Find the number of ways to reach Kth step in stair case
Last Updated :
01 Mar, 2022
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++
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int number_of_ways( int arr[], int n, int k)
{
if (k == 1)
return 1;
int dp[k + 1];
memset (dp, -1, sizeof dp);
for ( int i = 0; i < n; i++)
dp[arr[i]] = 0;
dp[0] = 1;
dp[1] = (dp[1] == -1) ? 1 : dp[1];
for ( int i = 2; i <= k; ++i) {
if (dp[i] == 0)
continue ;
dp[i] = dp[i - 1] + dp[i - 2];
dp[i] %= MOD;
}
return dp[k];
}
int main()
{
int arr[] = { 3 };
int n = sizeof (arr) / sizeof (arr[0]);
int k = 6;
cout << number_of_ways(arr, n, k);
return 0;
}
|
Java
class GFG
{
static final int MOD = 1000000007 ;
static int number_of_ways( int arr[],
int n, int k)
{
if (k == 1 )
return 1 ;
int dp[] = new int [k + 1 ];
int i;
for (i = 0 ; i < k + 1 ; i++)
dp[i] = - 1 ;
for (i = 0 ; i < n; i++)
dp[arr[i]] = 0 ;
dp[ 0 ] = 1 ;
dp[ 1 ] = (dp[ 1 ] == - 1 ) ? 1 : dp[ 1 ];
for (i = 2 ; i <= k; ++i)
{
if (dp[i] == 0 )
continue ;
dp[i] = dp[i - 1 ] + dp[i - 2 ];
dp[i] %= MOD;
}
return dp[k];
}
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));
}
}
|
Python3
MOD = 1000000007 ;
def number_of_ways(arr, n, k) :
if (k = = 1 ) :
return 1 ;
dp = [ - 1 ] * (k + 1 );
for i in range (n) :
dp[arr[i]] = 0 ;
dp[ 0 ] = 1 ;
dp[ 1 ] = 1 if (dp[ 1 ] = = - 1 ) else dp[ 1 ];
for i in range ( 2 , k + 1 ) :
if (dp[i] = = 0 ) :
continue ;
dp[i] = dp[i - 1 ] + dp[i - 2 ];
dp[i] % = MOD;
return dp[k];
if __name__ = = "__main__" :
arr = [ 3 ];
n = len (arr);
k = 6 ;
print (number_of_ways(arr, n, k));
|
C#
using System;
class GFG
{
static readonly int MOD = 1000000007;
static int number_of_ways( int []arr,
int n, int k)
{
if (k == 1)
return 1;
int []dp = new int [k + 1];
int i;
for (i = 0; i < k + 1; i++)
dp[i] = -1 ;
for (i = 0; i < n; i++)
dp[arr[i]] = 0;
dp[0] = 1;
dp[1] = (dp[1] == -1) ? 1 : dp[1];
for (i = 2; i <= k; ++i)
{
if (dp[i] == 0)
continue ;
dp[i] = dp[i - 1] + dp[i - 2];
dp[i] %= MOD;
}
return dp[k];
}
public static void Main (String[] args)
{
int []arr = { 3 };
int n = arr.Length;
int k = 6;
Console.WriteLine(number_of_ways(arr, n, k));
}
}
|
Javascript
<script>
let MOD = 1000000007;
function number_of_ways(arr, n, k)
{
if (k == 1)
return 1;
let dp = new Array(k + 1);
let i;
for (i = 0; i < k + 1; i++)
dp[i] = -1 ;
for (i = 0; i < n; i++)
dp[arr[i]] = 0;
dp[0] = 1;
dp[1] = (dp[1] == -1) ? 1 : dp[1];
for (i = 2; i <= k; ++i)
{
if (dp[i] == 0)
continue ;
dp[i] = dp[i - 1] + dp[i - 2];
dp[i] %= MOD;
}
return dp[k];
}
let arr = [ 3 ];
let n = arr.length;
let k = 6;
document.write(number_of_ways(arr, n, k));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(k)
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