Count 1’s in a sorted binary array

Last Updated : 12 Jul, 2024

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Given a binary array arr[] of size N, which is sorted in non-increasing order, count the number of 1’s in it.

Examples:

Input: arr[] = {1, 1, 0, 0, 0, 0, 0}
Output: 2

Input: arr[] = {1, 1, 1, 1, 1, 1, 1}
Output: 7

Input: arr[] = {0, 0, 0, 0, 0, 0, 0}
Output: 0

Naive approach:

A simple solution is to linearly traverse the array until we find the 1’s in the array and keep count of 1s. If the array element becomes 0 then return the count of 1’s.

Time Complexity: O(N).
Auxiliary Space: O(1)

Count 1’s in a sorted binary array using Binary search recursively:

We can use Binary Search to find count in O(Logn) time. The idea is to look for the last occurrence of 1 using Binary Search. Once we find the index’s last occurrence, we return index + 1 as count.

Follow the steps below to implement the above idea:

Do while low <= high:
- Calculate mid using low + (high – low) / 2.
- Check if the element at mid index is the last 1
- If the element is not last 1, move the low to right side recursively and return the result received from it.
- Otherwise, move the low to left recursively and return the result received from it.

The following is the implementation of the above idea.

C++

// C++ program to count one's in a boolean array
#include <bits/stdc++.h>
using namespace std;

/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */
int countOnes(bool arr[], int low, int high)
{
    if (high >= low) {
        // get the middle index
        int mid = low + (high - low) / 2;

        // check if the element at middle index is last 1
        if ((mid == high || arr[mid + 1] == 0)
            && (arr[mid] == 1))
            return mid + 1;

        // If element is not last 1, recur for right side
        if (arr[mid] == 1)
            return countOnes(arr, (mid + 1), high);

        // else recur for left side
        return countOnes(arr, low, (mid - 1));
    }
    return 0;
}

/* Driver Code */
int main()
{
    bool arr[] = { 1, 1, 1, 1, 0, 0, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Count of 1's in given array is "
         << countOnes(arr, 0, n - 1);
    return 0;
}

Java

// Java program to count 1's in a sorted array
class CountOnes {
    /* Returns counts of 1's in arr[low..high].  The
       array is assumed to be sorted in non-increasing
       order */
    int countOnes(int arr[], int low, int high)
    {
        if (high >= low) {
            // get the middle index
            int mid = low + (high - low) / 2;

            // check if the element at middle index is last
            // 1
            if ((mid == high || arr[mid + 1] == 0)
                && (arr[mid] == 1))
                return mid + 1;

            // If element is not last 1, recur for right
            // side
            if (arr[mid] == 1)
                return countOnes(arr, (mid + 1), high);

            // else recur for left side
            return countOnes(arr, low, (mid - 1));
        }
        return 0;
    }

    /* Driver code */
    public static void main(String args[])
    {
        CountOnes ob = new CountOnes();
        int arr[] = { 1, 1, 1, 1, 0, 0, 0 };
        int n = arr.length;
        System.out.println("Count of 1's in given array is "
                           + ob.countOnes(arr, 0, n - 1));
    }
}
/* This code is contributed by Rajat Mishra */

Python

# Python program to count one's in a boolean array

# Returns counts of 1's in arr[low..high].  The array is
# assumed to be sorted in non-increasing order


def countOnes(arr, low, high):

    if high >= low:

        # get the middle index
        mid = low + (high-low)//2

        # check if the element at middle index is last 1
        if ((mid == high or arr[mid+1] == 0) and (arr[mid] == 1)):
            return mid+1

        # If element is not last 1, recur for right side
        if arr[mid] == 1:
            return countOnes(arr, (mid+1), high)

        # else recur for left side
        return countOnes(arr, low, mid-1)

    return 0


# Driver Code
arr = [1, 1, 1, 1, 0, 0, 0]
print("Count of 1's in given array is", countOnes(arr, 0, len(arr)-1))

# This code is contributed by __Devesh Agrawal__

C#

// C# program to count 1's in a sorted array
using System;

class GFG {

    /* Returns counts of 1's in arr[low..high].
    The array is assumed to be sorted in
    non-increasing order */
    static int countOnes(int[] arr, int low, int high)
    {
        if (high >= low) {
            // get the middle index
            int mid = low + (high - low) / 2;

            // check if the element at middle
            // index is last 1
            if ((mid == high || arr[mid + 1] == 0)
                && (arr[mid] == 1))
                return mid + 1;

            // If element is not last 1, recur
            // for right side
            if (arr[mid] == 1)
                return countOnes(arr, (mid + 1), high);

