A car factory has two assembly lines, each with n stations. A station is denoted by S_{i,j} where i is either 1 or 2 and indicates the assembly line the station is on, and j indicates the number of the station. The time taken per station is denoted by a_{i,j}. Each station is dedicated to some sort of work like engine fitting, body fitting, painting and so on. So, a car chassis must pass through each of the n stations in order before exiting the factory. The parallel stations of the two assembly lines perform the same task. After it passes through station S_{i,j}, it will continue to station S_{i,j+1} unless it decides to transfer to the other line. Continuing on the same line incurs no extra cost, but transferring from line i at station j – 1 to station j on the other line takes time t_{i,j}. Each assembly line takes an entry time e_{i} and exit time x_{i} which may be different for the two lines. Give an algorithm for computing the minimum time it will take to build a car chassis.

The below figure presents the problem in a clear picture:

The following information can be extracted from the problem statement to make it simpler:

- Two assembly lines, 1 and 2, each with stations from 1 to n.
- A car chassis must pass through all stations from 1 to n in order(in any of the two assembly lines). i.e. it cannot jump from station i to station j if they are not at one move distance.
- The car chassis can move one station forward in the same line, or one station diagonally in the other line. It incurs an extra cost ti, j to move to station j from line i. No cost is incurred for movement in same line.
- The time taken in station j on line i is a
_{i, j}. - S
_{i, j}represents a station j on line i.

**Breaking the problem into smaller sub-problems:**

We can easily find the ith factorial if (i-1)th factorial is known. Can we apply the similar funda here?

If the minimum time taken by the chassis to leave station S_{i, j-1} is known, the minimum time taken to leave station S_{i, j} can be calculated quickly by combining a_{i, j} and t_{i, j}.

**T1(j)** indicates the minimum time taken by the car chassis to leave station j on assembly line 1.

**T2(j)** indicates the minimum time taken by the car chassis to leave station j on assembly line 2.

**Base cases:**

The entry time e_{i} comes into picture only when the car chassis enters the car factory.

Time taken to leave first station in line 1 is given by:

T1(1) = Entry time in Line 1 + Time spent in station S_{1,1}

T1(1) = e_{1} + a_{1,1}

Similarly, time taken to leave first station in line 2 is given by:

T2(1) = e_{2} + a_{2,1}

**Recursive Relations:**

If we look at the problem statement, it quickly boils down to the below observations:

The car chassis at station S_{1,j} can come either from station S_{1, j-1} or station S_{2, j-1}.

Case #1: Its previous station is S_{1, j-1}

The minimum time to leave station S_{1,j} is given by:

T1(j) = Minimum time taken to leave station S_{1, j-1} + Time spent in station S_{1, j}

T1(j) = T1(j-1) + a_{1, j}

Case #2: Its previous station is S_{2, j-1}

The minimum time to leave station S1, j is given by:

T1(j) = Minimum time taken to leave station S_{2, j-1} + Extra cost incurred to change the assembly line + Time spent in station S_{1, j}

T1(j) = T2(j-1) + t_{2, j} + a_{1, j}

The minimum time T1(j) is given by the minimum of the two obtained in cases #1 and #2.

T1(j) = min((T1(j-1) + a_{1, j}), (T2(j-1) + t_{2, j} + a_{1, j}))

Similarly the minimum time to reach station S2, j is given by:

T2(j) = min((T2(j-1) + a_{2, j}), (T1(j-1) + t_{1, j} + a_{2, j}))

The total minimum time taken by the car chassis to come out of the factory is given by:

Tmin = min(Time taken to leave station S_{i,n} + Time taken to exit the car factory)

Tmin = min(T1(n) + x_{1}, T2(n) + x_{2})

**Why dynamic programming?**

The above recursion exhibits overlapping sub-problems. There are two ways to reach station S_{1, j}:

- From station S
_{1, j-1} - From station S
_{2, j-1}

So, to find the minimum time to leave station S_{1, j} the minimum time to leave the previous two stations must be calculated(as explained in above recursion).

Similarly, there are two ways to reach station S_{2, j}:

- From station S
_{2, j-1} - From station S
_{1, j-1}

Please note that the minimum times to leave stations S_{1, j-1} and S_{2, j-1} have already been calculated.

So, we need two tables to store the partial results calculated for each station in an assembly line. The table will be filled in bottom-up fashion.

**Note:**

In this post, the word “leave” has been used in place of “reach” to avoid the confusion. Since the car chassis must spend a fixed time in each station, the word leave suits better.

