**Problem Statement:** Given an array that contains** only 1 and 0 **return the count of** maximum consecutive** ones in the array.

**Examples:**

Example 1:Input:prices = {1, 1, 0, 1, 1, 1}Output:3Explanation: There are two consecutive 1’s and three consecutive 1’s in the array out of which maximum is 3.Input:prices = {1, 0, 1, 1, 0, 1}Output:2Explanation:There are two consecutive 1's in the array.

**Solution**:

** Disclaimer**:

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

**Approach**: We maintain a** variable count** that keeps a track of the number of consecutive 1’s while traversing the array. The other variable max_count maintains the maximum number of 1’s, in other words, it maintains the answer.

We start traversing from the beginning of the array. Since we can encounter either a 1 or 0 there can be two situations:-

- If the value at the current index is equal to 1 we
**increase the value of count by one.**After updating the count variable if it becomes**more**than the max_count**update the max_count.** - If the value at the current index is equal to zero we make the
**variable count as 0**since there are**no more consecutive ones**.

** ***See the illustration below for a better understanding*

**Code:**

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int findMaxConsecutiveOnes(vector < int > & nums) {
int cnt = 0;
int maxi = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == 1) {
cnt++;
} else {
cnt = 0;
}
maxi = max(maxi, cnt);
}
return maxi;
}
};
int main() {
vector < int > nums = { 1, 1, 0, 1, 1, 1 };
Solution obj;
int ans = obj.findMaxConsecutiveOnes(nums);
cout << "The maximum consecutive 1's are " << ans;
return 0;
}
```

**Output: **The maximum consecutive 1’s are 3.

**Time Complexity: O(N) since the solution involves only a single pass.**

**Space Complexity: O(1) because no extra space is used.**

## Java Code

```
import java.util.*;
public class Main {
static int findMaxConsecutiveOnes(int nums[]) {
int cnt = 0;
int maxi = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 1) {
cnt++;
} else {
cnt = 0;
}
maxi = Math.max(maxi, cnt);
}
return maxi;
}
public static void main(String args[]) {
int nums[] = { 1, 1, 0, 1, 1, 1 };
int ans = findMaxConsecutiveOnes(nums);
System.out.println("The maximum consecutive 1's are " + ans);
}
}
```

**Output: **The maximum consecutive 1’s are 3.

**Time Complexity: O(N) since the solution involves only a single pass.**

**Space Complexity: O(1) because no extra space is used.**

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