Minimize maximum adjacent difference in a path from top-left to bottom-right

Given a matrix arr[][] of size M * N, where arr[i][j] represents the height of the cell (row, col). The task is to find a path from top-left to bottom-right such that the value of the maximum difference in height between adjacent cells is minimum for that path.

Note: A path energy is the maximum absolute difference in heights between two consecutive cells of the path.

Examples:

Input: 

Example matrix

Output: 1
Explanation: The path {1, 2, 1, 2, 2} has a maximum absolute difference of 1 in consecutive cells and this is better than all other possible paths

Input: 

Example matrix

Output: 1
Explanation: The highlighted path is the path with minimum value of maximum adjacent difference.

An approach using DFS + Binary Search:

The idea is to use Binary search to solve this problem, if we assume a threshold value as mid, and check whether there exist a path through which we can reach to end, then that might be my answer, else look for some bigger values.

• Initialize a variable start = 0 and end = maximum possible energy required and a variable result
• Do the following while start ≤ end
  • Calculate mid = (start + end) / 2
  • Check if mid energy is valid energy required by choosing any path into the matrix
    • Checking the valid energy using a valid function
      • Check if we go outside the matrix or if cell(i, j) is visited or absolute difference between consecutive cells is greater than our assumed maximum energy required
        • If true, then return false
      • Check if we reach the bottom-right cell
        • If true, then return true
      • Make the current cell(i, j) visited
      • Make all four direction calls and check if any path is valid
      • If true, then return true.
      • Otherwise, return false
  • If the value from a valid function returns true, then update the result and move the end to mid – 1
  • Otherwise, move the start to mid + 1
• Return the result

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach

#include <bits/stdc++.h>
using namespace std;
int m, n;

// Function to check if there is a valid path
bool isValid(int i, int j, int x,
             vector<vector<bool> >& visited,
             vector<vector<int> >& arr, long long parent)
{
    // Check if we go outside the matrix or
    // cell(i, j) is visited or absolute
    // difference between consecutive cell
    // is greater than our assumed maximum
    // energy required If true,
    // then return false
    if (i < 0 || j < 0 || i >= m || j >= n || visited[i][j]
        || abs((long long)arr[i][j] - parent) > x)
        return false;

    // Check if we reach at bottom-right
    // cell If true, then return true
    if (i == m - 1 && j == n - 1)
        return true;

    // Make the current cell(i, j) visited
    visited[i][j] = true;

    // Make all four direction call and
    // check if any path is valid If true,
    // then return true.
    if (isValid(i + 1, j, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i - 1, j, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i, j + 1, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i, j - 1, x, visited, arr, arr[i][j]))
        return true;

    // Otherwise, return false
    return false;
}

// Function to find the minimum value among
// the maximum adjacent differences
int minimumEnergyPath(vector<vector<int> >& arr)
{
    // Initialize a variable start = 0
    // and end = maximum possible
    // energy required
    int start = 0, end = 10000000;

    // Initialize a variable result
    int result = arr[0][0];

    // Loop to implement the binary search
    while (start <= end) {
        int mid = (start + end) / 2;

        // Initialize a visited array
        // of size (m * n)
        vector<vector<bool> > visited(
            m, vector<bool>(n, false));

        // Check if mid energy is valid
        // energy required by choosing any
        // path into the matrix If true,
        // update the result and
        // move end to mid - 1
        if (isValid(0, 0, mid, visited, arr, arr[0][0])) {

            result = mid;
            end = mid - 1;
        }

        // Otherwise, move start to mid + 1
        else {

            start = mid + 1;
        }
    }

    // Return the result
    return result;
}

// Driver code
int main()
{
    vector<vector<int> > arr = {
        { 1, 2, 1 },
        { 2, 8, 2 },
        { 2, 4, 2 },
    };
    m = arr.size(), n = arr[0].size();

    // Function Call
    cout << minimumEnergyPath(arr);

    return 0;
}

Java

// Java code to implement the above approach

import java.io.*;

class GFG {

    static int m, n;

    // Function to check if there is a valid path
    static boolean isValid(int i, int j, int x,
                           boolean[][] visited, int[][] arr,
                           long parent)
    {
        // Check if we go outside the matrix or
        // cell(i, j) is visited or absolute
        // difference between consecutive cell
        // is greater than our assumed maximum
        // energy required If true,
        // then return false
        if (i < 0 || j < 0 || i >= m || j >= n
            || visited[i][j]
            || Math.abs(arr[i][j] - parent) > x) {
            return false;
        }

        // Check if we reach at bottom-right
        // cell If true, then return true
        if (i == m - 1 && j == n - 1) {
            return true;
        }

