Count number of ways to cover a distance

Given a distance ‘dist’, count total number of ways to cover the distance with 1, 2 and 3 steps. 

Examples:

Input: n = 3
Output: 4
Explanation:
Below are the four ways
 1 step + 1 step + 1 step
 1 step + 2 step
 2 step + 1 step
 3 step

Input: n = 4
Output: 7
Explanation:
Below are the four ways
 1 step + 1 step + 1 step + 1 step
 1 step + 2 step + 1 step
 2 step + 1 step + 1 step 
 1 step + 1 step + 2 step
 2 step + 2 step
 3 step + 1 step
 1 step + 3 step

Recursive solution

• Approach: There are n stairs, and a person is allowed to next step, skip one position or skip two positions. So there are n positions. The idea is standing at the ith position the person can move by i+1, i+2, i+3 position. So a recursive function can be formed where at current index i the function is recursively called for i+1, i+2 and i+3 positions. 
  There is another way of forming the recursive function. To reach position i, a person has to jump either from i-1, i-2 or i-3 position where i is the starting position. 
  
• Algorithm:
  1. Create a recursive function (count(int n)) which takes only one parameter.
  2. Check the base cases. If the value of n is less than 0 then return 0, and if value of n is equal to zero then return 1 as it is the starting position.
  3. Call the function recursively with values n-1, n-2 and n-3 and sum up the values that are returned, i.e. sum = count(n-1) + count(n-2) + count(n-3).
  4. Return the value of sum.
• Implementation:

C++

// A naive recursive C++ program to count number of ways to cover
// a distance with 1, 2 and 3 steps
#include<iostream>
using namespace std;

// Returns count of ways to cover 'dist'
int printCountRec(int dist)
{
    // Base cases
    if (dist<0)      return 0;
    if (dist==0)  return 1;

    // Recur for all previous 3 and add the results
    return printCountRec(dist-1) +
           printCountRec(dist-2) +
           printCountRec(dist-3);
}

// driver program
int main()
{
    int dist = 4;
    cout << printCountRec(dist);
    return 0;
}

Java

// A naive recursive Java program to count number
// of ways to cover a distance with 1, 2 and 3 steps
import java.io.*;

class GFG
{
    // Function returns count of ways to cover 'dist'
    static int printCountRec(int dist)
    {
        // Base cases
        if (dist<0)
            return 0;
        if (dist==0)
            return 1;

        // Recur for all previous 3 and add the results
        return printCountRec(dist-1) +
               printCountRec(dist-2) +
               printCountRec(dist-3);
    }

    // driver program
    public static void main (String[] args)
    {
        int dist = 4;
        System.out.println(printCountRec(dist));
    }
}

// This code is contributed by Pramod Kumar

Python3

# A naive recursive Python3 program
# to count number of ways to cover
# a distance with 1, 2 and 3 steps

# Returns count of ways to
# cover 'dist'
def printCountRec(dist):

    # Base cases
    if dist < 0:
        return 0

    if dist == 0:
        return 1

    # Recur for all previous 3 and
   # add the results
    return (printCountRec(dist-1) +
            printCountRec(dist-2) +
            printCountRec(dist-3))

# Driver code
dist = 4
print(printCountRec(dist))
# This code is contributed by Anant Agarwal.

C#

// A naive recursive C# program to
// count number of ways to cover a
// distance with 1, 2 and 3 steps
using System;

class GFG {

    // Function returns count of
    // ways to cover 'dist'
    static int printCountRec(int dist)
    {
        // Base cases
        if (dist < 0)
            return 0;
        if (dist == 0)
            return 1;

        // Recur for all previous 3
        // and add the results
        return printCountRec(dist - 1) +
               printCountRec(dist - 2) +
               printCountRec(dist - 3);
    }

    // Driver Code
    public static void Main ()
    {
        int dist = 4;
        Console.WriteLine(printCountRec(dist));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// A naive recursive PHP program to
// count number of ways to cover
// a distance with 1, 2 and 3 steps

// Returns count of ways to cover 'dist'
function printCountRec( $dist)
{

    // Base cases
    if ($dist<0) return 0;
    if ($dist==0) return 1;

    // Recur for all previous 3
    // and add the results
    return printCountRec($dist - 1) +
           printCountRec($dist - 2) +
           printCountRec($dist - 3);
}

    // Driver Code
    $dist = 4;
    echo printCountRec($dist);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// A naive recursive javascript program to count number of ways to cover
// a distance with 1, 2 and 3 steps

// Returns count of ways to cover 'dist'
function printCountRec(dist)
{
    // Base cases
    if (dist<0)     return 0;
    if (dist==0) return 1;

    // Recur for all previous 3 and add the results
    return printCountRec(dist-1) +
        printCountRec(dist-2) +
        printCountRec(dist-3);
}

// driver program
var dist = 4;
document.write(printCountRec(dist));

</script>

Output:

7

• Complexity Analysis:
  • Time Complexity: O(3ⁿ). 
    The time complexity of the above solution is exponential, a close upper bound is O(3ⁿ). From each state 3, a recursive function is called. So the upper bound for n states is O(3ⁿ).
  • Space complexity: O(1). 
    No extra space is required.

