Find the number of ways to reach Kth step in stair case
Given an array arr[] of size N and an integer, K. Array represents the broken steps in a staircase. One can not reach a broken step. The task is to find the number of ways to reach the Kth step in the staircase starting from 0 when a step of maximum length 2 can be taken at any position. The answer can be very large. So, print the answer modulo 109 + 7.
Examples:
Input: arr[] = {3}, K = 6
Output: 4
0 -> 1 -> 2 -> 4 -> 5 -> 6
0 -> 1 -> 2 -> 4 -> 6
0 -> 2 -> 4 -> 5 -> 6
0 -> 2 -> 4 -> 6
Input: arr[] = {3, 4}, K = 6
Output: 0
Approach: This problem can be solved using dynamic programming. Create a dp[] array where dp[i] will store the number of ways to reach the ith step and the recurrence relation will be dp[i] = dp[i – 1] + dp[i – 2] only if the ith step is not broken otherwise 0. The final answer will be dp[K].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int MOD = 1000000007;// Function to return the number// of ways to reach the kth stepint number_of_ways(int arr[], int n, int k){ if (k == 1) return 1; // Create the dp array int dp[k + 1]; memset(dp, -1, sizeof dp); // Broken steps for (int i = 0; i < n; i++) dp[arr[i]] = 0; dp[0] = 1; dp[1] = (dp[1] == -1) ? 1 : dp[1]; // Calculate the number of ways for // the rest of the positions for (int i = 2; i <= k; ++i) { // If it is a blocked position if (dp[i] == 0) continue; // Number of ways to get to the ith step dp[i] = dp[i - 1] + dp[i - 2]; dp[i] %= MOD; } // Return the required answer return dp[k];}// Driver codeint main(){ int arr[] = { 3 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 6; cout << number_of_ways(arr, n, k); return 0;} |
Java
// Java implementation of the approach class GFG { static final int MOD = 1000000007; // Function to return the number // of ways to reach the kth step static int number_of_ways(int arr[], int n, int k) { if (k == 1) return 1; // Create the dp array int dp[] = new int[k + 1]; int i; for(i = 0; i < k + 1; i++) dp[i] = -1 ; // Broken steps for (i = 0; i < n; i++) dp[arr[i]] = 0; dp[0] = 1; dp[1] = (dp[1] == -1) ? 1 : dp[1]; // Calculate the number of ways for // the rest of the positions for (i = 2; i <= k; ++i) { // If it is a blocked position if (dp[i] == 0) continue; // Number of ways to get to the ith step dp[i] = dp[i - 1] + dp[i - 2]; dp[i] %= MOD; } // Return the required answer return dp[k]; } // Driver code public static void main (String[] args) { int arr[] = { 3 }; int n = arr.length; int k = 6; System.out.println(number_of_ways(arr, n, k)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach MOD = 1000000007; # Function to return the number # of ways to reach the kth step def number_of_ways(arr, n, k) : if (k == 1) : return 1; # Create the dp array dp = [-1] * (k + 1); # Broken steps for i in range(n) : dp[arr[i]] = 0; dp[0] = 1; dp[1] = 1 if (dp[1] == -1) else dp[1]; # Calculate the number of ways for # the rest of the positions for i in range(2, k + 1) : # If it is a blocked position if (dp[i] == 0) : continue; # Number of ways to get to the ith step dp[i] = dp[i - 1] + dp[i - 2]; dp[i] %= MOD; # Return the required answer return dp[k]; # Driver code if __name__ == "__main__" : arr = [ 3 ]; n = len(arr); k = 6; print(number_of_ways(arr, n, k)); # This code is contributed by kanugargng |
C#
// C# implementation of the approach using System;class GFG { static readonly int MOD = 1000000007; // Function to return the number // of ways to reach the kth step static int number_of_ways(int []arr, int n, int k) { if (k == 1) return 1; // Create the dp array int []dp = new int[k + 1]; int i; for(i = 0; i < k + 1; i++) dp[i] = -1 ; // Broken steps for (i = 0; i < n; i++) dp[arr[i]] = 0; dp[0] = 1; dp[1] = (dp[1] == -1) ? 1 : dp[1]; // Calculate the number of ways for // the rest of the positions for (i = 2; i <= k; ++i) { // If it is a blocked position if (dp[i] == 0) continue; // Number of ways to get to the ith step dp[i] = dp[i - 1] + dp[i - 2]; dp[i] %= MOD; } // Return the required answer return dp[k]; } // Driver code public static void Main (String[] args) { int []arr = { 3 }; int n = arr.Length; int k = 6; Console.WriteLine(number_of_ways(arr, n, k)); } }// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation of the approach let MOD = 1000000007; // Function to return the number // of ways to reach the kth step function number_of_ways(arr, n, k) { if (k == 1) return 1; // Create the dp array let dp = new Array(k + 1); let i; for(i = 0; i < k + 1; i++) dp[i] = -1 ; // Broken steps for (i = 0; i < n; i++) dp[arr[i]] = 0; dp[0] = 1; dp[1] = (dp[1] == -1) ? 1 : dp[1]; // Calculate the number of ways for // the rest of the positions for (i = 2; i <= k; ++i) { // If it is a blocked position if (dp[i] == 0) continue; // Number of ways to get to the ith step dp[i] = dp[i - 1] + dp[i - 2]; dp[i] %= MOD; } // Return the required answer return dp[k]; } let arr = [ 3 ]; let n = arr.length; let k = 6; document.write(number_of_ways(arr, n, k)); </script> |
Output:
4
Time Complexity: O(n)
Auxiliary Space: O(k)
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