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Minimum cost path in a Matrix by moving only on value difference of X
  • Difficulty Level : Expert
  • Last Updated : 17 Jun, 2021

Given a matrix mat[][] and an integer X, the task is to find the minimum number of operations required to reach  
(N, M)     from (1, 1)     . In each move, we can move either right or down in the matrix, but to move on the next cell in the matrix the value of the cell must be mat[i][j] + X     . In one operation the value at any cell can be decremented by 1.

Examples: 

Input: mat[][] = {{8, 10, 14}, {5, 41, 19}, {10, 2, 25}}, X = 3 
Output: 11 
Explanation: 
After performing the operations on the matrix: 
7 10 13 
5 41 16 
10 2 19
Here the minimum required operation is 11. 
8 => 7 = 1 
14 => 13 = 1 
19 => 16 = 3 
25 => 19 = 6
Path: 7 => 10 => 13 => 16 => 19

Input: mat[][] = {{15, 153}, {135, 17}}, X = 3 
Output: 125 
 

Approach: The idea is to use Dynamic Programming to solve this problem. The general idea is to iterate over every possible cell of the matrix and find the number of operations required if the value at the current cell is not changed in the final path. If the value at the current cell is mat[i][j]     , Then the required value at the cell is to be mat[i][j] - X * (i + j)     . Similarly, we can compute the number of operations required recursively.



Below is the implementation of the above approach:

C++




// C++ implementation to find the
// minimum number of operations
// required to move from
// (1, 1) to (N, M)
 
#include <bits/stdc++.h>
using namespace std;
 
const long long MAX = 1e18;
 
long long n, m;
vector<long long> v[151];
long long dp[151][151];
 
// Function to find the minimum
// operations required to move
// to bottom-right cell of matrix
long long min_operation(long long i,
    long long j, long long val, long long x) {
         
    // Condition to check if the
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1) {
        if (val > v[i][j]) {
            return dp[i][j] = MAX;
        }
        else {
            return dp[i][j] =
                v[i][j] - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of
    // matrix
    if (i == n || j == m) {
        return dp[i][j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already
    // computed
    if (dp[i][j] != -1) {
        return dp[i][j];
    }
     
    // Condition to check that the
    // movement with the current
    // value is not possible
    if (val > v[i][j]) {
        return dp[i][j] = MAX;
    }
    long long tmp = v[i][j] - val;
     
    // Recursive call to compute the
    // number of operation required
    // to compute the value
    tmp += min(min_operation(i + 1,
            j, val + x, x),
            min_operation(i,
            j + 1, val + x, x));
    return dp[i][j] = tmp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
long long solve(long long x)
{
    long long ans = INT64_MAX;
     
    // Loop to iterate over every
    // possible cell of the matrix
    for (long long i = 0;
        i < n; i++) {
 
        for (long long j = 0;
            j < m; j++) {
 
            long long val =
                v[i][j] - x * (i + j);
 
            memset(dp, -1,
                sizeof(dp));
 
            val = min_operation(
                0, 0, val, x);
 
            ans = min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    n = 2, m = 2;
    long long x = 3;
     
    v[0] = { 15, 153 };
    v[1] = { 135, 17 };
     
    // Function Call
    cout << solve(x) << endl;
    return 0;
}

Java




// Java implementation to find the 
// minimum number of operations
// required to move from 
// (1, 1) to (N, M)
import java.lang.*;
import java.util.*;
 
class GFG{
     
static final long MAX = (long)1e18;
static long n, m;
 
static List<List<Long>> v = new ArrayList<>(151);
static long[][] dp = new long[151][151];
 
// Function to find the minimum
// operations required to move
// to bottom-right cell of matrix
static long min_operation(long i, long j,
                          long val, long x)
{
     
    // Condition to check if the
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1)
    {
        if (val > v.get((int)i).get((int)j))
        {
            return dp[(int)i][(int)j] = MAX;
        }
        else
        {
            return dp[(int)i][(int)j] =
                v.get((int)i).get((int)j) - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of
    // matrix
    if (i == n || j == m)
    {
        return dp[(int)i][(int)j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already
    // computed
    if (dp[(int)i][(int)j] != -1)
    {
        return dp[(int)i][(int)j];
    }
     
