Count paths with distance equal to Manhattan distance

Last Updated : 13 Sep, 2022

Given two points (x1, y1) and (x2, y2) in 2-D coordinate system. The task is to count all the paths whose distance is equal to the Manhattan distance between both the given points.
Examples:

Input: x1 = 2, y1 = 3, x2 = 4, y2 = 5
Output: 6
Input: x1 = 2, y1 = 6, x2 = 4, y2 = 9
Output: 10

Approach: The Manhattan distance between the points (x1, y1) and (x2, y2) will be abs(x1 – x2) + abs(y1 – y2)
Let abs(x1 – x2) = m and abs(y1 – y2) = n
Every path with distance equal to the Manhattan distance will always have m + n edges, m horizontal and n vertical edges. Hence this is a basic case of Combinatorics which is based upon group formation. The idea behind this is the number of ways in which (m + n) different things can be divided into two groups, one containing m items and the other containing n items which is given by ^{m + n}C_n i.e. (m + n)! / m! * n!.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the value of nCk
ll binomialCoeff(int n, int k)
{
    ll res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] / 
    // [k * (k-1) *---* 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Function to return the number of paths
ll countPaths(int x1, int y1, int x2, int y2)
{
 
    // Difference between the 'x' 
    // coordinates of the given points
    int m = abs(x1 - x2);
 
    // Difference between the 'y' 
    // coordinates of the given points
    int n = abs(y1 - y2);
 
    return (binomialCoeff(m + n, n));
}
 
// Driver code
int main()
{
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5;
    cout << countPaths(x1, y1, x2, y2);
    return 0;
}

Java

// Java implementation of the approach 
class GfG 
{
     
//static long ll long long int 
 
// Function to return the value of nCk 
static long binomialCoeff(int n, int k) 
{ 
    long res = 1; 
 
    // Since C(n, k) = C(n, n-k) 
    if (k > n - k) 
        k = n - k; 
 
    // Calculate value of 
    // [n * (n-1) *---* (n-k+1)] / 
    // [k * (k-1) *---* 1] 
    for (int i = 0; i < k; ++i)
    { 
        res *= (n - i); 
        res /= (i + 1); 
    } 
 
    return res; 
} 
 
// Function to return the number of paths 
static long countPaths(int x1, int y1, int x2, int y2) 
{ 
 
    // Difference between the 'x' 
    // coordinates of the given points 
    int m = Math.abs(x1 - x2); 
 
    // Difference between the 'y' 
    // coordinates of the given points 
    int n = Math.abs(y1 - y2); 
 
    return (binomialCoeff(m + n, n)); 
} 
 
// Driver code 
public static void main(String[] args) 
{ 
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5; 
    System.out.println(countPaths(x1, y1, x2, y2)); 
}
} 
 
// This code is contributed by 
// Prerna Saini.

Python3

# Python3 implementation of the approach
 
# Function to return the value of nCk
def binomialCoeff(n, k):
 
    res = 1
 
    # Since C(n, k) = C(n, n-k)
    if (k > n - k):
        k = n - k
 
    # Calculate value of
    # [n * (n-1) *---* (n-k+1)] / 
    # [k * (k-1) *---* 1]
    for i in range(k):
        res *= (n - i)
        res //= (i + 1)
 
    return res
 
# Function to return the number
# of paths
def countPaths(x1, y1, x2, y2):
 
    # Difference between the 'x' 
    # coordinates of the given points
    m = abs(x1 - x2)
 
    # Difference between the 'y' 
    # coordinates of the given points
    n = abs(y1 - y2)
 
    return (binomialCoeff(m + n, n))
 
# Driver code
x1, y1, x2, y2 = 2, 3, 4, 5
print(countPaths(x1, y1, x2, y2))
 
# This code is contributed 
# by Mohit Kumar

C#

// C# implementation of the approach 
using System;
 
class GFG 
{
 
// Function to return the value of nCk 
static long binomialCoeff(int n, int k) 
{ 
    long res = 1; 
 
    // Since C(n, k) = C(n, n-k) 
    if (k > n - k) 
        k = n - k; 
 
    // Calculate value of 
    // [n * (n-1) *---* (n-k+1)] / 
    // [k * (k-1) *---* 1] 
    for (int i = 0; i < k; ++i)
    { 
        res *= (n - i); 
        res /= (i + 1); 
    } 
 
    return res; 
} 
 
// Function to return the number of paths 
static long countPaths(int x1, int y1,
                       int x2, int y2) 
{ 
 
    // Difference between the 'x' 
    // coordinates of the given points 
    int m = Math.Abs(x1 - x2); 
 
    // Difference between the 'y' 
    // coordinates of the given points 
    int n = Math.Abs(y1 - y2); 
 
    return (binomialCoeff(m + n, n)); 
} 
 
// Driver code 
public static void Main() 
{ 
    int x1 = 2, y1 = 3, x2 = 4, y2 = 5; 
    Console.Write(countPaths(x1, y1, x2, y2)); 
}
} 
 
// This code is contributed by 
// Akanksha Rai

PHP

<?php
// PHP implementation of the approach 
//static long ll long long int 
 
// Function to return the value of nCk 
function binomialCoeff($n, $k) 
{ 
    $res = 1; 
 
    // Since C(n, k) = C(n, n-k) 
    if ($k > $n - $k) 
        $k = $n - $k; 
 
    // Calculate value of 
    // [n * (n-1) *---* (n-k+1)] / 
    // [k * (k-1) *---* 1] 
    for ($i = 0; $i < $k; ++$i)
    { 
        $res *= ($n - $i); 
        $res /= ($i + 1); 
    } 
 
    return $res; 
} 
 
// Function to return the number of paths 
function countPaths($x1, $y1, $x2, $y2) 
{ 
 
    // Difference between the 'x' 
    // coordinates of the given points 
    $m =abs($x1 - $x2); 
 
    // Difference between the 'y' 
    // coordinates of the given points 
    $n = abs($y1 - $y2); 
 
    return (binomialCoeff($m + $n, $n)); 
} 
 
// Driver code 
{ 
    $x1 = 2; $y1 = 3; $x2 = 4; $y2 = 5; 
    echo(countPaths($x1, $y1, $x2, $y2)); 
}
 
// This code is contributed by 
// Code_Mech.

Javascript

<script>
 
// javascript implementation of the approach
 
// Function to return the value of nCk
function binomialCoeff(n, k)
{
    var res = 1;
    var i;
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n * (n-1) *---* (n-k+1)] / 
    // [k * (k-1) *---* 1]
    for (i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Function to return the number of paths
function countPaths(x1, y1, x2, y2)
{
 
    // Difference between the 'x' 
    // coordinates of the given points
    var m = Math.abs(x1 - x2);
 
    // Difference between the 'y' 
    // coordinates of the given points
    var n = Math.abs(y1 - y2);
 
    return (binomialCoeff(m + n, n));
}
 
// Driver code
    var x1 = 2, y1 = 3, x2 = 4, y2 = 5;
    document.write(countPaths(x1, y1, x2, y2));
 
</script>

Output:

Time complexity: O(k) where k=abs(y1 – y2)

Auxiliary space: O(1)

Pairs with same Manhattan and Euclidean distance

Mostafijur Rahaman

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Practice Tags :

Count paths with distance equal to Manhattan distance

C++

Java

Python3

C#

PHP

Javascript

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