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 one 2 times and by subtracting 2 one times. So the subproblem for 4 is called twice.
Since same suproblems are called again, this problem has Overlapping Subprolems 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

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

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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