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.
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The following is JAVA implementation of above idea.
java
/ Java program to count 1's in a sorted array
class CountOnes
{
/* Returns counts of 1's in arr[low..high]. The
array is assumed to be sorted in non-increasing
order */
int countOnes(int arr[], int low, int high)
{
if (high >= low)
{
// get the middle index
int mid = low + (high - low)/2;
// check if the element at middle index is last 1
if ( (mid == high || arr[mid+1] == 0) &&
(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 program to test above functions */
public static void main(String args[])
{
CountOnes ob = new CountOnes();
int arr[] = {1, 1, 1, 1, 0, 0, 0};
int n = arr.length;
System.out.println("Count of 1's in given array is " +
ob.countOnes(arr, 0, n-1) );
}
}
Output:
Count of 1's in given array is 4
Time complexity of the above solution is O(Logn)
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