  


Considering edge 2, 4 from the operation array, the shortest distance between 1 and 5 here is 3 


Considering the edge 1, 4 from the operation array, the shortest distance between 1 and 5 here is 2 


Considering the edge 1, 2 from the operation array the shortest distance between 1 and 5 here is 4 


Out of all these possible shortest distances, we have to maximize the shortest path or shortest distance. So the maximum of all the shortest distance that is max(2, 3, 4) is 4 so the answer for this graph is 4 units


This problem of maximize the shortest path can be solved in the series of steps as follows :

  1. The idea here is to calculate the minimum distance for every vertex from the first and the last vertex 
  2. Let the shortest distance for a random vertex v from first node be xv and from the last vertex be yv
  3. To calculate the minimum distance from a source vertex to another vertex we can use breadth-first algorithm or Dijkstra’s algorithm
  4. For this, maintain a 2D matrix having 2 rows and v columns to store the minimum distance of all the vertices from the starting vertex in the first row and from the last vertex in the second row 
  5.  Now, choose two operation vertices a and b from operation[] array to minimize the value of min(xa + yb, ya + xb)
  6. Let distance  = min(xa + yb, ya + xb)
  7. Fro minimizing this distance follow these rules
  8. Create a vector of pairs and store the value of (xv - yv) with their respective selected vertex and after storing all the values sort the vector of pairs
  9. Before traversing in the vector initialize the value of best to 0 and max to -INF
  10. Now traverse the above vector of pairs and for each selected node(say n) update the value of best to maximum of (best, max + dist[1][n]) and update max to maximum of (max, dist[0][n]).
  11. After the above operations the maximum of (dist[0][v-1] and best + 1) given the minimum shortest path which will ultimately solve the goal to maximize the shortest path
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#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
int v,e;
// To store graph as adjacency list
vector<int> graph[100000];
// To store the shortest path from first vertex and last vertex 
int dist[2][100000];
// Bfs traversal of finding shortest distance from source vertex
void bfs(int* dist, int starting_vertex)
    int queue[200000];
    // Filling initially each distance as infinity 
    fill(dist, dist + v, INF);
    int first = 0, second = 0;
    queue[first++] = starting_vertex;
    dist[starting_vertex] = 0;
    // Performing BFS 
    while (second < first) {
        int x = queue[second++];
        for (int y : graph[x]) {
            if (dist[y] == INF) {
                // Updating the distance and queue 
                dist[y] = dist[x] + 1;
                queue[first++] = y;
    Function that maximise the shortest path between source and destination 
    vertex by adding a single edge between given selected nodes
void shortestPathCost(int operation[], int K)
    vector<pair<int, int> > dp;
    // 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], v - 1);
    for (int i = 0; i < K; i++) {
        // Storing the values  of (x[i] - y[i]) 
                            - dist[1][operation[i]],
    sort(dp.begin(), dp.end());
    int best = 0;
    int MAX = -INF;
    // Traversing  dp array 
    for (auto it : dp) {
        int a = it.second;
        best = max(best,
                MAX + dist[1][a]);
             Maximise the value of (x[a] - y[b]) if a and b 
             are the current vertices in operations 
        MAX= max(MAX, dist[0][a]);
    // Printing the minimum cost
    cout<<min(dist[0][v - 1], best + 1);
int main()

    //Given total vertices and edges and operation array to choose edges 
    v = 5, e = 4;
    int K = 2;
    int operation[] = { 1, 2 ,4};
    // Sorting the nodes in operation
    sort(operation, operation + K);
    // Formation of graph
    // maximise the shortest path

    shortestPathCost(operation, K);
    return 0;




Time Complexity 

The time complexity for the solution of “maximize the shortest path” will be O(V * log V + E)  where ‘V’ is the number of vertices and ‘E’ is the number of edges so let's understand the time complexity for the given problem.

First we have to sort the array operation that will cost us (K * log K ) for using inbuilt sort function 

Second we are entering in shortestPathCost function here we are calling bfs for first vertex which will cost us (V + E) and again for last vertex which will cost ( V + E ) so the time complexity for these 2 bfs calls will be 2 * ( V + E) and simultaneously we are also storing the shortest distance in the dp array of size V then we are sorting the dp array which will cost V * log V


So the total time complexity after all the operations will be:

 (K * log K ) + 2 * (V + E) + V * log V


Here K is very small compared with V so neglect the factor of K in the driver code now the complexiy will become 

2 * (V + E) + V * log V

            V * (log V + 2 ) + E

~  O(V * log V + E) 


Space Complexity 

The space complexity for the problem “maximize the shortest path” is O(N) since we are using a vector of pairs to store  (xv – yv)  here ‘xv’ is the shortest distance of a particular vertex ‘V’ from the first node and similarly ‘yv’ is shortest distance for the same vertex from the last node

Frequently Asked Questions

  1. How Dijkstra is different from normal BFS?

Dijkstra and BFS are two different algorithms for calculating the shortest distance between two points. Dijkstra's approach is based on a priority queue, whereas regular bfs are based on a queue. we can use either of these algorithms to solve the above problem maximize the shortest path 

2. How to find the shortest path in the weighted graph?

There are two most famous algorithms for finding the shortest distance in the weighted graph that is using Bellman-Ford in O(VE). If there are no negative weight cycles we can also use the Dijkstra algorithm that will do the work for us in O(E + VLogV) 

3. Different ways to store graphs?

We can use an adjacency list, adjacency matrix, there is one more way here we can store Nodes as objects and edges as pointers.

Key Takeaways

In this blog, we solved the problem “maximize the shortest path” that is  maximizing the shortest distance between the first vertex and last vertex in the graph by adding a single edge from the operation array given.

Check out this problem - Root To Leaf Path

If you want to learn more about graphs and graph traversals and want to practice some questions which require you to take your basic knowledge on graphs a notch higher, then you can visit our Guided Path for graphs on  Coding Ninjas Studio.To be more confident in data structures and algorithms, try out our DS and Algo Course. Until then, All the best for your future endeavors, and Keep Coding.



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