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
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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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. 
?> 

Output:

The time complexity of above solution is exponential, a close upper bound is O(3ⁿ). If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, when we start from n = 6, we can reach 4 by subtracting 1 two times and by subtracting 2 one times. So the subproblem for 4 is called twice.
Since same subproblems are called again, this problem has Overlapping Subproblems property. So min square sum problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array count[] in bottom up manner.

Below is the implementation of above idea :

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,  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 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; 
        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 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
    count[1] = 1
    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. 
?> 

Output :

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

My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

Recommended Posts: