Minimum Cost to reach the end of the matrix

Last Updated : 31 Dec, 2023

Given a matrix of size N X N, such that each cell has an Uppercase Character in it, find the minimum cost to reach the cell (N-1, N-1) starting from (0, 0). From a cell (i, j), we can move Up(i-1, j), Down(i+1, j), Left (i, j-1) and Right(i, j+1). If the adjacent cell has the same character as the current cell then we can move to the adjacent cell with 0 cost, else the cost incurred is 1 coin.

Examples:

Input: m = 2, n = 3, mat = { { 'a', 'b', 'n' }, { 'd', 'e', 'f' } }
Output: 3
Explanation: One of the possible answers is to move from 'a' -> 'b' -> 'e' -> 'f' with the cost of 1 + 1 + 1 = 3

Input: m = 2, n = 2, mat = {{'a', 'a'}, {'a', 'a'}}
Output: 0
Explanation: We can take any path from the top left to the bottom right as the cost is always zero.

Approach: To solve the problem follow the below idea:

The idea is to use Breadth-First Search (BFS) to find the shortest path from the top-left corner to the bottom-right corner of a matrix. Determine the character in each cell, where moving to an adjacent cell with the same character incurs a 0 cost, and moving to a different character incurs a cost of 1 coin. The distance matrix is updated during BFS to keep track of the minimum cost to reach each cell. The code then initializes the matrix, performs BFS, and outputs the minimum cost to reach the destination.

Step-by-step approach:

Initialize a distance matrix (dis[][]) with large values (INF).
Start BFS from the top-left corner (cell (0, 0)).
- Use a deque (dq) to store cells to be processed.
- For each cell in the deque, explore its neighbors (up, down, left, right).
- Update the distance matrix dis[][] based on the cost of moving to the neighbor (0 or 1 coin).
- Push the neighbor to the front of the deque if the character of the neighbor is same as the current cell otherwise push it to the back of the deque.
Finally, return the answer as dis[m-1][n-1]

Below is the implementation of the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
#define INF 100000000
#define in_range(x, y, m, n)                               \
    (x >= 0 && x < m && y >= 0 && y < n)

char mat[1000][1000];
int dis[1000][1000];

// Initialize the distance matrix with INF values
void init(int m, int n)
{
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            dis[i][j] = INF;
}

// Perform Breadth-First Search to find the shortest path
void bfs(int startX, int startY, int m, int n)
{
    dis[startX][startY] = 0;
    deque<pair<int, int> > dq;
    pair<int, int> p;
    dq.push_front(make_pair(startX, startY));

    // BFS traversal
    while (!dq.empty()) {
        p = dq.front();
        dq.pop_front();
        int x = p.first;
        int y = p.second;
        int a[] = { 0, -1, 0, 1 }, b[] = { -1, 0, 1, 0 };

        // Explore neighbors
        for (int i = 0; i < 4; i++) {
            int tempX = x + a[i];
            int tempY = y + b[i];

            // Check if the neighbor is within the matrix
            // bounds
            if (in_range(tempX, tempY, m, n)) {

                // If the neighbor has the same character
                // and a shorter distance, update and add to
                // the front of the queue
                if (mat[tempX][tempY] == mat[x][y]
                    && dis[tempX][tempY] > dis[x][y]) {
                    dq.push_front(make_pair(tempX, tempY));
                    dis[tempX][tempY] = dis[x][y];
                }

                // If the neighbor has a different character
                // and a shorter distance, update and add to
                // the back of the queue
                else if (mat[tempX][tempY] != mat[x][y]) {
                    if (dis[tempX][tempY] > dis[x][y] + 1) {
                        dq.push_back(
                            make_pair(tempX, tempY));
                        dis[tempX][tempY] = dis[x][y] + 1;
                    }
                }
            }
        }
    }
}

// Display the matrix and distance matrix
void display(int m, int n)
{
    // Display the matrix
    cout << "Matrix:\n";
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }

    // Display the distance matrix
    cout << "Distance Matrix:\n";
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            cout << dis[i][j] << " ";
        }
        cout << endl;
    }
}

int main()
{
    // Static input
    int m = 2, n = 3;
    char matStatic[2][3]
        = { { 'a', 'b', 'n' }, { 'd', 'e', 'f' } };

    // Copy the static input to the global matrix
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            mat[i][j] = matStatic[i][j];
        }
    }

    // Initialize and perform BFS
    init(m, n);
    bfs(0, 0, m, n);

    // Output the result
    cout << "Shortest Path Distance: " << dis[m - 1][n - 1]
         << endl;

    return 0;
}

Java

import java.util.ArrayDeque;
import java.util.Deque;

class ShortestPathBFS {
    static final int INF = 100000000;

    static char[][] mat = new char[1000][1000];
    static int[][] dis = new int[1000][1000];

    static boolean inRange(int x, int y, int m, int n)
    {
        return x >= 0 && x < m && y >= 0 && y < n;
    }

    // Initialize the distance matrix with INF values
    static void init(int m, int n)
    {
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                dis[i][j] = INF;
    }

