Program to reverse a Subarray in a given Array.

Given an array Arr of N integers, the task is to reverse a subarray of that array. The range of this subarray is given by L and R.

Examples:

Input: arr = [1, 2, 3, 4, 5, 6, 7], L = 1, R = 5
Output: [1, 6, 5, 4, 3, 2, 7]

Input: arr = [10, 20, 30, 40, 50], L = 0, R = 2
Output: [30, 20, 10, 40, 50]

Approach: Follow the steps below to solve the problem:

If the size of the sub-list is 1, then return the array.
If the size of the sub list is 2, then swap if the range lies between them.
Else, Start swapping from L and R up to the mid of (L + R).

Below is the implementation of the above approach:

Python3

def subArrReverse(arr, L, R):

no_of_elements = R-L + 1

if(no_of_elements == 1):

return arr

elif(no_of_elements == 2):

arr[L], arr[R], = arr[R], arr[L]

return arr

else:

i = 0

while(i < no_of_elements//2):

arr[L + i], arr[R-i] = arr[R-i], arr[L + i]

if((i != L + i+1 and i != R-i-1) and (L + i != L + i+1 and R-i != R-i-1)):

arr[L + i+1], arr[R-i-1] = arr[R-i-1], arr[L + i+1]

i += 2

return arr

arr = [1, 2, 3, 4, 5, 6, 7]

L = 1

R = 5

print('Original Array :', arr)

print('Sub-Array reverse :', subArrReverse(arr, L, R))

Output

Original Array    : [1, 2, 3, 4, 5, 6, 7]
Sub-Array reverse : [1, 6, 5, 4, 3, 2, 7]

Time complexity: O(log2(n))
Auxiliary space: O(1)

Related articles:

Given an array Arr of N integers, the task is to reverse a subarray of that array. The range of this subarray is given by L and R.

Examples:

Input: arr = [1, 2, 3, 4, 5, 6, 7], L = 1, R = 5
Output: [1, 6, 5, 4, 3, 2, 7]

Input: arr = [10, 20, 30, 40, 50], L = 0, R = 2
Output: [30, 20, 10, 40, 50]

Approach: Follow the steps below to solve the problem:

If the size of the sub-list is 1, then return the array.
If the size of the sub list is 2, then swap if the range lies between them.
Else, Start swapping from L and R up to the mid of (L + R).

Below is the implementation of the above approach:

Python3

def subArrReverse(arr, L, R):

no_of_elements = R-L + 1

if(no_of_elements == 1):

return arr

elif(no_of_elements == 2):

arr[L], arr[R], = arr[R], arr[L]

return arr

else:

i = 0

while(i < no_of_elements//2):

arr[L + i], arr[R-i] = arr[R-i], arr[L + i]

if((i != L + i+1 and i != R-i-1) and (L + i != L + i+1 and R-i != R-i-1)):

arr[L + i+1], arr[R-i-1] = arr[R-i-1], arr[L + i+1]

i += 2

return arr

arr = [1, 2, 3, 4, 5, 6, 7]

L = 1

R = 5

print('Original Array :', arr)

print('Sub-Array reverse :', subArrReverse(arr, L, R))

Output

Original Array    : [1, 2, 3, 4, 5, 6, 7]
Sub-Array reverse : [1, 6, 5, 4, 3, 2, 7]

Time complexity: O(log2(n))
Auxiliary space: O(1)

Related articles:

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Program to reverse a Subarray in a given Array.

Python3

Python3

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