Minimum possible modifications in the matrix to reach destination

Difficulty Level : Hard
Last Updated : 15 Jun, 2021

Given a matrix of size N x M consisting of integers 1, 2, 3 and 4.
Each value represents the possible movement from that cell:

1 -> move left
2 -> move right
3 -> move up
4 -> move down.

The task is to find the minimum possible changes required in the matrix such that there exists a path from (0, 0) to (N-1, M-1).

Examples:

Input : mat[][] = {{2, 2, 4},
{1, 1, 1},
{3, 3, 2}};
Output : 1
Change the value of mat[1][2] to 4. So the sequence of moves will be
(0, 0) -> (0, 1) -> (0, 2) -> (1, 2) -> (2, 2)
Input : mat[][] = {{2, 2, 1},
{4, 2, 3},
{4, 3, 2}}
Output : 2

Prerequisites:
1. Djikstra’s Algorithm
2.0-1 BFS

Method 1

Let’s consider each cell of the 2D matrix as a node of the weighted graph and each node can has at most four connected nodes(possible four directions). Each edge is weighted as:
- weight(U, V) = 0, if the direction of movement of node U points to V, else
- weight(U, V) = 1
Now, this basically reduces to shortest path problem which can be computed using Djikstra’s Algorithm

Below is the implementation of the above approach:

C++

// CPP program to find minimum possible
// changes required in the matrix
#include <bits/stdc++.h>
using namespace std;
 
// Function to find next possible node
int nextNode(int x, int y, int dir, int N, int M)
{
    if (dir == 1)
        y--;
    else if (dir == 2)
        y++;
    else if (dir == 3)
        x--;
    else
        x++;
 
    // If node is out of matrix
    if (!(x >= 0 && x < N && y >= 0 && y < M))
        return -1;
    else
        return (x * N + y);
}
 
// Prints shortest paths from src
// to all other vertices
int dijkstra(vector<pair<int, int> >* adj,
             int src, int dest, int N, int M)
{
    // Create a set to store vertices
    // that are bein preprocessed
    set<pair<int, int> > setds;
 
    // Create a vector for distances
    // and initialize all distances
    // as infinite (large value)
    vector<int> dist(N * M, INT_MAX);
 
    // Insert source itself in Set
    // and initialize its distance as 0
    setds.insert(make_pair(0, src));
    dist[src] = 0;
 
    /* Looping till all shortest
        distance are finalized
        then setds will become empty */
    while (!setds.empty()) {
        // The first vertex in Set
        // is the minimum distance
        // vertex, extract it from set.
        pair<int, int> tmp = *(setds.begin());
        setds.erase(setds.begin());
 
        // vertex label is stored in second
        // of pair (it has to be done this
        // way to keep the vertices sorted
        // distance (distance must be
        // first item in pair)
        int u = tmp.second;
 
        // 'i' is used to get all adjacent
        // vertices of a vertex
        vector<pair<int, int> >::iterator i;
        for (i = adj[u].begin();
             i != adj[u].end(); ++i) {
            // Get vertex label and weight
            // of current adjacent of u.
            int v = (*i).first;
            int weight = (*i).second;
 
            // If there is shorter path from u to v
            if (dist[v] > dist[u] + weight) {
                // If distance of v is not
                // INF then it must be
                // in our set, so removing it
                // and inserting again with
                // updated less distance.
                // Note : We extract only
                // those vertices from Set
                // for which distance is
                // finalized. So for them,
                // we would never reach here
                if (dist[v] != INT_MAX)
                    setds.erase(setds.find(
                        { dist[v], v }));
 
                // Updating distance of v
                dist[v] = dist[u] + weight;
                setds.insert(make_pair(dist[v], v));
            }
        }
    }
 
    // Return the distance
    return dist[dest];
}
 
// Function to find minimum possible
// changes required in the matrix
int MinModifications(vector<vector<int> >& mat)
{
    int N = mat.size(), M = mat[0].size();
 
    // Converting given matrix to a graph
    vector<pair<int, int> > adj[N * M];
 
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Each cell is a node,
            // with label i*N + j
            for (int dir = 1; dir <= 4; dir++) {
                // Label of node if we
                // move in direction dir
                int nextNodeLabel
                    = nextNode(i, j, dir, N, M);
 
                // If invalid(out of matrix)
                if (nextNodeLabel == -1)
                    continue;
 
                // If direction is same as mat[i][j]
                if (dir == mat[i][j])
                    adj[i * N + j].push_back(
                        { nextNodeLabel, 0 });
                else
                    adj[i * N + j].push_back(
                        { nextNodeLabel, 1 });
            }
        }
    }
 
    // Applying djikstra's algorithm
    return dijkstra(adj, 0,
                    (N - 1) * N + M - 1, N, M);
}
 
// Driver code
int main()
{
    vector<vector<int> > mat = { { 2, 2, 1 },
                                 { 4, 2, 3 },
                                 { 4, 3, 2 } };
 
    // Function call
    cout << MinModifications(mat);
 
    return 0;
}

Output:

Time Complexity : $O(N * M * log(N * M))$

Method 2
Here, edge weights are 0, and 1 only i.e it’s a 0-1 graph. The shortest path in such graphs can be found using 0-1 BFS.

