Number of minimum length paths between 1 to N including each node

Given an undirected and unweighted graph of N nodes and M edges, the task is to count the minimum length paths between node 1 to N through each of the nodes. If there is doesn’t exist any such path, then print “-1”.

Note: The path can pass through a node any number of times.

Examples:

Input: N = 4, M= 4, edges = {{1, 2}, {2, 3}, {1, 3}, {2, 4}}
Output: 1 1 1 1
Explanation:

Total paths of minimum length from 1 to 4, passing from 1 is 1.
Total paths of minimum length from 1 to 4, passing from 2 is 1.
Total paths of minimum length from 1 to 4, passing from 3 is 1.
Total paths of minimum length from 1 to 4, passing from 4 is 1.

Input: N = 5, M = 5, edges = {{1, 2}, {1, 4}, {1 3}, {2, 5}, {2, 4}}
Output: 1 1 0 1 1

Approach: The given problem can be solved by performing two BFS, one from node 1 excluding node N and another from node N excluding node 1 to find the minimum distance of all the nodes from 1 and N, and the product of both the minimum distances will be the total count of minimum length paths from 1 to N including the node. Follow the steps below to solve the problem:

• Initialize a queue, say queue1 to perform BFS from node 1 and a queue queue2 to perform BFS from node N.
• Initialize arrays, say dist[] to store the shortest distance and ways[] to count the number of ways to reach that node.
• Perform two BFS and perform the following steps in each case:
  • Pop from the queue and store node in x and its distance in dis.
  • If dist[x] is smaller than dis then continue.
  • Traverse the adjacency list of x and for each child y, if dist[y] is greater than dis + 1 then update dist[y] equals dis + 1 and ways[y] equals ways[x]. Otherwise, if dist[y] equals dis +1 then add ways[x] to ways[y].
• Finally, iterate over the range N, and for each node print the count of minimum length paths as ways1[i]*ways2[i].

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to calculate the distances
// from node 1 to N
void countMinDistance(int n, int m,
                      int edges[][2])
{

    // Stores the number of edges
    vector<ll> g[10005];

    // Storing the edges in vector
    for (int i = 0; i < m; i++) {
        int a = edges[i][0] - 1;
        int b = edges[i][1] - 1;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    // Initialize queue
    queue<pair<ll, ll> > queue1;
    queue1.push({ 0, 0 });
    vector<int> dist(n, 1e9);
    vector<int> ways1(n, 0);
    dist[0] = 0;
    ways1[0] = 1;

    // BFS from 1st node using queue
    while (!queue1.empty()) {
        auto up = queue1.front();

        // Pop from queue
        queue1.pop();
        int x = up.first;
        int dis = up.second;
        if (dis > dist[x])
            continue;
        if (x == n - 1)
            continue;

        // Traversing the adjacency list
        for (ll y : g[x]) {
            if (dist[y] > dis + 1) {
                dist[y] = dis + 1;
                ways1[y] = ways1[x];
                queue1.push({ y, dis + 1 });
            }
            else if (dist[y] == dis + 1) {
                ways1[y] += ways1[x];
            }
        }
    }

    // Initialize queue
    queue<pair<ll, ll> > queue2;
    queue2.push({ n - 1, 0 });
    vector<int> dist1(n, 1e9);
    vector<int> ways2(n, 0);
    dist1[n - 1] = 0;
    ways2[n - 1] = 1;

    // BFS from last node
    while (!queue2.empty()) {
        auto up = queue2.front();

        // Pop from queue
        queue2.pop();
        int x = up.first;
        int dis = up.second;
        if (dis > dist1[x])
            continue;
        if (x == 0)
            continue;

        // Traverse the adjacency list
        for (ll y : g[x]) {
            if (dist1[y] > dis + 1) {
                dist1[y] = dis + 1;
                ways2[y] = ways2[x];
                queue2.push({ y, dis + 1 });
            }
            else if (dist1[y] == 1 + dis) {
                ways2[y] += ways2[x];
            }
        }
    }

    // Print the count of minimum
    // distance
    for (int i = 0; i < n; i++) {
        cout << ways1[i] * ways2[i] << " ";
    }
}

// Driver Code
int main()
{
    int N = 5, M = 5;
    int edges[M][2] = {
        { 1, 2 }, { 1, 4 }, { 1, 3 },
        { 2, 5 }, { 2, 4 }
    };
    countMinDistance(N, M, edges);

    return 0;
}

Python3

# Python 3 program for the above approach

# Function to calculate the distances
# from node 1 to N
def countMinDistance(n, m, edges):
    # Stores the number of edges
    g = [[] for i in range(10005)]

    # Storing the edges in vector
    for i in range(m):
        a = edges[i][0] - 1
        b = edges[i][1] - 1
        g[a].append(b)
        g[b].append(a)

