Count 1’s in a sorted binary array
Last Updated :
27 Dec, 2023
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++
#include <bits/stdc++.h>
using namespace std;
int countOnes( bool arr[], int low, int high)
{
if (high >= low) {
int mid = low + (high - low) / 2;
if ((mid == high || arr[mid + 1] == 0)
&& (arr[mid] == 1))
return mid + 1;
if (arr[mid] == 1)
return countOnes(arr, (mid + 1), high);
return countOnes(arr, low, (mid - 1));
}
return 0;
}
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
class CountOnes {
int countOnes( int arr[], int low, int high)
{
if (high >= low) {
int mid = low + (high - low) / 2 ;
if ((mid == high || arr[mid + 1 ] == 0 )
&& (arr[mid] == 1 ))
return mid + 1 ;
if (arr[mid] == 1 )
return countOnes(arr, (mid + 1 ), high);
return countOnes(arr, low, (mid - 1 ));
}
return 0 ;
}
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 ));
}
}
|
Python3
def countOnes(arr, low, high):
if high > = low:
mid = low + (high - low) / / 2
if ((mid = = high or arr[mid + 1 ] = = 0 ) and (arr[mid] = = 1 )):
return mid + 1
if arr[mid] = = 1 :
return countOnes(arr, (mid + 1 ), high)
return countOnes(arr, low, mid - 1 )
return 0
arr = [ 1 , 1 , 1 , 1 , 0 , 0 , 0 ]
print ( "Count of 1's in given array is" , countOnes(arr, 0 , len (arr) - 1 ))
|
C#
using System;
class GFG {
static int countOnes( int [] arr, int low, int high)
{
if (high >= low) {
int mid = low + (high - low) / 2;
if ((mid == high || arr[mid + 1] == 0)
&& (arr[mid] == 1))
return mid + 1;
if (arr[mid] == 1)
return countOnes(arr, (mid + 1), high);
return countOnes(arr, low, (mid - 1));
}
return 0;
}
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));
}
}
|
Javascript
<script>
function countOnes( arr, low, high)
{
if (high >= low)
{
let mid = Math.trunc(low + (high - low)/2);
if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1))
return mid+1;
if (arr[mid] == 1)
return countOnes(arr, (mid + 1), high);
return countOnes(arr, low, (mid -1));
}
return 0;
}
let arr = [ 1, 1, 1, 1, 0, 0, 0 ];
let n = arr.length;
document.write( "Count of 1's in given array is " +
countOnes(arr, 0, n-1));
</script>
|
PHP
<?php
function countOnes( $arr , $low , $high )
{
if ( $high >= $low )
{
$mid = $low + ( $high - $low )/2;
if ( ( $mid == $high or $arr [ $mid +1] == 0)
and ( $arr [ $mid ] == 1))
return $mid +1;
if ( $arr [ $mid ] == 1)
return countOnes( $arr , ( $mid + 1),
$high );
return countOnes( $arr , $low , ( $mid -1));
}
return 0;
}
$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);
?>
|
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;
int countOnes( bool arr[], int n)
{
int ans;
int low = 0, high = n - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < 1)
high = mid - 1;
else if (arr[mid] > 1)
low = mid + 1;
else
{
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
import java.io.*;
class GFG {
static int countOnes( int arr[], int n)
{
int ans;
int low = 0 , high = n - 1 ;
while (low <= high) {
int mid = (low + high) / 2 ;
if (arr[mid] < 1 )
high = mid - 1 ;
else if (arr[mid] > 1 )
low = mid + 1 ;
else
{
if (mid == n - 1 || arr[mid + 1 ] != 1 )
return mid + 1 ;
else
low = mid + 1 ;
}
}
return 0 ;
}
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));
}
}
|
Python3
def countOnes(arr, n):
low = 0
high = n - 1
while (low < = high):
mid = (low + high) / / 2
if (arr[mid] < 1 ):
high = mid - 1
elif (arr[mid] > 1 ):
low = mid + 1
else :
if (mid = = n - 1 or arr[mid + 1 ] ! = 1 ):
return mid + 1
else :
low = mid + 1
return 0
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))
|
C#
using System;
public class GFG {
static int countOnes( int [] arr, int n)
{
int low = 0, high = n - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] < 1)
high = mid - 1;
else if (arr[mid] > 1)
low = mid + 1;
else
{
if (mid == n - 1 || arr[mid + 1] != 1)
return mid + 1;
else
low = mid + 1;
}
}
return 0;
}
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));
}
}
|
Javascript
<script>
function countOnes(arr,n)
{
let ans;
let low = 0, high = n - 1;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
if (arr[mid] < 1)
high = mid - 1;
else if (arr[mid] > 1)
low = mid + 1;
else
{
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;
document.write( "Count of 1's in given array is " + countOnes(arr, n));
</script>
|
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 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]);
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;
long total = Arrays.stream(arr)
.filter(i -> i == 1 )
.count();
System.out.println( "Count of 1's in given array is "
+ total);
}
}
|
Python3
arr = [ 1 , 1 , 1 , 1 , 0 , 0 , 0 , 0 , 0 ]
print ( "Count of 1's in given array is " ,arr.count( 1 ))
|
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 };
total = colors.Count(c => c == 1);
Console.WriteLine( "Count of 1's in given array is " +total);
}
}
|
Javascript
const arr = [1, 1, 1, 1, 0, 0, 0, 0, 0];
const size = arr.length;
const ptr = arr.findIndex((el) => el === 0);
console.log(`Count of 1 's in given array is ${ptr}`);
|
Output
Count of 1's in given array is 4
Time Complexity: O(n)
Auxiliary Space: O(1)
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