    // Perform Breadth-First Search to find the shortest
    // path
    static void bfs(int startX, int startY, int m, int n)
    {
        dis[startX][startY] = 0;
        Deque<int[]> dq = new ArrayDeque<>();
        int[] p = { startX, startY };
        dq.addFirst(p);

        // BFS traversal
        while (!dq.isEmpty()) {
            p = dq.pollFirst();
            int x = p[0];
            int y = p[1];
            int[] a = { 0, -1, 0, 1 }, b = { -1, 0, 1, 0 };

            // Explore neighbors
            for (int i = 0; i < 4; i++) {
                int tempX = x + a[i];
                int tempY = y + b[i];

                // Check if the neighbor is within the
                // matrix bounds
                if (inRange(tempX, tempY, m, n)) {
                    // If the neighbor has the same
                    // character and a shorter distance,
                    // update and add to the front of the
                    // queue
                    if (mat[tempX][tempY] == mat[x][y]
                        && dis[tempX][tempY] > dis[x][y]) {
                        dq.addFirst(
                            new int[] { tempX, tempY });
                        dis[tempX][tempY] = dis[x][y];
                    }
                    // If the neighbor has a different
                    // character and a shorter distance,
                    // update and add to the back of the
                    // queue
                    else if (mat[tempX][tempY]
                             != mat[x][y]) {
                        if (dis[tempX][tempY]
                            > dis[x][y] + 1) {
                            dq.addLast(
                                new int[] { tempX, tempY });
                            dis[tempX][tempY]
                                = dis[x][y] + 1;
                        }
                    }
                }
            }
        }
    }

    // Display the matrix and distance matrix
    static void display(int m, int n)
    {
        System.out.println("Matrix:");
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(mat[i][j] + " ");
            }
            System.out.println();
        }

        System.out.println("Distance Matrix:");
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(dis[i][j] + " ");
            }
            System.out.println();
        }
    }

    public static void main(String[] args)
    {
        int m = 2, n = 3;
        char[][] matStatic
            = { { 'a', 'b', 'n' }, { 'd', 'e', 'f' } };

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                mat[i][j] = matStatic[i][j];
            }
        }

        // Initialize and perform BFS
        init(m, n);
        bfs(0, 0, m, n);

        // Output the result
        System.out.println("Shortest Path Distance: "
                           + dis[m - 1][n - 1]);
    }
}

Python3

from collections import deque

INF = 10**9

# Initialize the distance matrix with INF values
def init(m, n):
    return [[INF for _ in range(n)] for _ in range(m)]

# Perform Breadth-First Search to find the shortest path
def bfs(startX, startY, mat, dis):
    dis[startX][startY] = 0
    dq = deque([(startX, startY)])
    dirs = [(0, -1), (-1, 0), (0, 1), (1, 0)]

    # BFS traversal
    while dq:
        x, y = dq.popleft()

        # Explore neighbors
        for dx, dy in dirs:
            tempX, tempY = x + dx, y + dy

            # Check if the neighbor is within the matrix bounds
            if 0 <= tempX < len(mat) and 0 <= tempY < len(mat[0]):

                # If the neighbor has the same character and a shorter distance,
                # update and add to the front of the queue
                if mat[tempX][tempY] == mat[x][y] and dis[tempX][tempY] > dis[x][y]:
                    dq.appendleft((tempX, tempY))
                    dis[tempX][tempY] = dis[x][y]

                # If the neighbor has a different character and a shorter distance,
                # update and add to the back of the queue
                elif mat[tempX][tempY] != mat[x][y]:
                    if dis[tempX][tempY] > dis[x][y] + 1:
                        dq.append((tempX, tempY))
                        dis[tempX][tempY] = dis[x][y] + 1

# Display the matrix and distance matrix
def display(mat, dis):
    # Display the matrix
    print("Matrix:")
    for row in mat:
        print(" ".join(row))

    # Display the distance matrix
    print("Distance Matrix:")
    for row in dis:
        print(" ".join(map(str, row)))

# Driver code
if __name__ == "__main__":
    # Static input
    m, n = 2, 3
    mat = [['a', 'b', 'n'], ['d', 'e', 'f']]

    # Initialize and perform BFS
    dis = init(m, n)
    bfs(0, 0, mat, dis)

    # Output the result
    print("Shortest Path Distance:", dis[m - 1][n - 1])

using System;
using System.Collections.Generic;

class Program
{
    const int INF = 100000000;

    static char[,] mat = new char[1000, 1000];
    static int[,] dis = new int[1000, 1000];

    // Initialize the distance matrix with INF values
    static void Init(int m, int n)
    {
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                dis[i, j] = INF;
            }
        }
    }

    // Perform Breadth-First Search to find the shortest path
    static void BFS(int startX, int startY, int m, int n)
    {
        dis[startX, startY] = 0;
        Queue<Tuple<int, int>> dq = new Queue<Tuple<int, int>>();
        Tuple<int, int> p = Tuple.Create(startX, startY);
        dq.Enqueue(p);

        // BFS traversal
        while (dq.Count > 0)
        {
            p = dq.Dequeue();
            int x = p.Item1;
            int y = p.Item2;
            int[] a = { 0, -1, 0, 1 };
            int[] b = { -1, 0, 1, 0 };

