Given a 2D matrix consisting of positive integers, the task is to find the minimum number of steps required to reach the end(leftmost-bottom cell) of the matrix. If we are at cell (i, j) we can go to cells (i, j+arr[i][j]) or (i+arr[i][j], j). We can not go out of bounds. If no path exists, print -1.
Examples:
Input : mat[][] = {{2, 1, 2}, {1, 1, 1}, {1, 1, 1}} Output : 2 Explanation : The path will be {0, 0} -> {0, 2} -> {2, 2} Thus, we are reaching end in two steps. Input : mat[][] = {{1, 1, 2}, {1, 1, 1}, {2, 1, 1}} Output : 3
A simple solution is to explore all possible solutions. This will take exponential time.
Better approach: We can use dynamic programming to solve this problem in polynomial time.
Let’s decide the states of ‘dp’. We will build up our solution on 2d DP.
Let’s say we are at cell {i, j}. We will try to find the minimum number of steps required to reach the cell (n-1, n-1) from this cell.
We only have two possible paths i.e. to cells {i, j+arr[i][j]} or {i+arr[i][j], j}.
A simple recurrence relation will be:
dp[i][j] = 1 + min(dp[i+arr[i][j]][j], dp[i][j+arr[i][j]])
Below is the implementation of the above idea:
C++
// C++ program to implement above approach #include <bits/stdc++.h> #define n 3 using namespace std; // 2d array to store // states of dp int dp[n][n]; // array to determine whether // a state has been solved before int v[n][n]; // Function to find the minimum number of // steps to reach the end of matrix int minSteps( int i, int j, int arr[][n]) { // base cases if (i == n - 1 and j == n - 1) return 0; if (i > n - 1 || j > n - 1) return 9999999; // if a state has been solved before // it won't be evaluated again. if (v[i][j]) return dp[i][j]; v[i][j] = 1; // recurrence relation dp[i][j] = 1 + min(minSteps(i + arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr)); return dp[i][j]; } // Driver Code int main() { int arr[n][n] = { { 2, 1, 2 }, { 1, 1, 1 }, { 1, 1, 1 } }; int ans = minSteps(0, 0, arr); if (ans >= 9999999) cout << -1; else cout << ans; return 0; } |
Java
// Java program to implement above approach class GFG { static int n = 3 ; // 2d array to store // states of dp static int dp[][] = new int [n][n]; // array to determine whether // a state has been solved before static int [][] v = new int [n][n]; // Function to find the minimum number of // steps to reach the end of matrix static int minSteps( int i, int j, int arr[][]) { // base cases if (i == n - 1 && j == n - 1 ) { return 0 ; } if (i > n - 1 || j > n - 1 ) { return 9999999 ; } // if a state has been solved before // it won't be evaluated again. if (v[i][j] == 1 ) { return dp[i][j]; } v[i][j] = 1 ; // recurrence relation dp[i][j] = 1 + Math.min(minSteps(i + arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr)); return dp[i][j]; } // Driver Code public static void main(String[] args) { int arr[][] = { { 2 , 1 , 2 }, { 1 , 1 , 1 }, { 1 , 1 , 1 } }; int ans = minSteps( 0 , 0 , arr); if (ans >= 9999999 ) { System.out.println(- 1 ); } else { System.out.println(ans); } } } // This code contributed by Rajput-Ji |
Python3
# Python3 program to implement above approach import numpy as np; n = 3 # 2d array to store # states of dp dp = np.zeros((n, n)); # array to determine whether # a state has been solved before v = np.zeros((n, n)); # Function to find the minimum number of # steps to reach the end of matrix def minSteps(i, j, arr) : # base cases if (i = = n - 1 and j = = n - 1 ) : return 0 ; if (i > n - 1 or j > n - 1 ) : return 9999999 ; # if a state has been solved before # it won't be evaluated again. if (v[i][j]) : return dp[i][j]; v[i][j] = 1 ; # recurrence relation dp[i][j] = 1 + min (minSteps(i + arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr)); return dp[i][j]; # Driver Code arr = [ [ 2 , 1 , 2 ], [ 1 , 1 , 1 ], [ 1 , 1 , 1 ] ]; ans = minSteps( 0 , 0 , arr); if (ans > = 9999999 ) : print ( - 1 ); else : print (ans); # This code is contributed by AnkitRai01 |
C#
// C# program to implement above approach using System; class GFG { static int n = 3; // 2d array to store // states of dp static int [,]dp = new int [n, n]; // array to determine whether // a state has been solved before static int [,] v = new int [n, n]; // Function to find the minimum number of // steps to reach the end of matrix static int minSteps( int i, int j, int [,]arr) { // base cases if (i == n - 1 && j == n - 1) { return 0; } if (i > n - 1 || j > n - 1) { return 9999999; } // if a state has been solved before // it won't be evaluated again. if (v[i, j] == 1) { return dp[i, j]; } v[i, j] = 1; // recurrence relation dp[i, j] = 1 + Math.Min(minSteps(i + arr[i,j], j, arr), minSteps(i, j + arr[i,j], arr)); return dp[i, j]; } // Driver Code static public void Main () { int [,]arr = { { 2, 1, 2 }, { 1, 1, 1 }, { 1, 1, 1 } }; int ans = minSteps(0, 0, arr); if (ans >= 9999999) { Console.WriteLine(-1); } else { Console.WriteLine(ans); } } } // This code contributed by ajit. |
Javascript
<script> // Javascript program to implement // above approach let n = 3; // 2d array to store // states of dp let dp = new Array(n); for (let i = 0; i < n; i++) { dp[i] = new Array(n); } // array to determine whether // a state has been solved before let v = new Array(n); for (let i = 0; i < n; i++) { v[i] = new Array(n); } // Function to find the minimum number of // steps to reach the end of matrix function minSteps(i, j, arr) { // base cases if (i == n - 1 && j == n - 1) { return 0; } if (i > n - 1 || j > n - 1) { return 9999999; } // if a state has been solved before // it won't be evaluated again. if (v[i][j] == 1) { return dp[i][j]; } v[i][j] = 1; // recurrence relation dp[i][j] = 1 + Math.min(minSteps(i + arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr)); return dp[i][j]; } let arr = [ [ 2, 1, 2 ], [ 1, 1, 1 ], [ 1, 1, 1 ] ]; let ans = minSteps(0, 0, arr); if (ans >= 9999999) { document.write(-1); } else { document.write(ans); } </script> |
2
Time Complexity: O(N2).
Auxiliary Space: O(N2)
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