        // Make the current cell(i, j) visited
        visited[i][j] = true;

        // Make all four direction call and
        // check if any path is valid If true,
        // then return true.
        if (isValid(i + 1, j, x, visited, arr, arr[i][j]))
            return true;
        if (isValid(i - 1, j, x, visited, arr, arr[i][j]))
            return true;
        if (isValid(i, j + 1, x, visited, arr, arr[i][j]))
            return true;
        if (isValid(i, j - 1, x, visited, arr, arr[i][j]))
            return true;

        // Otherwise, return false
        return false;
    }

    // Function to find the minimum value among
    // the maximum adjacent differences
    static int minimumEnergyPath(int[][] arr)
    {
        // Initialize a variable start = 0
        // and end = maximum possible
        // energy required
        int start = 0, end = 10000000;

        // Initialize a variable result
        int result = arr[0][0];

        // Loop to implement the binary search
        while (start <= end) {
            int mid = (start + end) / 2;

            // Initialize a visited array
            // of size (m * n)
            boolean[][] visited = new boolean[m][n];

            // Check if mid energy is valid
            // energy required by choosing any
            // path into the matrix If true,
            // update the result and
            // move end to mid - 1
            if (isValid(0, 0, mid, visited, arr,
                        arr[0][0])) {
                result = mid;
                end = mid - 1;
            }

            // Otherwise, move start to mid + 1
            else {
                start = mid + 1;
            }
        }

        // Return the result
        return result;
    }

    public static void main(String[] args)
    {
        int[][] arr
            = { { 1, 2, 1 }, { 2, 8, 2 }, { 2, 4, 2 } };

        m = arr.length;
        n = arr[0].length;
        // Function call
        System.out.print(minimumEnergyPath(arr));
    }
}

// This code is contributed by lokesh.

Python3

# Python code to implement the above approach

m,n=0,0
# Function to check if there is a valid path
def isValid(i,j,x,visited,arr,parent):
    # Check if we go outside the matrix or
    # cell(i, j) is visited or absolute
    # difference between consecutive cell
    # is greater than our assumed maximum
    # energy required If true,
    # then return false
    if(i<0 or j<0 or i>=m or j>=n or visited[i][j] or abs(arr[i][j]-parent)>x):
        return False
    
    # Check if we reach at bottom-right
    # cell If true, then return true
    if(i==m-1 and j==n-1):
        return True
    
    # Make the current cell(i, j) visited
    visited[i][j]=True
    
    # Make all four direction call and
    # check if any path is valid If true,
    # then return true.
    if(isValid(i+1,j,x,visited,arr,arr[i][j])):
        return True
    if(isValid(i-1,j,x,visited,arr,arr[i][j])):
        return True
    if(isValid(i,j+1,x,visited,arr,arr[i][j])):
        return True
    if(isValid(i,j-1,x,visited,arr,arr[i][j])):
        return True
    
    # Otherwise, return false
    return False
    
# Function to find the minimum value among
# the maximum adjacent differences
def minimumEnergyPath(arr):
    # Initialize a variable start = 0
    # and end = maximum possible
    # energy required
    start,end=0,10000000
    
    # Initialize a variable result
    result=arr[0][0]
    
    # Loop to implement the binary search
    while(start<=end):
        mid=(start+end)//2
        
        # Initialize a visited array
        # of size (m * n)
        visited=[[False for i in range(n)] for j in range(m)]
        
        # Check if mid energy is valid
        # energy required by choosing any
        # path into the matrix If true,
        # update the result and
        # move end to mid - 1
        if(isValid(0,0,mid,visited,arr,arr[0][0])):
            result=mid
            end=mid-1
        
        # Otherwise, move start to mid + 1
        else:
            start=mid+1
    
    # Return the result
    return result
    
# Driver Code 
arr=[[1,2,1],[2,8,2],[2,4,2]]
m=len(arr)
n=len(arr[0])

# Function Call
print(minimumEnergyPath(arr))

# This code is contributed by Pushpesh Raj.

C#

// C# implementation
using System;

public class GFG {
  static int m, n;

  // Function to check if there is a valid path
  public static bool isValid(int i, int j, int x,
                             bool[, ] visited,
                             int[, ] arr, int parent)
  {
    
    // Check if we go outside the matrix or
    // cell(i, j) is visited or absolute
    // difference between consecutive cell
    // is greater than our assumed maximum
    // energy required If true,
    // then return false
    if ((i < 0) || (j < 0) || (i >= m) || (j >= n)
        || (visited[i, j] == true)
        || (Math.Abs(arr[i, j] - parent) > x))
      return false;

