Counts paths from a point to reach Origin : Memory Optimized

Given two integers N and M, which denotes the position of in the coordinate system, the task is to count path from this point to reach origin by taking steps either left or right.
Examples:

Input: 3 6 
Output: Number of Paths 84
Input: 3 3 
Output: Number of Paths 20

We have already discussed this problem in this article – 
Counts paths from a point to reach Origin
Approach: The idea is to use the Dynamic programming paradigm to solve this problem. In this problem, there is a restriction to move only left or down in a path, due to which for any point number of paths is equal to the sum of the number of paths for the left point and the number of paths for the right point and in this way we compute the path for every point in a bottom-up manner from (0, 0) to (N, M). The key observation in this problem to reduce the space is for computing the solution for a point say (i, j) we only need the value of (i-1, j) and (i, j-1), that is we don’t need the portion of (i-1, j-1) computed values once they are used.

Below is the implementation of the above approach:

C++

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// C++ implementation to count the
// paths from a points to origin
  
#include <iostream>
#include<cstring>
  
using namespace std;
  
// Function to count the paths from
// a point to the origin
long Count_Paths(int x,int y)
{
    // Base Case when the point
    // is already at origin
    if(x == 0 && y == 0) 
        return 0;
      
    // Base Case when the point
    // is on the x or y-axis
    if(x == 0 || y == 0)
        return 1;
      
    long dp[max(x,y)+1],
     p=max(x,y),q=min(x,y);
      
    // Loop to fill all the 
    // position as 1 in the array
    for(int i=0;i<=p;i++)
        dp[i]=1;
      
    // Loop to count the number of
    // paths from a point to origin
    // in bottom-up manner
    for(int i=1;i<=q;i++)
        for(int j=1;j<=p;j++)
            dp[j]+=dp[j-1];
      
    return dp[p];
}
  
// Driver Code
int main()
{
    int x = 3,y = 3;
      
    // Function Call
    cout<< "Number of Paths "
        << Count_Paths(x,y);
    return 0;
}

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Java

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// Java implementation to count the
// paths from a points to origin
  
  
class GFG {
  
    // Function to count the paths from
    // a point to the origin
    static long Count_Paths(int x, int y) {
        // Base Case when the point
        // is already at origin
        if (x == 0 && y == 0)
            return 0;
  
        // Base Case when the point
        // is on the x or y-axis
        if (x == 0 || y == 0)
            return 1;
  
        int[] dp = new int[Math.max(x, y) + 1];
        int p = Math.max(x, y), q = Math.min(x, y);
  
        // Loop to fill all the
        // position as 1 in the array
        for (int i = 0; i <= p; i++)
            dp[i] = 1;
  
        // Loop to count the number of
        // paths from a point to origin
        // in bottom-up manner
        for (int i = 1; i <= q; i++)
            for (int j = 1; j <= p; j++)
                dp[j] += dp[j - 1];
  
        return dp[p];
    }
  
    // Driver Code
    public static void main(String[] args) {
        int x = 3, y = 3;
  
        // Function Call
        System.out.print("Number of Paths " + Count_Paths(x, y));
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation to count the 
# paths from a points to origin 
  
# Function to count the paths from 
# a point to the origin 
def Count_Paths(x, y): 
      
    # Base Case when the point
    # is already at origin 
    if (x == 0 and y == 0):
        return 0
      
    # Base Case when the point
    # is on the x or y-axis 
    if (x == 0 and y == 0):
        return 1
      
    dp = [0] * (max(x, y) + 1)
      
    p = max(x, y)
    q = min(x, y) 
      
    # Loop to fill all the 
    # position as 1 in the array 
    for i in range(p + 1): 
        dp[i] = 1
      
    # Loop to count the number of 
    # paths from a point to origin 
    # in bottom-up manner 
    for i in range(1, q + 1):
        for j in range(1, p + 1): 
            dp[j] += dp[j - 1
      
    return dp[p] 
  
# Driver Code 
x = 3
y = 3
      
# Function call 
print("Number of Paths ", Count_Paths(x, y))
  
# This code is contributed by sanjoy_62

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C#

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// C# implementation to count the
// paths from a points to origin 
using System;
  
class GFG {
   
    // Function to count the paths from
    // a point to the origin
    static long Count_Paths(int x, int y) {
        // Base Case when the point
        // is already at origin
        if (x == 0 && y == 0)
            return 0;
   
        // Base Case when the point
        // is on the x or y-axis
        if (x == 0 || y == 0)
            return 1;
   
        int[] dp = new int[Math.Max(x, y) + 1];
        int p = Math.Max(x, y), q = Math.Min(x, y);
   
        // Loop to fill all the
        // position as 1 in the array
        for (int i = 0; i <= p; i++)
            dp[i] = 1;
   
        // Loop to count the number of
        // paths from a point to origin
        // in bottom-up manner
        for (int i = 1; i <= q; i++)
            for (int j = 1; j <= p; j++)
                dp[j] += dp[j - 1];
   
        return dp[p];
    }
   
    // Driver Code
    public static void Main(String[] args) {
        int x = 3, y = 3;
   
        // Function Call
        Console.Write("Number of Paths " + Count_Paths(x, y));
    }
}
  
// This code is contributed by sapnasingh4991

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Output:

Number of Paths  20

Performance Analysis:

  • Time Complexity: In the above-given approach, there are two loops of which takes O(N*M) time in the worst case. Therefore, the time complexity for this approach will be O(N*M).
  • Space Complexity: In the above-given approach, there is a single array of the size maximum value of N and M. Therefore the space complexity for the above approach will be O(max(N, M))

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