# Program to find longest subarray of 1s after deleting one element using Python

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Suppose we have a binary array called nums, we can delete one element from it. We have to find the size of the longest non-empty subarray which is containing only 1's in the resulting array. If there is no such subarray, then return 0.

So, if the input is like nums = [1,0,1,1,1,0,1,1,0], then the output will be 5 because by removing 0 from position 5, we can get a subarray [1,1,1,1,1] there are five 1s.

To solve this, we will follow these steps −

• if 0 is not in nums, then

• return size of nums - 1

• if 1 is not in nums, then

• return 0

• a := a new list

• cnt := 0

• for each i in nums, do

• if i is same as 0, then

• if cnt is not same as 0, then

• insert cnt at the end of a

• cnt := 0

• insert i at the end of a

• otherwise,

• cnt := cnt + 1

• if cnt is not same as 0, then

• insert cnt at the end of a

• Max := 0

• for i in range 0 to size of a, do

• if a[i] is not same as 0, then

• go for next iteration

• if a[i] is same as 0 and i is same as size of a - 1, then

• Max := maximum of Max and a[i-1]

• otherwise when a[i] is same as 0 and i is same as 0, then

• Max := maximum of Max and a[i+1]

• otherwise when a[i] is same as 0, then

• Max := maximum of Max and (a[i+1]+a[i-1])

• return Max

## Example

Live Demo

def solve(nums):
if 0 not in nums:
return len(nums)-1
if 1 not in nums:
return 0
a = []
cnt = 0
for i in nums:
if i == 0:
if cnt != 0:
a.append(cnt)
cnt = 0
a.append(i)
else:
cnt += 1
if cnt!=0:
a.append(cnt)
Max = 0
for i in range(len(a)):
if a[i] != 0:
continue
if a[i] == 0 and i == len(a)-1:
Max = max(Max,a[i-1])
elif a[i] == 0 and i == 0:
Max = max(Max,a[i+1])
elif a[i] == 0:
Max = max(Max,a[i+1]+a[i-1])
return Max
nums = [1,0,1,1,1,0,1,1,0]
print(solve(nums))

## Input

[1,0,1,1,1,0,1,1,0]

## Output

5
Updated on 29-May-2021 13:58:34