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 

Approach: An approach to finding 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++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the count of
// maximum consecutive 1s in the
// binary representation 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 representation 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

// Java implementation of the approach
class GFG
{

// Function to return the count of
// maximum consecutive 1s in the
// binary representation 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 representation 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

# Python3 implementation of the approach

# Function to return the count of
# maximum consecutive 1s in the
# binary representation 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 representation 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#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the count of
// maximum consecutive 1s in the
// binary representation 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 representation 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

Javascript

<script>

// javascript implementation of the approach

// Function to return the count of
// maximum consecutive 1s in the
// binary representation of x
function maxConsecutiveOnes(x)
{
    // Initialize result
    var 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 representation among all
// the elements of arr
function maxOnes(arr , n)
{
    // To store the answer
    var ans = 0;

    // For every element of the array
    for (i = 0; i < n; i++)
    {

        // Count of maximum consecutive 1s in
        // the binary representation of
        // the current element
        var currMax = maxConsecutiveOnes(arr[i]);

        // Update the maximum count so far
        ans = Math.max(ans, currMax);
    }
    return ans;
}

// Driver code
var arr = [ 1, 2, 3, 4 ];
var n = arr.length;

document.write(maxOnes(arr, n));

// This code contributed by Princi Singh

</script>

Output

2

Time Complexity: O(N*log(maxArr)), as we are using a loop to traverse N times and in each traversal, we are calling the function maxConsecutiveOnes which will cost log(maxArr). Where N is the number of elements in the array and maxArr is the element with maximum value in the array.
Auxiliary Space: O(1), as we are not using any extra space.