# Maximum number of consecutive 1’s in binary representation of all the array elements

Given an array arr[] of N elements, the task is to find the maximum number of consecutive 1’s in the binary representation of an element among all the elements of the given array.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 2
Binary(1) = 01
Binary(2) = 10
Binary(3) = 11
Binary(4) = 100

Input: arr[] = {10, 15, 37, 89}
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: An approach to find the count of maximum consecutive 1s in the binary representation of a number has been discussed in this article. The same approach can be used to find the same for all the elements of the given array and the maximum among those values is the required answer.

Below is the implementation of the above approach:

 // C++ implementation of the approach #include using namespace std;    // Function to return the count of // maximum consecutive 1s in the // binary represntation of x int maxConsecutiveOnes(int x) {     // Initialize result     int count = 0;        // Count the number of iterations to     // reach x = 0.     while (x != 0) {         // This operation reduces length         // of every sequence of 1s by one         x = (x & (x << 1));            count++;     }        return count; }    // Function to return the count of // maximum consecutive 1s in the // binary represntation among all // the elements of arr[] int maxOnes(int arr[], int n) {     // To store the answer     int ans = 0;        // For every element of the array     for (int i = 0; i < n; i++) {            // Count of maximum consecutive 1s in         // the binary representation of         // the current element         int currMax = maxConsecutiveOnes(arr[i]);            // Update the maximum count so far         ans = max(ans, currMax);     }        return ans; }    // Driver code int main() {     int arr[] = { 1, 2, 3, 4 };     int n = sizeof(arr) / sizeof(int);        cout << maxOnes(arr, n);        return 0; }

 // Java implementation of the approach class GFG {    // Function to return the count of // maximum consecutive 1s in the // binary represntation of x static int maxConsecutiveOnes(int x) {     // Initialize result     int count = 0;        // Count the number of iterations to     // reach x = 0.     while (x != 0)      {         // This operation reduces length         // of every sequence of 1s by one         x = (x & (x << 1));            count++;     }     return count; }    // Function to return the count of // maximum consecutive 1s in the // binary represntation among all // the elements of arr[] static int maxOnes(int arr[], int n) {     // To store the answer     int ans = 0;        // For every element of the array     for (int i = 0; i < n; i++)      {            // Count of maximum consecutive 1s in         // the binary representation of         // the current element         int currMax = maxConsecutiveOnes(arr[i]);            // Update the maximum count so far         ans = Math.max(ans, currMax);     }     return ans; }    // Driver code public static void main(String []args) {     int arr[] = { 1, 2, 3, 4 };     int n = arr.length;        System.out.println(maxOnes(arr, n)); } }    // This code is contributed by 29AjayKumar

 # Python3 implementation of the approach     # Function to return the count of  # maximum consecutive 1s in the  # binary represntation of x  def maxConsecutiveOnes(x) :         # Initialize result      count = 0;         # Count the number of iterations to      # reach x = 0.      while (x != 0) :                    # This operation reduces length          # of every sequence of 1s by one          x = (x & (x << 1));             count += 1;             return count;     # Function to return the count of  # maximum consecutive 1s in the  # binary represntation among all  # the elements of arr[]  def maxOnes(arr, n) :         # To store the answer      ans = 0;         # For every element of the array      for i in range(n) :            # Count of maximum consecutive 1s in          # the binary representation of          # the current element          currMax = maxConsecutiveOnes(arr[i]);             # Update the maximum count so far          ans = max(ans, currMax);         return ans;     # Driver code  if __name__ == "__main__" :         arr = [ 1, 2, 3, 4 ];      n = len(arr);         print(maxOnes(arr, n));     # This code is contributed by AnkitRai01

 // C# implementation of the approach using System;                        class GFG {    // Function to return the count of // maximum consecutive 1s in the // binary represntation of x static int maxConsecutiveOnes(int x) {     // Initialize result     int count = 0;        // Count the number of iterations to     // reach x = 0.     while (x != 0)      {         // This operation reduces length         // of every sequence of 1s by one         x = (x & (x << 1));            count++;     }     return count; }    // Function to return the count of // maximum consecutive 1s in the // binary represntation among all // the elements of arr[] static int maxOnes(int []arr, int n) {     // To store the answer     int ans = 0;        // For every element of the array     for (int i = 0; i < n; i++)      {            // Count of maximum consecutive 1s in         // the binary representation of         // the current element         int currMax = maxConsecutiveOnes(arr[i]);            // Update the maximum count so far         ans = Math.Max(ans, currMax);     }     return ans; }    // Driver code public static void Main(String []args) {     int []arr = { 1, 2, 3, 4 };     int n = arr.Length;        Console.WriteLine(maxOnes(arr, n)); } }    // This code is contributed by 29AjayKumar

Output:
2

Third year Department of Information Technology Jadavpur University

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Improved By : AnkitRai01, 29AjayKumar

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