            // else recur for left side
            return countOnes(arr, low, (mid - 1));
        }

        return 0;
    }

    /* Driver code */
    public static void Main()
    {
        int[] arr = { 1, 1, 1, 1, 0, 0, 0 };
        int n = arr.Length;

        Console.WriteLine("Count of 1's in given "
                          + "array is "
                          + countOnes(arr, 0, n - 1));
    }
}

// This code is contributed by Sam007

JavaScript

// Javascript program to count one's in a boolean array

/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */
function countOnes( arr, low, high)
{
  if (high >= low)
  {
    // get the middle index
    let mid = Math.trunc(low + (high - low)/2);

    // check if the element at middle index is last 1
    if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1))
      return mid+1;

    // If element is not last 1, recur for right side
    if (arr[mid] == 1)
      return countOnes(arr, (mid + 1), high);

    // else recur for left side
    return countOnes(arr, low, (mid -1));
  }
  return 0;
}
    // Driver program 
    
   let arr = [ 1, 1, 1, 1, 0, 0, 0 ];
   let n = arr.length;
  console.log("Count of 1's in given array is " + 
                    countOnes(arr, 0, n-1));

PHP

<?php
// PHP program to count one's in a
// boolean array

/* Returns counts of 1's in arr[low..high].
The array is assumed to be sorted in 
non-increasing order */
function countOnes( $arr, $low, $high)
{
    if ($high >= $low)
    {
        // get the middle index
        $mid = $low + ($high - $low)/2;
    
        // check if the element at middle
        // index is last 1
        if ( ($mid == $high or $arr[$mid+1] == 0) 
                           and ($arr[$mid] == 1))
            return $mid+1;
    
        // If element is not last 1, recur for 
        // right side
        if ($arr[$mid] == 1)
            return countOnes($arr, ($mid + 1),
                                          $high);
    
        // else recur for left side
        return countOnes($arr, $low, ($mid -1));
    }
    
    return 0;
}

/* Driver code */
$arr = array(1, 1, 1, 1, 0, 0, 0);
$n = count($arr);
echo "Count of 1's in given array is " , 
                      countOnes($arr, 0, $n-1);

// This code is contributed by anuj_67.
?>

Output

Count of 1's in given array is 4

Time complexity: O(Log(N))
Auxiliary Space: O(log(N))

Count 1’s in a sorted binary array using binary search iteratively:

Follow the steps below for the implementation:

Do while low <= high
- Calculate the middle index say mid
- Check if arr[mid] is less than 1 then move the high to left side (i.e, high = mid – 1)
- If the element is not last 1 then move the low to the right side (i.e, low = mid + 1)
- Check if the element at the middle index is last 1 then return mid + 1
- Otherwise move to low to right (i.e, low = mid + 1)

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */

int countOnes(bool arr[], int n)
{
    int ans;
    int low = 0, high = n - 1;
    while (low <= high) { // get the middle index
        int mid = (low + high) / 2;

        // else recur for left side
        if (arr[mid] < 1)
            high = mid - 1;
        // If element is not last 1, recur for right side
        else if (arr[mid] > 1)
            low = mid + 1;
        else
        // check if the element at middle index is last 1
        {
            if (mid == n - 1 || arr[mid + 1] != 1)
                return mid + 1;
            else
                low = mid + 1;
        }
    }
}

int main()
{
    bool arr[] = { 1, 1, 1, 1, 0, 0, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Count of 1's in given array is "
         << countOnes(arr, n);
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {

    static int countOnes(int arr[], int n)
    {
        int ans;
        int low = 0, high = n - 1;
        while (low <= high) { // get the middle index
            int mid = (low + high) / 2;

            // else recur for left side
            if (arr[mid] < 1)
                high = mid - 1;

            // If element is not last 1, recur for right
            // side
            else if (arr[mid] > 1)
                low = mid + 1;
            else

            // check if the element at middle index is last
            // 1
            {
                if (mid == n - 1 || arr[mid + 1] != 1)
                    return mid + 1;
                else
                    low = mid + 1;
            }
        }
        return 0;
    }

    // Driver code
    public static void main(String[] args)
    {

        int arr[] = { 1, 1, 1, 1, 0, 0, 0 };
        int n = arr.length;

        System.out.println("Count of 1's in given array is "
                           + countOnes(arr, n));
    }
}

// This code is contributed by patel2127.