    // Condition to check that the
    // movement with the current
    // value is not possible
    if (val > v.get((int)i).get((int)j))
    {
        return dp[(int)i][(int)j] = MAX;
    }
    long tmp = v.get((int)i).get((int)j) - val;
     
    // Recursive call to compute the
    // number of operation required
    // to compute the value
    tmp += Math.min(min_operation(i + 1,
                             j, val + x, x),
                    min_operation(i, j + 1,
                                   val + x, x));
                                    
    return dp[(int)i][(int)j] = tmp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
static long solve(long x)
{
    long ans = Long.MAX_VALUE;
     
    // Loop to iterate over every
    // possible cell of the matrix
    for(long i = 0; i < n; i++)
    {
        for(long j = 0; j < m; j++)
        {
            long val = v.get((int)i).get((int)j) -
                       x * (i + j);
     
            for(int k = 0; k < dp.length; k++)
                for(int l = 0; l < dp[k].length; l++)
                    dp[k][l] = -1;
                     
            val = min_operation(0l, 0l, val, x);
            ans = Math.min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
    n = 2; m = 2;
    long x = 3;
     
    v.add(Arrays.asList(15l, 153l));
    v.add(Arrays.asList(135l, 17l));
     
    // Function call
    System.out.println(solve(x));
}
}
 
// This code is contributed by offbeat

Python3




# Python3 implementation to find the
# minimum number of operations
# required to move from
# (1, 1) to (N, M)
MAX = 1e18
v = [[ 0, 0]] * (151)
dp = [[-1 for i in range(151)]
        for i in range(151)]
 
# Function to find the minimum
# operations required to move
# to bottom-right cell of matrix
def min_operation(i, j, val, x):
 
    # Condition to check if the
    # current cell is the bottom-right
    # cell of the matrix
    if (i == n - 1 and j == m - 1):
        if (val > v[i][j]):
            dp[i][j] = MAX
            return MAX
 
        else:
            dp[i][j] = v[i][j] - val
            return dp[i][j]
 
    # Condition to check if the
    # current cell is out of
    # matrix
    if (i == n or j == m):
        dp[i][j] = MAX
        return MAX
 
    # Condition to check if the
    # current indices is already
    # computed
    if (dp[i][j] != -1):
        return dp[i][j]
 
    # Condition to check that the
    # movement with the current
    # value is not possible
    if (val > v[i][j]):
        dp[i][j] = MAX
        return MAX
 
    tmp = v[i][j] - val
 
    # Recursive call to compute the
    # number of operation required
    # to compute the value
    tmp += min(min_operation(i + 1, j,
                        val + x, x),
            min_operation(i, j + 1,
                            val + x, x))
    dp[i][j] = tmp
 
    return tmp
 
# Function to find the minimum
# number of operations required
# to reach the bottom-right cell
def solve(x):
     
    ans = 10 ** 19
 
    # Loop to iterate over every
    # possible cell of the matrix
    for i in range(n):
        for j in range(m):
            val = v[i][j] - x * (i + j)
 
            for ii in range(151):
                for jj in range(151):
                    dp[ii][jj] = -1
 
            val = min_operation(0, 0, val, x)
            ans = min(ans, val)
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    n = 2
    m = 2
    x = 3
 
    v[0] = [ 15, 153 ]
    v[1] = [ 135, 17 ]
 
    # Function Call
    print(solve(x))
 
# This code is contributed by mohit kumar 29    

C#




// C# implementation to find the  
// minimum number of operations 
// required to move from  
// (1, 1) to (N, M) 
using System;
using System.Collections.Generic;
 
class GFG{
     
static long MAX = (long)1e18;
static long n, m;
static List<List<long>> v = new List<List<long>>();
 
static long[,] dp = new long[151, 151];
 
// Function to find the minimum 
// operations required to move
// to bottom-right cell of matrix
static long min_operation(long i, long j,
                          long val, long x)
{
     