Efficient solution

• Approach: The idea is similar, but it can be observed that there are n states but the recursive function is called 3 ^ n times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done in two ways. 
  • The first way is to keep the recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing (Top-Down Approach).
  • The second way is to take an extra space of size n and start computing values of states from 1, 2 .. to n, i.e. compute values of i, i+1, i+2 and then use them to calculate the value of i+3 (Bottom-Up Approach).
  • Overlapping Subproblems in Dynamic Programming.
  • Optimal substructure property in Dynamic Programming.
  • Dynamic Programming(DP) problems
• Algorithm:
  1. Create an array of size n + 1 and initialize the first 3 variables with 1, 1, 2. The base cases.
  2. Run a loop from 3 to n.
  3. For each index i, compute value of ith position as dp[i] = dp[i-1] + dp[i-2] + dp[i-3].
  4. Print the value of dp[n], as the Count of number of ways to cover a distance.
• Implementation:

C++

// A Dynamic Programming based C++ program to count number of ways
// to cover a distance with 1, 2 and 3 steps
#include<iostream>
using namespace std;

int printCountDP(int dist)
{
    int count[dist+1];

    // Initialize base values. There is one way to cover 0 and 1
    // distances and two ways to cover 2 distance
     count[0] = 1;
     if(dist >= 1)
            count[1] = 1;
     if(dist >= 2)
              count[2] = 2;

    // Fill the count array in bottom up manner
    for (int i=3; i<=dist; i++)
       count[i] = count[i-1] + count[i-2] + count[i-3];

    return count[dist];
}

// driver program
int main()
{
    int dist = 4;
    cout << printCountDP(dist);
    return 0;
}

Java

// A Dynamic Programming based Java program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps
import java.io.*;

class GFG
{
    // Function returns count of ways to cover 'dist'
    static int printCountDP(int dist)
    {
        int[] count = new int[dist+1];

        // Initialize base values. There is one way to
        // cover 0 and 1 distances and two ways to
        // cover 2 distance
        count[0] = 1;
          if(dist >= 1)
            count[1] = 1;
        if(dist >= 2)
              count[2] = 2;

        // Fill the count array in bottom up manner
        for (int i=3; i<=dist; i++)
            count[i] = count[i-1] + count[i-2] + count[i-3];

        return count[dist];
    }

    // driver program
    public static void main (String[] args)
    {
        int dist = 4;
        System.out.println(printCountDP(dist));
    }
}

// This code is contributed by Pramod Kumar

Python3

# A Dynamic Programming based on Python3
# program to count number of ways to
# cover a distance with 1, 2 and 3 steps

def printCountDP(dist):
    count = [0] * (dist + 1)

    # Initialize base values. There is
    # one way to cover 0 and 1 distances
    # and two ways to cover 2 distance
    count[0] = 1
    if dist >= 1 :
        count[1] = 1
    if dist >= 2 :
        count[2] = 2

    # Fill the count array in bottom
    # up manner
    for i in range(3, dist + 1):
        count[i] = (count[i-1] +
                   count[i-2] + count[i-3])

    return count[dist];

# driver program
dist = 4;
print( printCountDP(dist))

# This code is contributed by Sam007.

C#

// A Dynamic Programming based C# program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps
using System;

class GFG {

    // Function returns count of ways
    // to cover 'dist'
    static int printCountDP(int dist)
    {
        int[] count = new int[dist + 1];

        // Initialize base values. There is one
        // way to cover 0 and 1 distances
        // and two ways to cover 2 distance
        count[0] = 1;
        count[1] = 1;
        count[2] = 2;

        // Fill the count array
        // in bottom up manner
        for (int i = 3; i <= dist; i++)
            count[i] = count[i - 1] +
                       count[i - 2] +
                       count[i - 3];

        return count[dist];
    }

    // Driver Code
    public static void Main ()
    {
        int dist = 4;
        Console.WriteLine(printCountDP(dist));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// A Dynamic Programming based PHP program
// to count number of ways to cover a
// distance with 1, 2 and 3 steps

function printCountDP( $dist)
{
    $count = array();

    // Initialize base values. There is
    // one way to cover 0 and 1 distances
    // and two ways to cover 2 distance
    $count[0] = 1; $count[1] = 1;
    $count[2] = 2;

