Maximum neighbor element in a matrix within distance K

Given a matrix mat[][] and an integer K, the task is to find the maximum neighbor within an absolute distance of K for each element of the matrix.

In other words for each Matrix[i][j] find maximum Matrix[p][q] such that abs (i-p) + abs (j-q) ≤ K.

Examples:

Input: mat[][] = {{1, 2}, {4, 5}}, K = 1
Output: {{4, 5}, {5, 5}}
Explanation:
Maximum neighbour to the element mat[0][0] = 4 (Distance = 1)
Maximum neighbour to the element mat[0][1] = 5 (Distance = 1)
Maximum neighbour to the element mat[1][0] = 5 (Distance = 1)
Maximum neighbour to the element mat[1][1] = 5 (Distance = 0)

Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, K = 2
Output: {{7, 8, 9}, {8, 9, 9}, {9, 9, 9}}



Approach: The idea is to iterate over a loop from 1 to K, to choose the element from neighbors with distance less than or equal to K. Each time, Iterate over the matrix to find the maximum adjacent element for each element of the matrix.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// maximum neighbor element within
// the distance of less than K
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to print the matrix
void printMatrix(vector<vector<int> > A)
{
    // Loop to iterate over the matrix
    for (int i = 0; i < A.size(); i++) {
        for (int j = 0; j < A[0].size(); j++)
            cout << A[i][j] << ' ';
        cout << '\n';
    }
}
  
// Function to find the maximum
// neighbor within the distance
// of less than equal to K
vector<vector<int> > getMaxNeighbour(
        vector<vector<int> >& A, int K)
{
    vector<vector<int> > ans = A;
      
    // Loop to find the maximum
    // element within the distance
    // of less than K
    for (int q = 1; q <= K; q++) {
        for (int i = 0; i < A.size(); i++) {
            for (int j = 0; j < A[0].size(); j++) {
                int maxi = ans[i][j];
                if (i > 0)
                    maxi = max(
                        maxi, ans[i - 1][j]);
                if (j > 0)
                    maxi = max(
                        maxi, ans[i][j - 1]);
                if (i < A.size() - 1)
                    maxi = max(
                        maxi, ans[i + 1][j]);
                if (j < A[0].size() - 1)
                    maxi = max(
                        maxi, ans[i][j + 1]);
                A[i][j] = maxi;
            }
        }
        ans = A;
    }
    return ans;
}
  
// Driver Code
int main()
{
    vector<vector<int> > B = { { 1, 2, 3 }, 
                  { 4, 5, 6 }, { 7, 8, 9 } };
    printMatrix(getMaxNeighbour(B, 2));
    return 0;
}

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Java

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// Java implementation to find the 
// maximum neighbor element within
// the distance of less than K
import java.util.*;
  
class GFG{
  
// Function to print the matrix
static void printMatrix(int [][] A)
{
      
    // Loop to iterate over the matrix
    for(int i = 0; i < A.length; i++)
    {
       for(int j = 0; j < A[0].length; j++)
          System.out.print(A[i][j] + " ");
       System.out.print('\n');
    }
}
  
// Function to find the maximum
// neighbor within the distance
// of less than equal to K
static int [][] getMaxNeighbour(int [][] A,
                                int K)
{
    int [][] ans = A;
      
    // Loop to find the maximum
    // element within the distance
    // of less than K
    for(int q = 1; q <= K; q++) 
    {
       for(int i = 0; i < A.length; i++)
       {
          for(int j = 0; j < A[0].length; j++) 
          {
             int maxi = ans[i][j];
             if (i > 0)
                 maxi = Math.max(maxi, 
                                 ans[i - 1][j]);
             if (j > 0)
                 maxi = Math.max(maxi, 
                                 ans[i][j - 1]);
             if (i < A.length - 1)
                 maxi = Math.max(maxi, 
                                 ans[i + 1][j]);
             if (j < A[0].length - 1)
                 maxi = Math.max(maxi,
                                 ans[i][j + 1]);
             A[i][j] = maxi;
          }
       }
       ans = A;
    }
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int [][] B = { { 1, 2, 3 }, 
                   { 4, 5, 6 },
                   { 7, 8, 9 } };
                     
    printMatrix(getMaxNeighbour(B, 2));
}
}
  
// This code is contributed by Princi Singh

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C#

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// C# implementation to find the 
// maximum neighbor element within
// the distance of less than K
using System;
  
class GFG{
  
// Function to print the matrix
static void printMatrix(int [,] A)
{
      
    // Loop to iterate over the matrix
    for(int i = 0; i < A.GetLength(0); i++)
    {
       for(int j = 0; j < A.GetLength(1); j++)
          Console.Write(A[i, j] + " ");
       Console.Write('\n');
    }
}
  
// Function to find the maximum
// neighbor within the distance
// of less than equal to K
static int [,] getMaxNeighbour(int [,] A,
                               int K)
{
    int [,] ans = A;
      
    // Loop to find the maximum
    // element within the distance
    // of less than K
    for(int q = 1; q <= K; q++) 
    {
       for(int i = 0; i < A.GetLength(0); i++)
       {
          for(int j = 0; j < A.GetLength(1); j++)
          {
             int maxi = ans[i, j];
             if (i > 0)
                 maxi = Math.Max(maxi, 
                                 ans[i - 1, j]);
             if (j > 0)
                 maxi = Math.Max(maxi, 
                                 ans[i, j - 1]);
             if (i < A.GetLength(0) - 1)
                 maxi = Math.Max(maxi, 
                                 ans[i + 1, j]);
             if (j < A.GetLength(0) - 1)
                 maxi = Math.Max(maxi,
                                 ans[i, j + 1]);
             A[i, j] = maxi;
          }
       }
       ans = A;
    }
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    int [,] B = { { 1, 2, 3 }, 
                  { 4, 5, 6 },
                  { 7, 8, 9 } };
                      
    printMatrix(getMaxNeighbour(B, 2));
}
}
  
// This code is contributed by Princi Singh

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Output:

7 8 9 
8 9 9 
9 9 9

Time Complexity: O(M*N*K)
Space Complexity: O(M*N)

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