    # Initialize queue
    queue1 = []
    queue1.append([0, 0])
    dist = [1e9 for i in range(n)]
    ways1 = [0 for i in range(n)]
    dist[0] = 0
    ways1[0] = 1

    # BFS from 1st node using queue
    while (len(queue1)>0):
        up = queue1[0]

        # Pop from queue
        queue1 = queue1[:-1]
        x = up[0]
        dis = up[1]
        if (dis > dist[x]):
            continue
        if (x == n - 1):
            continue

        # Traversing the adjacency list
        for y in g[x]:
            if (dist[y] > dis + 1):
                dist[y] = dis + 1
                ways1[y] = ways1[x]
                queue1.append([y, dis + 1])

            elif(dist[y] == dis + 1):
                ways1[y] += ways1[x]

    # Initialize queue
    queue2 = []
    queue2.append([n - 1, 0])
    dist1 = [1e9 for i in range(n)]
    ways2 = [0 for i in range(n)]
    dist1[n - 1] = 0
    ways2[n - 1] = 1

    # BFS from last node
    while(len(queue2)>0):
        up = queue2[0]

        # Pop from queue
        queue2 = queue2[:-1]
        x = up[0]
        dis = up[1]
        if (dis > dist1[x]):
            continue
        if (x == 0):
            continue

        # Traverse the adjacency list
        for y in g[x]:
            if (dist1[y] > dis + 1):
                dist1[y] = dis + 1
                ways2[y] = ways2[x]
                queue2.append([y, dis + 1])

            elif(dist1[y] == 1 + dis):
                ways2[y] += ways2[x]

    # Print the count of minimum
    # distance
    ways1[n-1] = 1
    ways2[n-1] = 1
    for i in range(n):
        print(ways1[i] * ways2[i],end = " ")


# Driver Code
if __name__ == '__main__':
    N = 5
    M = 5
    edges = [[1, 2],[1, 4],[1, 3],[2, 5],[2, 4]]
    countMinDistance(N, M, edges)

    # This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>
// Javascript program for the above approach

// Function to calculate the distances
// from node 1 to N
function countMinDistance(n, m, edges) {
  // Stores the number of edges
  let g = new Array(10005).fill(0).map(() => []);

  // Storing the edges in vector
  for (let i = 0; i < m; i++) {
    let a = edges[i][0] - 1;
    let b = edges[i][1] - 1;
    g[a].push(b);
    g[b].push(a);
  }

  // Initialize queue
  let queue1 = [];
  queue1.push([0, 0]);
  let dist = new Array(n).fill(1e9);
  let ways1 = new Array(n).fill(0);
  dist[0] = 0;
  ways1[0] = 1;

  // BFS from 1st node using queue
  while (queue1.length > 0) {
    let up = queue1[0];

    // Pop from queue
    queue1.pop();
    let x = up[0];
    let dis = up[1];
    if (dis > dist[x]) continue;
    if (x == n - 1) continue;

    // Traversing the adjacency list
    for (let y of g[x]) {
      if (dist[y] > dis + 1) {
        dist[y] = dis + 1;
        ways1[y] = ways1[x];
        queue1.push([y, dis + 1]);
      } else if (dist[y] == dis + 1) ways1[y] += ways1[x];
    }
  }
  // Initialize queue
  let queue2 = [];
  queue2.push([n - 1, 0]);
  let dist1 = new Array(n).fill(1e9);
  let ways2 = new Array(n).fill(0);
  dist1[n - 1] = 0;
  ways2[n - 1] = 1;

  // BFS from last node
  while (queue2.length > 0) {
    let up = queue2[0];

    // Pop from queue
    queue2.pop();
    let x = up[0];
    let dis = up[1];
    if (dis > dist1[x]) continue;
    if (x == 0) continue;

    // Traverse the adjacency list
    for (let y of g[x]) {
      if (dist1[y] > dis + 1) {
        dist1[y] = dis + 1;
        ways2[y] = ways2[x];
        queue2.push([y, dis + 1]);
      } else if (dist1[y] == 1 + dis) ways2[y] += ways2[x];
    }
  }
  // Print the count of minimum
  // distance
  ways1[n - 1] = 1;
  ways2[n - 1] = 1;
  for (let i = 0; i < n; i++) document.write(ways1[i] * ways2[i] + " ");
}

// Driver Code

let N = 5;
let M = 5;
let edges = [
  [1, 2],
  [1, 4],
  [1, 3],
  [2, 5],
  [2, 4],
];
countMinDistance(N, M, edges);

// This code is contributed by gfgking

</script>

Output:

1 1 0 1 1

Time Complexity: O(N + M)
Auxiliary Space: O(N)