    // Check if we reach at bottom-right
    // cell If true, then return true
    if (i == m - 1 && j == n - 1)
      return true;

    // Make the current cell(i, j) visited
    visited[i, j] = true;

    // Make all four direction call and
    // check if any path is valid If true,
    // then return true.
    if (isValid(i + 1, j, x, visited, arr, arr[i, j]))
      return true;
    if (isValid(i - 1, j, x, visited, arr, arr[i, j]))
      return true;
    if (isValid(i, j + 1, x, visited, arr, arr[i, j]))
      return true;
    if (isValid(i, j - 1, x, visited, arr, arr[i, j]))
      return true;

    // Otherwise, return false
    return false;
  }

  // Function to find the minimum value among
  // the maximum adjacent differences
  public static int minimumEnergyPath(int[, ] arr)
  {
    // Initialize a variable start = 0
    // and end = maximum possible
    // energy required
    int start = 0, end = 10000000;

    // Initialize a variable result
    int result = arr[0, 0];

    // Loop to implement the binary search
    while (start <= end) {
      int mid = (int)((start + end) / 2);

      // Initialize a visited array
      // of size (m * n)
      bool[, ] visited = new bool[m, n];
      for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
          visited[i, j] = false;
        }
      }

      // Check if mid energy is valid
      // energy required by choosing any
      // path into the matrix If true,
      // update the result and
      // move end to mid - 1
      if (isValid(0, 0, mid, visited, arr, arr[0, 0])
          == true) {

        result = mid;
        end = mid - 1;
      }

      // Otherwise, move start to mid + 1
      else {

        start = mid + 1;
      }
    }

    // Return the result
    return result;
  }

  static public void Main()
  {
    int[, ] arr = {
      { 1, 2, 1 },
      { 2, 8, 2 },
      { 2, 4, 2 },
    };
    m = arr.GetLength(0);
    n = arr.GetLength(1);

    // Function Call
    Console.WriteLine(minimumEnergyPath(arr));
  }
}

// This code is contributed by ksam24000

Javascript

// JavaScript code for the above approach

let m, n;

// Function to check if there is a valid path
function isValid(i, j, x, visited, arr, parent)
{
    // Check if we go outside the matrix or
    // cell(i, j) is visited or absolute
    // difference between consecutive cell
    // is greater than our assumed maximum
    // energy required If true,
    // then return false
    if (i < 0 || j < 0 || i >= m || j >= n || visited[i][j]
        || Math.abs(arr[i][j] - parent) > x)
        return false;

    // Check if we reach at bottom-right
    // cell If true, then return true
    if (i == m - 1 && j == n - 1)
        return true;

    // make the current cell(i, j) visited
    visited[i][j] = true;

    // make all four direction call and
    // check if any path is valid If true,
    // then return true.
    if (isValid(i + 1, j, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i - 1, j, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i, j + 1, x, visited, arr, arr[i][j]))
        return true;
    if (isValid(i, j - 1, x, visited, arr, arr[i][j]))
        return true;

    // Otherwise, return false
    return false;
}

// Function to find the minimum value among
// the maximum adjacent differences
function minimumEnergyPath(arr)
{
    // Initialize a variable start = 0
    // and end = maximum possible
    // energy required
    let start = 0, end = 10000000;

    // Initialize a variable result
    let result = arr[0][0];

    // Loop to implement the binary search
    while(start <= end) {
        let mid = Math.floor((start + end) / 2);

        // Initialize a visited array
        // of size (m * n)
        let visited
            = new Array(m) 
        for(let i = 0; i < m; i++)
        {
            visited[i] = new Array(n).fill(0)
        }

        // Check if mid energy is valid
        // energy required by choosing any
        // path into the matrix If true,
        // update the result and
        // move end to mid - 1
        if(isValid(0, 0, mid, visited, arr, arr[0][0])) {

            result = mid;
            end = mid - 1;
        }

        // Otherwise, move start to mid + 1
        else {

            start = mid + 1;
        }
    }

    // Return the result
    return result;
}

// Driver code

let arr = [
    [ 1, 2, 1 ],
    [ 2, 8, 2 ],
    [ 2, 4, 2 ],
];
m = arr.length, n = arr[0].length;

// Function Call
console.log(minimumEnergyPath(arr));

// This code is contributed by Potta Lokesh

Output

1

Time Complexity: O(log₂(K) * (M*N)), where K is the maximum element in the matrix and M, and N are the number of rows and columns in the given matrix respectively.
Auxiliary Space: O(M * N)