Distance Traveled by Two Trains together in the same Direction
Last Updated :
31 Jan, 2023
Given two arrays A[] and B[], each consisting of N integers, containing the speeds of two trains travelling in the same direction, at each time unit, the task is to find the total distance travelled by the two trains together(side by side) throughout the journey.
Examples:
Input: A[] = {1, 2, 3, 2, 4}, B[] = {2, 1, 3, 1, 4}
Output: 3
Explanation :
Since A[1] + A[0] = B[0] + B[1], both the trains have travelled same distance after 2 units of time.
Now, since A[2] = B[2] = 3, both the trains have traveled this distance together.
After the 3rd unit of time, the speed of the trains are different.
Therefore, the total distance traveled by the two trains together is 3.
Input: A[] = {1, 1, 3, 2, 4}, B[] = {3, 1, 2, 1, 4}
Output:
Approach:
Follow the steps below to solve the problem:
- Traverse both the arrays simultaneously.
- For every ith index check if sum(A[0] .. A[i – 1]) is equal to sum(B[0] .. B[i – 1]) as well as if A[i] and B[i] are equal or not.
- If the above two conditions are satisfied, add A[i] to the answer.
- Finally, after traversal of the complete array, print answer.
Below is the implementation of above approach:
C++
#include <iostream>
using namespace std;
int calc_distance(int A[], int B[], int n)
{
int distance_traveled_A = 0;
int distance_traveled_B = 0;
int answer = 0;
for (int i = 0; i < 5; i++) {
distance_traveled_A += A[i];
distance_traveled_B += B[i];
if ((distance_traveled_A
== distance_traveled_B)
&& (A[i] == B[i])) {
answer += A[i];
}
}
return answer;
}
int main()
{
int A[5] = { 1, 2, 3, 2, 4 };
int B[5] = { 2, 1, 3, 1, 4 };
int N = sizeof(A) / sizeof(A[0]);
cout << calc_distance(A, B, N);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
class GFG{
static int calc_distance(int A[], int B[], int n)
{
int distance_traveled_A = 0;
int distance_traveled_B = 0;
int answer = 0;
for(int i = 0; i < 5; i++)
{
distance_traveled_A += A[i];
distance_traveled_B += B[i];
if ((distance_traveled_A ==
distance_traveled_B) &&
(A[i] == B[i]))
{
answer += A[i];
}
}
return answer;
}
public static void main (String[] args)
{
int A[] = { 1, 2, 3, 2, 4 };
int B[] = { 2, 1, 3, 1, 4 };
int N = A.length;
System.out.println(calc_distance(A, B, N));
}
}
|
Python3
def calc_distance(A, B, n):
distance_traveled_A = 0
distance_traveled_B = 0
answer = 0
for i in range(5):
distance_traveled_A += A[i]
distance_traveled_B += B[i]
if ((distance_traveled_A ==
distance_traveled_B) and
(A[i] == B[i])):
answer += A[i]
return answer
A = [ 1, 2, 3, 2, 4 ]
B = [ 2, 1, 3, 1, 4 ]
N = len(A)
print(calc_distance(A, B, N))
|
C#
using System;
class GFG{
static int calc_distance(int []A, int []B,
int n)
{
int distance_traveled_A = 0;
int distance_traveled_B = 0;
int answer = 0;
for(int i = 0; i < 5; i++)
{
distance_traveled_A += A[i];
distance_traveled_B += B[i];
if ((distance_traveled_A ==
distance_traveled_B) &&
(A[i] == B[i]))
{
answer += A[i];
}
}
return answer;
}
public static void Main(string []s)
{
int []A = { 1, 2, 3, 2, 4 };
int []B = { 2, 1, 3, 1, 4 };
int N = A.Length;
Console.Write(calc_distance(A, B, N));
}
}
|
Javascript
<script>
function calc_distance(A , B , n)
{
var distance_traveled_A = 0;
var distance_traveled_B = 0;
var answer = 0;
for (i = 0; i < 5; i++) {
distance_traveled_A += A[i];
distance_traveled_B += B[i];
if ((distance_traveled_A == distance_traveled_B) && (A[i] == B[i])) {
answer += A[i];
}
}
return answer;
}
var A = [ 1, 2, 3, 2, 4 ];
var B = [ 2, 1, 3, 1, 4 ];
var N = A.length;
document.write(calc_distance(A, B, N));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
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