            // Explore neighbors
            for (int i = 0; i < 4; i++)
            {
                int tempX = x + a[i];
                int tempY = y + b[i];

                // Check if the neighbor is within the matrix bounds
                if (InRange(tempX, tempY, m, n))
                {
                    // If the neighbor has the same character and a shorter distance, update and add to
                    // the front of the queue
                    if (mat[tempX, tempY] == mat[x, y] && dis[tempX, tempY] > dis[x, y])
                    {
                        dq.Enqueue(Tuple.Create(tempX, tempY));
                        dis[tempX, tempY] = dis[x, y];
                    }
                    // If the neighbor has a different character and a shorter distance, update and add to
                    // the back of the queue
                    else if (mat[tempX, tempY] != mat[x, y])
                    {
                        if (dis[tempX, tempY] > dis[x, y] + 1)
                        {
                            dq.Enqueue(Tuple.Create(tempX, tempY));
                            dis[tempX, tempY] = dis[x, y] + 1;
                        }
                    }
                }
            }
        }
    }

    // Check if the position is within the matrix bounds
    static bool InRange(int x, int y, int m, int n)
    {
        return x >= 0 && x < m && y >= 0 && y < n;
    }

    // Display the matrix and distance matrix
    static void Display(int m, int n)
    {
        // Display the matrix
        Console.WriteLine("Matrix:");
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                Console.Write(mat[i, j] + " ");
            }
            Console.WriteLine();
        }

        // Display the distance matrix
        Console.WriteLine("Distance Matrix:");
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                Console.Write(dis[i, j] + " ");
            }
            Console.WriteLine();
        }
    }

    static void Main()
    {
        // Static input
        int m = 2, n = 3;
        char[,] matStatic = { { 'a', 'b', 'n' }, { 'd', 'e', 'f' } };

        // Copy the static input to the global matrix
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                mat[i, j] = matStatic[i, j];
            }
        }

        // Initialize and perform BFS
        Init(m, n);
        BFS(0, 0, m, n);

        // Output the result
        Console.WriteLine("Shortest Path Distance: " + dis[m - 1, n - 1]);
    }
}

// This code is contributed by shivamgupta0987654321

JavaScript

// javaScript code for the above approach

const INF = 100000000;

// Function to check if a cell is within the matrix bounds
function inRange(x, y, m, n) {
    return x >= 0 && x < m && y >= 0 && y < n;
}

const mat = [];
const dis = [];

// Initialize the distance matrix with INF values
function init(m, n) {
    for (let i = 0; i < m; i++) {
        mat.push(new Array(n).fill(''));
        dis.push(new Array(n).fill(INF));
    }
}

// Perform Breadth-First Search to find the shortest path
function bfs(startX, startY, m, n) {
    dis[startX][startY] = 0;
    const dq = [];
    dq.push([startX, startY]);

    // BFS traversal
    while (dq.length > 0) {
        const [x, y] = dq.shift();
        const a = [0, -1, 0, 1];
        const b = [-1, 0, 1, 0];

        // Explore neighbors
        for (let i = 0; i < 4; i++) {
            const tempX = x + a[i];
            const tempY = y + b[i];

            // Check if the neighbor is within the matrix bounds
            if (inRange(tempX, tempY, m, n)) {

                // If the neighbor has the same character
                // and a shorter distance, update and add to
                // the front of the queue
                if (mat[tempX][tempY] === mat[x][y] && dis[tempX][tempY] > dis[x][y]) {
                    dq.unshift([tempX, tempY]);
                    dis[tempX][tempY] = dis[x][y];
                }
    
                // If the neighbor has a different character
                // and a shorter distance, update and add to
                // the back of the queue
                else if (mat[tempX][tempY] !== mat[x][y]) {
                    if (dis[tempX][tempY] > dis[x][y] + 1) {
                        dq.push([tempX, tempY]);
                        dis[tempX][tempY] = dis[x][y] + 1;
                    }
                }
            }
        }
    }
}

// Display the matrix and distance matrix
function display(m, n) {
    // Display the matrix
    console.log("Matrix:");
    for (let i = 0; i < m; i++) {
        console.log(mat[i].join(' '));
    }

    // Display the distance matrix
    console.log("Distance Matrix:");
    for (let i = 0; i < m; i++) {
        console.log(dis[i].join(' '));
    }
}

// Driver code
// Static input
const m = 2, n = 3;
const matStatic = [
    ['a', 'b', 'n'],
    ['d', 'e', 'f']
];

// Initialize and perform BFS
init(m, n);

// Copy the static input to the global matrix
for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
        mat[i][j] = matStatic[i][j];
    }
}

bfs(0, 0, m, n);

// Output the result
console.log("Shortest Path Distance:", dis[m - 1][n - 1]);

Output

Shortest Path Distance: 3

Time Complexity: O(m * n), where 'm' is the number of rows and 'n' is the number of columns in the matrix.
Auxiliary Space: O(m * n)

Find minimum steps required to reach the end of a matrix | Set - 1

anshu_1230

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