    // Condition to check if the 
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1)
    {
        if (val > v[(int)i][(int)j])
        {
            return dp[(int)i, (int)j] = MAX;
        }
        else
        {
            return dp[(int)i, (int)j] = v[(int)i][(int)j] - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of 
    // matrix
    if (i == n || j == m) 
    {
        return dp[(int)i, (int)j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already 
    // computed 
    if (dp[(int)i, (int)j] != -1)
    {
        return dp[(int)i, (int)j];
    }
     
    // Condition to check that the 
    // movement with the current 
    // value is not possible 
    if (val > v[(int)i][(int)j])
    {
        return dp[(int)i, (int)j] = MAX;
         
    }
     
    long temp = v[(int)i][(int)j] - val;
     
    // Recursive call to compute the 
    // number of operation required
    // to compute the value
    temp += Math.Min(min_operation(i + 1, j, val + x, x),
                     min_operation(i, j + 1, val + x, x));
    return dp[(int)i, (int)j] = temp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
static long solve(long x)
{
     
    long ans = Int64.MaxValue;
     
    // Loop to iterate over every 
    // possible cell of the matrix
    for(long i = 0; i < n; i++)
    {
        for(long j = 0; j < m; j++)
        {
            long val = v[(int)i][(int)j] - x * (i + j);
            for(long k = 0; k < dp.GetLength(0); k++)
            {
                for(long l = 0; l < dp.GetLength(1); l++)
                {
                    dp[k, l] = -1;
                }
            }
            val = min_operation(0, 0, val, x);
            ans = Math.Min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
static public void Main()
{
    for(int i = 0; i < 151; i++)
    {
        v.Add(new List<long>());
        v[i].Add(0);
        v[i].Add(0);
    }
    v[0][0] = 15;
    v[0][1] = 153;
    v[1][0] = 135;
    v[1][1] = 17;
    n = 2; m = 2;
    long x = 3;
     
    // Function call
    Console.WriteLine(solve(x));
}
}
 
// This code is contributed by avanitrachhadiya2155

Javascript




<script>
 
// Javascript implementation to find the
// minimum number of operations
// required to move from
// (1, 1) to (N, M)
let MAX = 1e18;
 
let n, m;
let v = new Array(151);
for(let i = 0; i < 151; i++)
{
    v[i] = [0, 0];
}
let dp = new Array(151);
for(let i = 0; i < 151; i++)
{
    dp[i] = new Array(151);
}
 
// Function to find the minimum
// operations required to move
// to bottom-right cell of matrix
function min_operation(i, j, val, x)
{
     
    // Condition to check if the
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1)
    {
        if (val > v[i][j])
        {
            return dp[i][j] = MAX;
        }
        else
        {
            return dp[i][j] = v[i][j] - val;
        }
    }
      
    // Condition to check if the
    // current cell is out of
    // matrix
    if (i == n || j == m)
    {
        return dp[i][j] = MAX;
    }
      
    // Condition to check if the
    // current indices is already
    // computed
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
      
    // Condition to check that the
    // movement with the current
    // value is not possible
    if (val > v[i][j])
    {
        return dp[i][j] = MAX;
    }
    let tmp = v[i][j] - val;
      
    // Recursive call to compute the
    // number of operation required
    // to compute the value
    tmp += Math.min(min_operation(i + 1,
                             j, val + x, x),
                    min_operation(i, j + 1,
                                   val + x, x));
                                     
    return dp[i][j] = tmp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
function solve(x)
{
    let ans = Number.MAX_VALUE;
      
    // Loop to iterate over every
    // possible cell of the matrix
    for(let i = 0; i < n; i++)
    {
        for(let j = 0; j < m; j++)
        {
            let val = v[i][j] -
                       x * (i + j);
      
            for(let k = 0; k < dp.length; k++)
                for(let l = 0; l < dp[k].length; l++)
                    dp[k][l] = -1;
                      
            val = min_operation(0, 0, val, x);
            ans = Math.min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
let x = 3;
n = 2; m = 2;
      
v[0] = [ 15, 153 ];
v[1] = [ 135, 17 ];
  
// Function call
document.write(solve(x)+"<br>");
 
// This code is contributed by unknown2108
 
</script>
Output: 
125

 

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