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

• Last Updated : 22 Jun, 2022

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:
Binary(1) = 01
Binary(2) = 10
Binary(3) = 11
Binary(4) = 100

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

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 ``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

 ``

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.

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