TCS Coding Questions 2022 Day 1 Slot 2

Coding Question 2 for 2022 (September slot)

In this article, we will discuss about the TCS Coding Question which is asked in the TCS placement test. This type of Coding Questions will help you to crack your upcoming TCS exam as well as during your inteview process.

TCS Coding Question 2 Day 1 slot 2

TCS Coding Question Day 1 Slot 2 – Question 2

Given N gold wires, each wire has a length associated with it. At a time, only two adjacent small wires are assembled at the end of a large wire and the cost of forming is the sum of their length. Find the minimum cost when all wires are assembled to form a single wire.


For Example:

Suppose, Arr[]={7,6,8,6,1,1,}

{7,6,8,6,1,1}-{7,6,8,6,2} , cost =2

{7,6,8,6,2}- {7,6,8,8}, cost = 8

{7,6,8,8} – {13,8,8}, cost=13

{13,8,8} -{13,16}, cost=16

{13, 16} – {29}, cost =29

2+8+13+16+29=68


Hence , the minimum cost to assemble all gold wires is 68.


Constraints

  • 1<=N<=30
  • 1<= Arr[i]<=100

Example 1:

Input 

6  -> Value of N, represent size of Arr

7  -> Value of Arr[0], represent length of 1st wire

6 -> Value of Arr[1], represent length of 2nd wire

8 -> Value of Arr[2] , represent length of 3rd wire

6 -> Value of Arr[3], represent length of 4th wire

1 -> Value of Arr[4], represent length of 5th wire

1 -> Value of Arr[5], represent length of 6th wire


Output :

68 


Example 2:

Input 

4   -> Value of N, represents size of Arr

12  -> Value of Arr[0], represents length of 1st wire 

2   -> Value of Arr[1], represent length of 2nd wire

2   -> Value of Arr[2], represent length of 3rd wire

5  -> Value of Arr[3], represent length of 4th wire


Output :

34

One comment on “TCS Coding Questions 2022 Day 1 Slot 2”


  • Lohith

    Code written in python

    def mini(l1,n):
    count=n-1
    n1=0
    lis=l1
    while True:
    N=[lis[i]+lis[i+1] for i in range(len(lis)) if i+1<len(lis)]
    c=sum(N)
    j=0
    for i in range(len(N)):
    if N[i]<c:
    c=N[i]
    j=N.index(N[i])
    n1+=N[j]
    lis=lis[0:j]+[N[j]]+lis[j+2:n]
    print(lis)
    count-=1
    if count==0:
    break
    return n1
    n=int(input())
    l1=list(map(int,input().split()))
    print(mini(l1,n))