Below is the implementation of the above approach:

C++

// C++ program to find minimum
// possible changes required
// in the matrix
#include <bits/stdc++.h>
using namespace std;
 
// Function to find next possible node
int nextNode(int x, int y, int dir,
             int N, int M)
{
    if (dir == 1)
        y--;
    else if (dir == 2)
        y++;
    else if (dir == 3)
        x--;
    else
        x++;
 
    // If node is out of matrix
    if (!(x >= 0 && x < N && y >= 0 && y < M))
        return -1;
    else
        return (x * N + y);
}
 
// Prints shortest distance
// from given source to
// every other vertex
int zeroOneBFS(vector<pair<int, int> >* adj,
               int src, int dest, int N, int M)
{
    // Initialize distances
    // from given source
    int dist[N * M];
    for (int i = 0; i < N * M; i++)
        dist[i] = INT_MAX;
 
    // Double ended queue to do BFS.
    deque<int> Q;
    dist[src] = 0;
    Q.push_back(src);
 
    while (!Q.empty()) {
        int v = Q.front();
        Q.pop_front();
 
        for (auto i : adj[v]) {
            // Checking for the optimal distance
            if (dist[i.first] > dist[v]
                                    + i.second) {
                dist[i.first] = dist[v]
                                + i.second;
 
                // Put 0 weight edges to front
                // and 1 weight edges to back
                // so that vertices are processed
                // in increasing order of weights.
                if (i.second == 0)
                    Q.push_front(i.first);
                else
                    Q.push_back(i.first);
            }
        }
    }
 
    // Shortest distance to
    // reach destination
    return dist[dest];
}
 
// Function to find minimum possible
// changes required in the matrix
int MinModifications(vector<vector<int> >& mat)
{
    int N = mat.size(), M = mat[0].size();
 
    // Converting given matrix to a graph
    vector<pair<int, int> > adj[N * M];
 
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Each cell is a node
            // with label i*N + j
            for (int dir = 1; dir <= 4; dir++) {
                // Label of node if we
                // move in direction dir
                int nextNodeLabel = nextNode(i, j,
                                             dir, N, M);
 
                // If invalid(out of matrix)
                if (nextNodeLabel == -1)
                    continue;
 
                // If direction is same as mat[i][j]
                if (dir == mat[i][j])
                    adj[i * N + j].push_back(
                        { nextNodeLabel, 0 });
                else
                    adj[i * N + j].push_back(
                        { nextNodeLabel, 1 });
            }
        }
    }
 
    // Applying djikstra's algorithm
    return zeroOneBFS(adj, 0,
                      (N - 1) * N + M - 1, N, M);
}
 
// Driver code
int main()
{
    vector<vector<int> > mat = { { 2, 2, 1 },
                                 { 4, 2, 3 },
                                 { 4, 3, 2 } };
 
    // Function call
    cout << MinModifications(mat);
 
    return 0;
}

Python3

# Python3 program to find minimum
# possible changes required
# in the matrix
from collections import deque
 
# Function to find next possible node
def nextNode(x, y, dir, N, M):
    if (dir == 1):
        y -= 1
    elif (dir == 2):
        y += 1
    elif (dir == 3):
        x -= 1
    else:
        x += 1
 
    # If node is out of matrix
    if (not (x >= 0 and x < N and y >= 0 and y < M)):
        return -1
    else:
        return (x * N + y)
 
# Prints shortest distance
# from given source to
# every other vertex
def zeroOneBFS(adj, src, dest, N, M):
   
    # Initialize distances
    # from given source
    dist = [10**8] *(N * M)
 
    # Double ended queue to do BFS.
    Q = deque()
    dist[src] = 0
    Q.append(src)
 
    while (len(Q) > 0):
        v = Q.popleft()
 
        for i in adj[v]:
           
            # print(i)
            # Checking for the optimal distance
            if (dist[i[0]] > dist[v] + i[1]):
                dist[i[0]] = dist[v] + i[1]
                 
