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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 −

2 | 1 | 2 |

1 | 1 | 1 |

1 | 1 | 1 |

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]])*

#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; }

Number of steps: 2

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