In this post, we will see the solution of the Python Coding Question which is already asked in the TCS NQT Exam.

Problem Statement: A factory is packing candies into the packets. The candies packets here represent an array of N number of integer values. The task is to find the empty packets(0) of candies and push it to the end of the conveyor belt(array).

Example 1 :

N=10 and arr = [40,50,0,10,90,0,50,0,65,0].

There are 4 empty packets in the given set. These 4 empty packets represented as 0 should be pushed towards the end of the array

Input :

10 – Value of N

[40,50,0,10,90,0,50,0,65,0] – Element of arr[O] to arr[N-1]

Output:

40 50 10 90 50 65 0 0 0 0

Example 2:

N=6 and arr = [6,0,1,8,0,2].

There are 2 empty packets in the given set. These 2 empty packets represented as 0 should be pushed towards the end of the array

Input:

6 — Value of N.

[6,0,1,8,0,2] – Element of arr[0] to arr[N-1]

Output:

6 1 8 2 0 0

Solution 1:

# taking inputs
N = int(input())
arr = list(map(int, input().split()))

# empty list created
non_zero_arr = []
zero_arr = []

# Iterate through the list to collect all non-zero
# and zero elements separately in the list
for packet in arr :

    # non-zero element check
    if packet :
        non_zero_arr.append(packet)
    else :
        zero_arr.append(packet)

# concatenating both the list
result_arr = non_zero_arr + zero_arr

print(result_arr)

Output:

8
40 50 0 10 90 0 50 0
[40, 50, 10, 90, 50, 0, 0, 0]

Solution 2:

# taking inputs
N = int(input())
arr = list(map(int, input().split()))

# Iterate through the list to collect all non-zero element
non_zero_arr = [packet for packet in arr if packet]

# number of zeros count 
zeros_count = arr.count(0)

# concatenating both the list
result_arr = non_zero_arr + [0] * zeros_count

print(result_arr)

Output:

8
40 50 0 1 9 0 5 0
[40, 50, 1, 9, 5, 0, 0, 0]

Both time and space complexity for the above solutions is O(N).