Given a binary array sorted in non-increasing order, count the number of 1’s in it.
Examples:
Input: arr[] = {1, 1, 0, 0, 0, 0, 0}
Output: 2
Input: arr[] = {1, 1, 1, 1, 1, 1, 1}
Output: 7
Input: arr[] = {0, 0, 0, 0, 0, 0, 0}
Output: 0
A simple solution is to linearly traverse the array. The time complexity of the simple solution is O(n). We can use Binary Search to find count in O(Logn) time. The idea is to look for last occurrence of 1 using Binary Search. Once we find the index last occurrence, we return index + 1 as count.
The following is the implementation of above idea.
Python
# Python program to count one's in a boolean array # Returns counts of 1's in arr[low..high]. The array is # assumed to be sorted in non-increasing order def countOnes(arr,low,high): if high>=low: # get the middle index mid = low + (high-low)//2 # check if the element at middle index is last 1 if ((mid == high or arr[mid+1]==0) and (arr[mid]==1)): return mid+1 # If element is not last 1, recur for right side if arr[mid]==1: return countOnes(arr, (mid+1), high) # else recur for left side return countOnes(arr, low, mid-1) return 0 # Driver Code arr=[1, 1, 1, 1, 0, 0, 0] print ("Count of 1's in given array is",countOnes(arr, 0 , len(arr)-1)) # This code is contributed by __Devesh Agrawal__ |
Output
Count of 1's in given array is 4
Time complexity of the above solution is O(Logn)
Space complexity o(log n) (function call stack)
Please refer complete article on Count 1’s in a sorted binary array for more details!
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