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 from . 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 . 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 , Then the required value at the cell is to be . 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 using namespace std;   const long long MAX = 1e18;   long long n, m; vector v; long long dp;   // 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 = { 15, 153 };     v = { 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> v = new ArrayList<>(151); static long[][] dp = new long;   // 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 = [ 15, 153 ]     v = [ 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> v = new List>();   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());         v[i].Add(0);         v[i].Add(0);     }     v = 15;     v = 153;     v = 135;     v = 17;     n = 2; m = 2;     long x = 3;           // Function call     Console.WriteLine(solve(x)); } }   // This code is contributed by avanitrachhadiya2155

Javascript



Output:

125

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