**Implementation:**

## C++

// A C++ program to find minimum possible

// time by the car chassis to complete

#include

using namespace std;

#define NUM_LINE 2

#define NUM_STATION 4

// Utility function to find minimum of two numbers

int min(int a, int b)

{

return a < b ? a : b;
}
int carAssembly(int a[][NUM_STATION],
int t[][NUM_STATION],
int *e, int *x)
{
int T1[NUM_STATION], T2[NUM_STATION], i;
// time taken to leave first station in line 1
T1[0] = e[0] + a[0][0];
// time taken to leave first station in line 2
T2[0] = e[1] + a[1][0];
// Fill tables T1[] and T2[] using the
// above given recursive relations
for (i = 1; i < NUM_STATION; ++i)
{
T1[i] = min(T1[i - 1] + a[0][i],
T2[i - 1] + t[1][i] + a[0][i]);
T2[i] = min(T2[i - 1] + a[1][i],
T1[i - 1] + t[0][i] + a[1][i]);
}
// Consider exit times and retutn minimum
return min(T1[NUM_STATION - 1] + x[0],
T2[NUM_STATION - 1] + x[1]);
}
// Driver Code
int main()
{
int a[][NUM_STATION] = {{4, 5, 3, 2},
{2, 10, 1, 4}};
int t[][NUM_STATION] = {{0, 7, 4, 5},
{0, 9, 2, 8}};
int e[] = {10, 12}, x[] = {18, 7};
cout << carAssembly(a, t, e, x);
return 0;
}
// This is code is contributed by rathbhupendra
[tabby title ="C"]

`// A C program to find minimum possible time by the car chassis to complete ` `#include <stdio.h> ` `#define NUM_LINE 2 ` `#define NUM_STATION 4 ` ` ` `// Utility function to find minimum of two numbers ` `int` `min(` `int` `a, ` `int` `b) { ` `return` `a < b ? a : b; } ` ` ` `int` `carAssembly(` `int` `a[][NUM_STATION], ` `int` `t[][NUM_STATION], ` `int` `*e, ` `int` `*x) ` `{ ` ` ` `int` `T1[NUM_STATION], T2[NUM_STATION], i; ` ` ` ` ` `T1[0] = e[0] + a[0][0]; ` `// time taken to leave first station in line 1 ` ` ` `T2[0] = e[1] + a[1][0]; ` `// time taken to leave first station in line 2 ` ` ` ` ` `// Fill tables T1[] and T2[] using the above given recursive relations ` ` ` `for` `(i = 1; i < NUM_STATION; ++i) ` ` ` `{ ` ` ` `T1[i] = min(T1[i-1] + a[0][i], T2[i-1] + t[1][i] + a[0][i]); ` ` ` `T2[i] = min(T2[i-1] + a[1][i], T1[i-1] + t[0][i] + a[1][i]); ` ` ` `} ` ` ` ` ` `// Consider exit times and retutn minimum ` ` ` `return` `min(T1[NUM_STATION-1] + x[0], T2[NUM_STATION-1] + x[1]); ` `} ` ` ` `int` `main() ` `{ ` ` ` `int` `a[][NUM_STATION] = {{4, 5, 3, 2}, ` ` ` `{2, 10, 1, 4}}; ` ` ` `int` `t[][NUM_STATION] = {{0, 7, 4, 5}, ` ` ` `{0, 9, 2, 8}}; ` ` ` `int` `e[] = {10, 12}, x[] = {18, 7}; ` ` ` ` ` `printf` `(` `"%d"` `, carAssembly(a, t, e, x)); ` ` ` ` ` `return` `0; ` `} ` |

Output:

35

## Java

`// A java program to find minimum possible ` `// time by the car chassis to complete ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `NUM_LINE = ` `2` `; ` ` ` `static` `int` `NUM_STATION = ` `4` `; ` ` ` ` ` `// Utility function to find minimum of two numbers ` ` ` `static` `int` `min(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `return` `a < b ? a : b; ` ` ` ` ` `} ` ` ` ` ` `static` `int` `carAssembly(` `int` `a[][], ` `int` `t[][], ` `int` `e[], ` `int` `x[]) ` ` ` `{ ` ` ` `int` `T1[]= ` `new` `int` `[NUM_STATION]; ` ` ` `int` `T2[] =` `new` `int` `[NUM_STATION] ; ` ` ` `int` `i; ` ` ` ` ` `// time taken to leave first station in line 1 ` ` ` `T1[` `0` `] = e[` `0` `] + a[` `0` `][` `0` `]; ` ` ` ` ` `// time taken to leave first station in line 2 ` ` ` `T2[` `0` `] = e[` `1` `] + a[` `1` `][` `0` `]; ` ` ` ` ` `// Fill tables T1[] and T2[] using ` ` ` `// the above given recursive relations ` ` ` `for` `(i = ` `1` `; i < NUM_STATION; ++i) ` ` ` `{ ` ` ` `T1[i] = min(T1[i - ` `1` `] + a[` `0` `][i], ` ` ` `T2[i - ` `1` `] + t[` `1` `][i] + a[` `0` `][i]); ` ` ` `T2[i] = min(T2[i - ` `1` `] + a[` `1` `][i], ` ` ` `T1[i - ` `1` `] + t[` `0` `][i] + a[` `1` `][i]); ` ` ` `} ` ` ` ` ` `// Consider exit times and retutn minimum ` ` ` `return` `min(T1[NUM_STATION-` `1` `] + x[` `0` `], ` ` ` `T2[NUM_STATION-` `1` `] + x[` `1` `]); ` ` ` `} ` ` ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `a[][] = {{` `4` `, ` `5` `, ` `3` `, ` `2` `}, ` ` ` `{` `2` `, ` `10` `, ` `1` `, ` `4` `}}; ` ` ` `int` `t[][] = {{` `0` `, ` `7` `, ` `4` `, ` `5` `}, ` ` ` `{` `0` `, ` `9` `, ` `2` `, ` `8` `}}; ` ` ` `int` `e[] = {` `10` `, ` `12` `}, x[] = {` `18` `, ` `7` `}; ` ` ` ` ` `System.out.println(carAssembly(a, t, e, x)); ` ` ` ` ` `} ` `} ` `// This code is contributed by vt_m ` |

## Python3

`# Python program to find minimum possible ` `# time by the car chassis to complete ` ` ` `def` `carAssembly (a, t, e, x): ` ` ` ` ` `NUM_STATION ` `=` `len` `(a[` `0` `]) ` ` ` `T1 ` `=` `[` `0` `for` `i ` `in` `range` `(NUM_STATION)] ` ` ` `T2 ` `=` `[` `0` `for` `i ` `in` `range` `(NUM_STATION)] ` ` ` ` ` `T1[` `0` `] ` `=` `e[` `0` `] ` `+` `a[` `0` `][` `0` `] ` `# time taken to leave ` ` ` `# first station in line 1 ` ` ` `T2[` `0` `] ` `=` `e[` `1` `] ` `+` `a[` `1` `][` `0` `] ` `# time taken to leave ` ` ` `# first station in line 2 ` ` ` ` ` `# Fill tables T1[] and T2[] using ` ` ` `# above given recursive relations ` ` ` `for` `i ` `in` `range` `(` `1` `, NUM_STATION): ` ` ` `T1[i] ` `=` `min` `(T1[i` `-` `1` `] ` `+` `a[` `0` `][i], ` ` ` `T2[i` `-` `1` `] ` `+` `t[` `1` `][i] ` `+` `a[` `0` `][i]) ` ` ` `T2[i] ` `=` `min` `(T2[i` `-` `1` `] ` `+` `a[` `1` `][i], ` ` ` `T1[i` `-` `1` `] ` `+` `t[` `0` `][i] ` `+` `a[` `1` `][i] ) ` ` ` ` ` `# consider exit times and return minimum ` ` ` `return` `min` `(T1[NUM_STATION ` `-` `1` `] ` `+` `x[` `0` `], ` ` ` `T2[NUM_STATION ` `-` `1` `] ` `+` `x[` `1` `]) ` ` ` `a ` `=` `[[` `4` `, ` `5` `, ` `3` `, ` `2` `], ` ` ` `[` `2` `, ` `10` `, ` `1` `, ` `4` `]] ` `t ` `=` `[[` `0` `, ` `7` `, ` `4` `, ` `5` `], ` ` ` `[` `0` `, ` `9` `, ` `2` `, ` `8` `]] ` `e ` `=` `[` `10` `, ` `12` `] ` `x ` `=` `[` `18` `, ` `7` `] ` ` ` `print` `(carAssembly(a, t, e, x)) ` ` ` `# This code is contributed by Soumen Ghosh ` |

