Find minimum steps required to reach the end of a matrix in C++

C++Server Side ProgrammingProgramming

Suppose we have a 2D matrix with positive integers. We have to find the minimum steps required to move from to the end of the matrix (rightmost bottom cell), If we are at cell (i, j), we can go to the cell (i, j+mat[i, j]) or (i+mat[i, j], j), We cannot cross the bounds. So if the matrix is like −

212
111
111

The output will be 2. Path will be (0, 0) →(0, 2) → (2, 2)

Here we will use the Dynamic programming approach to solve this. Suppose we are at cell (i, j), we want to reach (n-1, n-1) cell. We can use the recurrence relation like below −

DP[i, j]=1+min ⁡(DP [i+arr [i , j] , j], DP[i , j+arr [ i , j]])

Example

#include<iostream>
#define N 3
using namespace std;
int table[N][N];
bool temp_val[N][N];
int countSteps(int i, int j, int arr[][N]) {
   if (i == N - 1 and j == N - 1)
      return 0;
   if (i > N - 1 || j > N - 1)
      return INT_MAX;
   if (temp_val[i][j])
      return table[i][j];
   temp_val[i][j] = true;
   table[i][j] = 1 + min(countSteps(i + arr[i][j], j, arr), countSteps(i, j + arr[i][j], arr));
   return table[i][j];
}
int main() {
   int arr[N][N] = { { 2, 1, 2 }, { 1, 1, 1 }, { 1, 1, 1 } };
   int ans = countSteps(0, 0, arr);
   if (ans >= INT_MAX)
      cout << -1;
   else
      cout <<"Number of steps: "<< ans;
}

Output

Number of steps: 2
raja
Published on 18-Dec-2019 11:34:20
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