                # Put 0 weight edges to front
                # and 1 weight edges to back
                # so that vertices are processed
                # in increasing order of weights.
                if (i[1] == 0):
                    Q.appendleft(i[0])
                else:
                    Q.append(i[0])
 
    # Shortest distance to
    # reach destination
    return dist[dest]
 
# Function to find minimum possible
# changes required in the matrix
def MinModifications(mat):
    N, M = len(mat), len(mat[0])
 
    # Converting given matrix to a graph
    adj = [[] for i in range(N * M)]
 
    for i in range(N):
        for j in range(M):
           
            # Each cell is a node
            # with label i*N + j
            for dir in range(1, 5):
               
                # Label of node if we
                # move in direction dir
                nextNodeLabel = nextNode(i, j, dir, N, M)
 
                # If invalid(out of matrix)
                if (nextNodeLabel == -1):
                    continue
 
                # If direction is same as mat[i][j]
                if (dir == mat[i][j]):
                    adj[i * N + j].append([nextNodeLabel, 0])
                else:
                    adj[i * N + j].append([nextNodeLabel, 1])
 
    # Applying djikstra's algorithm
    return zeroOneBFS(adj, 0, (N - 1) * N + M - 1, N, M)
 
# Driver code
if __name__ == '__main__':
    mat = [ [ 2, 2, 1 ],
             [ 4, 2, 3 ],
             [ 4, 3, 2 ] ]
 
    # Function call
    print (MinModifications(mat))
 
# This code is contributed by mohit kumar 29.

Javascript

<script>
 
// Javascript program to find minimum
// possible changes required in the matrix
 
// Function to find next possible node
function nextNode(x, y, dir, N, M)
{
    if (dir == 1)
        y--;
    else if (dir == 2)
        y++;
    else if (dir == 3)
        x--;
    else
        x++;
  
    // If node is out of matrix
    if (!(x >= 0 && x < N && y >= 0 && y < M))
        return -1;
    else
        return (x * N + y);
}
 
// Prints shortest distance
// from given source to
// every other vertex
function zeroOneBFS(adj, src, dest, N, M)
{
     
    // Initialize distances
    // from given source
    let dist = new Array(N * M);
    for(let i = 0; i < N * M; i++)
        dist[i] = Number.MAX_VALUE;
  
    // Double ended queue to do BFS.
    let Q = [];
    dist[src] = 0;
    Q.push(src);
  
    while (Q.length != 0)
    {
        let v = Q.shift();
  
        for(let i = 0; i < adj[v].length; i++)
        {
             
            // Checking for the optimal distance
            if (dist[adj[v][i][0]] > dist[v] +
                     adj[v][i][1])
            {
                dist[adj[v][i][0]] = dist[v] +
                                      adj[v][i][1];
  
                // Put 0 weight edges to front
                // and 1 weight edges to back
                // so that vertices are processed
                // in increasing order of weights.
                if (adj[v][i][1] == 0)
                    Q.push(adj[v][i][0]);
                else
                    Q.push(adj[v][i][0]);
            }
        }
    }
  
    // Shortest distance to
    // reach destination
    return dist[dest];
}
 
// Function to find minimum possible
// changes required in the matrix
function MinModifications(mat)
{
    let N = mat.length, M = mat[0].length;
  
    // Converting given matrix to a graph
    let adj = new Array(N * M);
     for(let i = 0; i < (N * M); i++)
    {
        adj[i] = [];
    }
     
    for(let i = 0; i < N; i++)
    {
        for(let j = 0; j < M; j++)
        {
             
            // Each cell is a node
            // with label i*N + j
            for(let dir = 1; dir <= 4; dir++)
            {
                 
                // Label of node if we
                // move in direction dir
                let nextNodeLabel = nextNode(i, j,
                                             dir, N, M);
  
                // If invalid(out of matrix)
                if (nextNodeLabel == -1)
                    continue;
  
                // If direction is same as mat[i][j]
                if (dir == mat[i][j])
                    adj[i * N + j].push(
                        [nextNodeLabel, 0]);
                else
                    adj[i * N + j].push(
                        [nextNodeLabel, 1]);
            }
        }
    }
  
    // Applying djikstra's algorithm
    return zeroOneBFS(adj, 0, (N - 1) * N + M - 1,
                      N, M);
}
 
// Driver code
let mat = [ [ 2, 2, 1 ],
            [ 4, 2, 3 ],
            [ 4, 3, 2 ] ];
             
document.write(MinModifications(mat));
 
// This code is contributed by unknown2108
 
</script>

Output:

Time Complexity : $O(N * M)$

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