    // Fill the count array
    // in bottom up manner
    for ( $i = 3; $i <= $dist; $i++)
    $count[$i] = $count[$i - 1] +
                 $count[$i - 2] +
                 $count[$i - 3];

    return $count[$dist];
}

// Driver Code
$dist = 4;
echo printCountDP($dist);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// A Dynamic Programming based Javascript program
// to count number of ways to cover a distance
// with 1, 2 and 3 steps

// Function returns count of ways
// to cover 'dist'
function printCountDP(dist)
{
    let count = new Array(dist + 1);

    // Initialize base values. There is one
    // way to cover 0 and 1 distances
    // and two ways to cover 2 distance
    count[0] = 1;
    if (dist >= 1)
        count[1] = 1;
    if (dist >= 2)
        count[2] = 2;

    // Fill the count array
    // in bottom up manner
    for(let i = 3; i <= dist; i++)
        count[i] = count[i - 1] +
                   count[i - 2] +
                   count[i - 3];

    return count[dist];
}

// Driver code
let dist = 4;
document.write(printCountDP(dist));

// This code is contributed by divyeshrabadiya07

</script>

Output :

7

• Complexity Analysis:
  • Time Complexity: O(n). 
    Only one traversal of the array is needed. So Time Complexity is O(n)
  • Space complexity: O(n). 
    To store the values in a DP O(n) extra space is needed.

More Optimal Solution

Approach: Instead of using array of size n+1 we can use array of size 3 because for calculating no of ways for a particular step we need only last     3 steps no of ways.

    Algorithm:

1. Create an array of size 3 and initialize the values for step 0,1,2 as 1,1,2 (Base cases).
2. Run a loop from 3 to n(dist).
3. For each index compute the value as ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3] and store its value at i%3 index of array ways. If we are computing value for index 3 then the computed value will go at index 0 because for larger indices(4 ,5,6…..) we don’t need the value of index 0.
4. Return the value of ways[n%3].

C++

// A Dynamic Programming based C++ program to count number of ways
#include<iostream>
using namespace std;

int printCountDP(int dist)
{
        //Create the array of size 3.
        int  ways[3] , n = dist;

        //Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;

        //Run a loop from 3 to n
        //Bottom up approach to fill the array
        for(int i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];

        return ways[n%3];
}

// driver program
int main()
{
    int dist = 4;
    cout << printCountDP(dist);
    return 0;
}

Java

// A Dynamic Programming based Java program to count number of ways
import java.util.*;

class GFG {

static int printCountDP(int dist)
{
        // Create the array of size 3.
        int []ways = new int[3];
        int n = dist;


        // Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;

        // Run a loop from 3 to n
        // Bottom up approach to fill the array
        for(int i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];

        return ways[n%3];
}

// driver program
public static void main(String arg[])
    {
    int dist = 4;
    System.out.print(printCountDP(dist));
    }
}

// this code is contributed by shivanisinghss2110

Python3

# A Dynamic Programming based C++ program to count number of ways
def prCountDP( dist):

        # Create the array of size 3.
        ways = [0]*3
        n = dist

        # Initialize the bases cases
        ways[0] = 1
        ways[1] = 1
        ways[2] = 2

        # Run a loop from 3 to n
        # Bottom up approach to fill the array
        for i in range(3, n + 1):
            ways[i % 3] = ways[(i - 1) % 3] + ways[(i - 2) % 3] + ways[(i - 3) % 3]

        return ways[n % 3]

# driver program
dist = 4
print(prCountDP(dist))

# This code is contributed by shivanisinghss2110

C#

// A Dynamic Programming based C#
// program to count number of ways
using System;

class GFG{

static int printCountDP(int dist)
{
    // Create the array of size 3.
    int []ways = new int[3];
    int n = dist;

    // Initialize the bases cases
    ways[0] = 1;
    ways[1] = 1;
    ways[2] = 2;

    // Run a loop from 3 to n
    // Bottom up approach to fill the array
    for(int i = 3; i <= n; i++)
        ways[i % 3] = ways[(i - 1) % 3] +
                      ways[(i - 2) % 3] +
                      ways[(i - 3) % 3];

    return ways[n % 3];
}

// Driver code
public static void Main(String []arg)
{
    int dist = 4;

    Console.Write(printCountDP(dist));
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>
// A Dynamic Programming based javascript program to count number of ways

function printCountDP( dist)
{
        //Create the array of size 3.
        var  ways= [] , n = dist;
        ways.length = 3 ;

        //Initialize the bases cases
        ways[0] = 1;
        ways[1] = 1;
        ways[2] = 2;

        //Run a loop from 3 to n
        //Bottom up approach to fill the array
        for(var i=3 ;i<=n ;i++)
            ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];

        return ways[n%3];
}

// driver code

    var dist = 4;
    document.write(printCountDP(dist));
</script>

Output : 

7

Time Complexity : O(n)

Space Complexity : O(1)

This article is contributed by Vignesh Venkatesan. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above