**Problem Statement:** Given a sorted binary array (consisting of only 0’s and 1’s), the task is to find the total number of 1’s in the given array.

**Examples:**

Example 1:Input:[0,0,0,1,1]Output:2Explanation:Total number of 1’s in given array is 2, so answer will be 2Example 2:Input:[0,0,0,0,0,0,1]Output:1Explanation:Total number of 1’s in given array is 1Example 3:Input:[0,0,0,0,0,0,0]Output:0Explanation:There are no 1’s in the given array So, output will be 0

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

### Solution 1:

**Brute force:**

A brute force approach is to traverse through the array, and while traversing if 1 is encountered for the first time in the array that means all the elements to the right-hand side of this first occurrence of 1 will be 1’s only, as the array is sorted.

- Traverse through array
- Find the index where you are getting the first occurrence of 1 then return the answer as
**Length of the array – index of the first occurrence of 1**

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
int totalOnes(int arr[],int n) {
int ans = 0;
//traversing through array
for (int i = 0; i < n; i++) {
//found 1 in array for the first time
if (arr[i] == 1) {
ans = n - i;
break; // exit from loop as we found answer
}
}
return ans;
}
int main()
{
int binaryArray[] = {0,0,0,1,1,1};
cout<<"Total number of 1's: "<<totalOnes(binaryArray,6);
}
```

**Output:** Total number of 1’s: 3

**Time complexity:** O(n)

**Space complexity:** O(1)

## Java Code

```
public class TUF {
public static void main(String[] args) {
int[] binaryArray = {0,0,0,1,1,1};
System.out.println("Total number of 1's: " + totalOnes(binaryArray));
}
private static int totalOnes(int[] arr) {
int n = arr.length;
int ans = 0;
//traversing through array
for (int i = 0; i < arr.length; i++) {
//found 1 in array for the first time
if (arr[i] == 1) {
ans = n - i;
break; // exit from loop as we found answer
}
}
return ans;
}
}
```

**Output:** Total number of 1’s: 3

**Time complexity:** O(n)

**Space complexity:** O(1)

**Solution 2: Optimized approach**

**Intuition & Approach:**

Find the first occurrence index of 1 and all the elements on the right side of this index will be 1 only.

Since the array is sorted we can find the first occurrence of 1 using the **Binary Search algorithm**

- If the element at the middle index is 0 then make start index = middle index +1
- Else if the element at middle index is 1 then update answer as the middle index ( because the middle index element may be the first occurrence of 1) and make the end index = middle index – 1
- Return ans

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
int totalOnes(int arr[],int n) {
int start = 0, end = n - 1;
int ans = 0;
while (start <= end) {
int mid = start + (end - start) / 2;
if (arr[mid] == 0) {
start = mid + 1;
}
else if (arr[mid] == 1) {
ans = n - mid;
end = mid - 1;
}
}
return ans;
}
int main()
{
int binaryArray[] = {0,0,0,1,1,1};
cout<<"Total number of 1's: "<<totalOnes(binaryArray,6);
}
```

**Output:** Total number of 1’s: 3

**Time complexity:** O(logn) because the binary search algorithm is used

**Space complexity:** O(1) as extra space is not used.

## Java Code

```
public class TUF {
public static void main(String[] args) {
int[] binaryArray = {0,0,0,1,1,1};
System.out.println("Total number of 1's: " + totalOnes(binaryArray));
}
private static int totalOnes(int[] arr) {
int start = 0, end = arr.length - 1;
int ans = 0, n = arr.length;
while (start <= end) {
int mid = start + (end - start) / 2;
if (arr[mid] == 0) {
start = mid + 1;
}
else if (arr[mid] == 1) {
ans = n - mid;
end = mid - 1;
}
}
return ans;
}
}
```

**Output:** Total number of 1’s: 3

**Time complexity:** O(logn) because the binary search algorithm is used

**Space complexity:** O(1) as extra space is not used.

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