## C#

`// A C# program to find minimum possible ` `// time by the car chassis to complete ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `NUM_STATION = 4; ` ` ` ` ` `// Utility function to find minimum ` ` ` `// of two numbers ` ` ` `static` `int` `min(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `return` `a < b ? a : b; ` ` ` ` ` `} ` ` ` ` ` `static` `int` `carAssembly(` `int` `[,]a, ` `int` `[,]t, ` ` ` `int` `[]e, ` `int` `[]x) ` ` ` `{ ` ` ` `int` `[]T1= ` `new` `int` `[NUM_STATION]; ` ` ` `int` `[]T2 =` `new` `int` `[NUM_STATION] ; ` ` ` `int` `i; ` ` ` ` ` `// time taken to leave first station ` ` ` `// in line 1 ` ` ` `T1[0] = e[0] + a[0,0]; ` ` ` ` ` `// time taken to leave first station ` ` ` `// in line 2 ` ` ` `T2[0] = e[1] + a[1,0]; ` ` ` ` ` `// Fill tables T1[] and T2[] using ` ` ` `// the above given recursive relations ` ` ` `for` `(i = 1; i < NUM_STATION; ++i) ` ` ` `{ ` ` ` `T1[i] = min(T1[i - 1] + a[0,i], ` ` ` `T2[i - 1] + t[1,i] + a[0,i]); ` ` ` `T2[i] = min(T2[i - 1] + a[1,i], ` ` ` `T1[i - 1] + t[0,i] + a[1,i]); ` ` ` `} ` ` ` ` ` `// Consider exit times and retutn ` ` ` `// minimum ` ` ` `return` `min(T1[NUM_STATION-1] + x[0], ` ` ` `T2[NUM_STATION-1] + x[1]); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `int` `[,]a = { {4, 5, 3, 2}, ` ` ` `{2, 10, 1, 4} }; ` ` ` ` ` `int` `[,]t = { {0, 7, 4, 5}, ` ` ` `{0, 9, 2, 8} }; ` ` ` ` ` `int` `[]e = {10, 12}; ` ` ` `int` `[]x = {18, 7}; ` ` ` ` ` `Console.Write(carAssembly(a, t, e, x)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by nitin mittal. ` |

## PHP

`<?php ` `// A PHP program to find minimum ` `// possible time by the car chassis ` `// to complete ` ` ` `$NUM_LINE` `= 2; ` `$NUM_STATION` `= 4; ` ` ` `// Utility function to find ` `// minimum of two numbers ` `function` `carAssembly(` `$a` `, ` `$t` `, ` ` ` `$e` `, ` `$x` `) ` `{ ` ` ` `global` `$NUM_LINE` `, ` ` ` `$NUM_STATION` `; ` ` ` `$T1` `= ` `array` `(); ` ` ` `$T2` `= ` `array` `(); ` ` ` `$i` `; ` ` ` ` ` `$T1` `[0] = ` `$e` `[0] + ` `$a` `[0][0]; ` `// time taken to leave ` ` ` `// first station in line 1 ` ` ` `$T2` `[0] = ` `$e` `[1] + ` `$a` `[1][0]; ` `// time taken to leave ` ` ` `// first station in line 2 ` ` ` ` ` `// Fill tables T1[] and T2[] ` ` ` `// using the above given ` ` ` `// recursive relations ` ` ` `for` `(` `$i` `= 1; ` ` ` `$i` `< ` `$NUM_STATION` `; ++` `$i` `) ` ` ` `{ ` ` ` `$T1` `[` `$i` `] = min(` `$T1` `[` `$i` `- 1] + ` `$a` `[0][` `$i` `], ` ` ` `$T2` `[` `$i` `- 1] + ` `$t` `[1][` `$i` `] + ` ` ` `$a` `[0][` `$i` `]); ` ` ` `$T2` `[` `$i` `] = min(` `$T2` `[` `$i` `- 1] + ` `$a` `[1][` `$i` `], ` ` ` `$T1` `[` `$i` `- 1] + ` `$t` `[0][` `$i` `] + ` ` ` `$a` `[1][` `$i` `]); ` ` ` `} ` ` ` ` ` `// Consider exit times ` ` ` `// and return minimum ` ` ` `return` `min(` `$T1` `[` `$NUM_STATION` `- 1] + ` `$x` `[0], ` ` ` `$T2` `[` `$NUM_STATION` `- 1] + ` `$x` `[1]); ` `} ` ` ` `// Driver Code ` `$a` `= ` `array` `(` `array` `(4, 5, 3, 2), ` ` ` `array` `(2, 10, 1, 4)); ` `$t` `= ` `array` `(` `array` `(0, 7, 4, 5), ` ` ` `array` `(0, 9, 2, 8)); ` `$e` `= ` `array` `(10, 12); ` `$x` `= ` `array` `(18, 7); ` ` ` `echo` `carAssembly(` `$a` `, ` `$t` `, ` `$e` `, ` `$x` `); ` ` ` `// This code is contributed ` `// by anuj_67. ` `?> ` |

**Output:**

35

The bold line shows the path covered by the car chassis for given input values.

**Exercise:**

Extend the above algorithm to print the path covered by the car chassis in the factory.

This article is compiled by **Aashish Barnwal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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