Python

'''package whatever #do not write package name here '''


def countOnes(arr, n):
    low = 0
    high = n - 1
    while (low <= high):  # get the middle index
        mid = (low + high) // 2

        # else recur for left side
        if (arr[mid] < 1):
            high = mid - 1

        # If element is not last 1, recur for right side
        elif(arr[mid] > 1):
            low = mid + 1
        else:

            # check if the element at middle index is last 1
            if (mid == n - 1 or arr[mid + 1] != 1):
                return mid + 1
            else:
                low = mid + 1

    return 0


# Driver code
if __name__ == '__main__':

    arr = [1, 1, 1, 1, 0, 0, 0]
    n = len(arr)

    print("Count of 1's in given array is ", countOnes(arr, n))

# This code is contributed by umadevi9616

C#

/*package whatever //do not write package name here */
using System;
public class GFG {

    static int countOnes(int[] arr, int n)
    {

        int low = 0, high = n - 1;
        while (low <= high) { // get the middle index
            int mid = (low + high) / 2;

            // else recur for left side
            if (arr[mid] < 1)
                high = mid - 1;

            // If element is not last 1, recur for right
            // side
            else if (arr[mid] > 1)
                low = mid + 1;
            else

            // check if the element at middle index is last
            // 1
            {
                if (mid == n - 1 || arr[mid + 1] != 1)
                    return mid + 1;
                else
                    low = mid + 1;
            }
        }
        return 0;
    }

    // Driver code
    public static void Main(String[] args)
    {

        int[] arr = { 1, 1, 1, 1, 0, 0, 0 };
        int n = arr.Length;

        Console.WriteLine("Count of 1's in given array is "
                          + countOnes(arr, n));
    }
}

// This code is contributed by umadevi9616

JavaScript

/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */

function countOnes(arr,n)
{
    let ans;
    let low = 0, high = n - 1;
    while (low <= high) { // get the middle index
        let mid = Math.floor((low + high) / 2);
 
        // else recur for left side
        if (arr[mid] < 1)
            high = mid - 1;
        // If element is not last 1, recur for right side
        else if (arr[mid] > 1)
            low = mid + 1;
        else
        // check if the element at middle index is last 1
        {
            if (mid == n - 1 || arr[mid + 1] != 1)
                return mid + 1;
            else
                low = mid + 1;
        }
    }
}

let arr=[ 1, 1, 1, 1, 0, 0, 0];
let n = arr.length;
console.log( "Count of 1's in given array is "+ countOnes(arr, n));


// This code is contributed by unknown2108

Output

Count of 1's in given array is 4

Time complexity: O(Log(N))
Auxiliary Space: O(1)

Count 1’s in a sorted binary array using inbuilt functions:

Below is the implementation using inbuilt functions:

C++

#include <bits/stdc++.h>
using namespace std;

int main()
{

    int arr[] = { 1, 1, 1, 1, 0, 0, 0, 0, 0 };
    int size = sizeof(arr) / sizeof(arr[0]);

    // Pointer to the first occurrence of one

    auto ptr
        = upper_bound(arr, arr + size, 1, greater<int>());
    cout << "Count of 1's in given array is "
         << (ptr - arr);

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static void main(String[] args)
    {
        int[] arr = { 1, 1, 1, 1, 0, 0, 0, 0, 0 };
        int size = arr.length;

        // Counting the number of 1's in the array
        long total = Arrays.stream(arr)
                         .filter(i -> i == 1)
                         .count();

        System.out.println("Count of 1's in given array is "
                           + total);
    }
}
// This code is contributed by prajwal kandekar

Python

# code
arr = [1, 1, 1, 1, 0, 0, 0, 0, 0 ]

print("Count of 1's in given array is ",arr.count(1))

# This code is contributed by Pranay Arora

C#

using System;
using System.Linq;
 
class GFG{
 
static public void Main()
{
    var total = 0;
    int[] colors = { 1, 1, 1, 1, 0, 0, 0, 0, 0 };
    // Count function to count the specifically 
      // 1 from the array
    total = colors.Count(c => c == 1);
  
    Console.WriteLine("Count of 1's in given array is "+total);
}
}

// This code is contributed by Prince Kumar

JavaScript

const arr = [1, 1, 1, 1, 0, 0, 0, 0, 0];
const size = arr.length;

// Pointer to the first occurrence of one
const ptr = arr.findIndex((el) => el === 0);

//If Array doesnot have 0's then findIndex will return -1
//In that case the number of 1's will be equal to size of array not -1
if(ptr === -1)
console.log(`Count of 1 's in given array is ${arr.length}`);
else
  console.log(`Count of 1 's in given array is ${ptr}`);
// This code is contributed by Prajwal Kandekar

Output

Count of 1's in given array is 4

Time Complexity: O(n)
Auxiliary Space: O(1)

Two Sum - Pair Closest to 0

GeeksforGeeks

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