You are given an array of 0s and 1s in random order. Segregate 0s on left side and 1s on right side of the array. Traverse array only once.
Example:
Input array = [0, 1, 0, 1, 0, 0, 1, 1, 1, 0] Output array = [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]
Method #1: Using sort() function
Python3
# driver program to testarr = [0, 1, 0, 1, 1, 1]arr.sort()print("Array after segregation is", end=' ')print(*arr)# This code is contributed by vikkycirus |
Array after segregation is 0 0 1 1 1 1
Time Complexity: O(n*logn)
Auxiliary Space: O(1)
Method 2: (Count 0s or 1s)
- Count the number of 0s. Let count be C.
- Once we have count, we can put C 0s at the beginning and 1s at the remaining n – C positions in array.
Time Complexity : O(n)
Auxiliary Space: O(1)
Thanks to Naveen for suggesting this method.
Method 3: (Use two indexes to traverse)
Method 2 traverses the array two times. Method 3 does the same in a single pass.
Maintain two indexes. Initialize first index left as 0 and second index right as n-1.
Do the following while left < right
a) Keep incrementing index left while there are 0s at it
b) Keep decrementing index right while there are 1s at it
c) If left < right then exchange arr[left] and arr[right]
Implementation:
Python
# Python program to sort a binary array in one pass# Function to put all 0s on left and all 1s on rightdef segregate0and1(arr, size): # Initialize left and right indexes left, right = 0, size-1 while left < right: # Increment left index while we see 0 at left while arr[left] == 0 and left < right: left += 1 # Decrement right index while we see 1 at right while arr[right] == 1 and left < right: right -= 1 # If left is smaller than right then there is a 1 at left # and a 0 at right. Exchange arr[left] and arr[right] if left < right: arr[left] = 0 arr[right] = 1 left += 1 right -= 1 return arr# driver program to testarr = [0, 1, 0, 1, 1, 1]arr_size = len(arr)print("Array after segregation")print(segregate0and1(arr, arr_size))# This code is contributed by Pratik Chhajer |
Array after segregation [0, 0, 1, 1, 1, 1]
Time Complexity: O(n)
Auxiliary Space: O(1)
Another approach :
1. Take two pointer type0(for element 0) starting from beginning (index = 0) and type1(for element 1) starting from end (index = array.length-1).
Initialize type0 = 0 and type1 = array.length-1
2. It is intended to Put 1 to the right side of the array. Once it is done, then 0 will definitely towards left side of array.
Please refer complete article on Segregate 0s and 1s in an array for more details!
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