Given a sorted binary array, efficiently find the number of 1’s in it.

For example,

**Input: **A[] = [0, 0, 0, 0, 1, 1, 1]

**Output: **Total number of 1’s present are 3

**Input: **A[] = [0, 0, 1, 1, 1, 1, 1]

**Output: **Total number of 1’s present are 5

A simple solution would be to run a **linear search** on the array and find the first occurrence of one. Then the output will be number of elements in the array minus index of the first occurrence of one. The problem with this approach is that its worst case time complexity is O(n).

We can easily solve this problem in O(log(n)) time by using **recursion**. The idea is to take advantage of the fact that the input is sorted (i.e. all 0’s followed by all 1’s). We split the array to two halves and recuse for both the halves. If the last element of the sub-array is 0, then all 0’s are present in it since it is sorted and we return 0 from the function. If the first element of the array is 1, then all its elements are 1’s only since array is sorted and we return number of elements in that partition.

**C++ implementation –**

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#include <iostream> using namespace std; // Function to find number of 1's in a sorted binary array int count(bool arr[], int n) { // if last element of the array is 0, no ones can // be present in it since it is sorted if (arr[n - 1] == 0) return 0; // if first element of the array is 1, all its elements // are ones only since it is sorted if (arr[0]) return n; // divide array into left and right sub-array and recuse return count(arr, n/2) + count(arr + n/2, n - n/2); } // main function int main() { bool arr[] = { 0, 0, 0, 0, 1, 1, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Total number of 1's present are " << count(arr, n); return 0; } |

**Output: **

Total number of 1’s present are 3

**Time complexity** of above solution is O(log(n)).

**Auxiliary space** used by the program is O(1).

**Thanks for reading.**

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