Count & Toggle queries on Binary Array

Count & Toggle queries on Binary Array
Count & Toggle queries on Binary Array

This is the concept where both count and toggle queries are applied to the binary array.

Count Query is used when there is a need to calculate a number of specific bits it appears in any piece of code. We can define it as count (start, end) which will count frequency of 1’s at intervals from start to end.

For example:
Let’s assume a binary number: 010011 which is equivalent to 19
After applying query count(1,3) it will give output as 1
Explanation: from range 1-3 only 1 1’s is present.

Other Example:

Input: arr [ ] = {1, 1, 1, 0, 0, 1, 0}
Output: 4

Input: arr  [ ] = {1, 1, 1, 0, 1, 1, 1}
Output: 6

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

C++ Implementation:-

Function for finding the number of 1’s in a sorted binary array

def countOnes(Arr, left_element, right_element):

# if last element in the array is 0, no 1’s can
	# be present in the array because it is already sorted
	if Arr[right_element] == 0:
		return 0
	# if first element in the array is 1, we know all its elements are
	# are 1’s as the array is already sorted
	if Arr[left_element] == 1:
		return right - left + 1
	# divide array into left and right subarray and recursion
	middle = (left_element + right_element) // 2
	return countOnes(Arr, left_element, middle) + countOnes(Arr, middle + 1, right_element)
if __name__ == '__main__':
	Arr = [0, 0, 0, 0, 1, 1, 1]
	print(countOnes(Arr, 0, len(Arr) - 1))

Toggle Query:

  • Toggle Query is used when there is a need to flip the bits in a specific range of code.
  • We can define it as a toggle (start, end) which will flip the bits in given range mentioned from start to end.

For example:

Let us assume binary no as considered above: 010011 which is equivalent to 19. After applying query toggle(1,3) it will give output as 101011.

Explanation: The query will flip the bits as 0 to 1 and 1 to 0 in specific range as given in query by the user. Hence, it flips the bits from 1 to 3 and gives the output as 101011 which is equivalent to 43.

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C Implementation:-

#include <stdio.h>
void binary(unsigned int n)
{
	unsigned int mask = 0x80000000;
    while(mask) {
	  printf("%d", (n&mask)?1:0);
	  mask = mask >> 1;
	}
}
void main()
{
    unsigned int num = 0;
    printf("Enter an integer: ");
    scanf("%d", &num);
    printf("Binary represntation of the input number:\n");
    binary(num); 
    num = ~num;
    printf("\n\nAfter toggling all bits of the number:\n");
    binary(num);
    printf("\n");
}

Now when we are aware of count and toggle query, its time to implement them together. Consider an example where binary no is: 0110001

And the instructions are:

Toggle (0, 3)
Toggle (4, 6)
Count (1, 5)
Toggle (2, 4)
Count (1, 2)
Here, the range is as (L, R)

Step1: toggle(0,3) will give output as 1000001
Step2: toggle(4,6) will give output as 1001111
Step3: count(1,5) will give output as 3
Step4: toggle(2,4) will give output as 1110111
Step5: count(1,2) will give output as 2

Now let’s try to understand toggle and count queries on a binary array together. You are given a binary array which has a number of 0’s and 1’s and q queries. q can be large! For each query, you are given a certain range [L, R] where L represents the starting range i.e. left and R represents ending range i.e. right.

suppose arr [ ] = {1,0,0,1}
L = 1 , R = 3 .
do toggling , resultant array = {0,1,1,1}
L = 1 , R =4
count all one's , ans = 3.

So, now we have to keep this thing in mind that we have to either toggle bits in the given range i.e make 0 = 1 and 1 = 0 on in simple words alternate it and on another query in which we have to count all 1’s in the range of [Left, Right].

1.	#include<bits/stdc++.h> 
2.	#define ll long long 
3.	#define mod 1000000007 
4.	#define mp make_pair 
5.	#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); 
6.	#define pb push_back 
7.	#define endl '\n' 
8.	#define pii pair<int,int> 
9.	#define all(v) v.begin(),v.end() 
10.	ll poww(ll x,ll y) {ll res=1;x%=mod; for(;y;y>>=1){if(y&1)res=res*x%mod;x=x*x%mod;}return res;} 
11.	using namespace std; 
12.	 
13.	int a[25][100005],seg[25][400005],n,lazy[25][400005]; 
14.	 
15.	void build(int i,int l,int r,int x){ 
16.	 
17.	if(l>r) 
18.	return; 
19.	if(l==r){ 
20.	seg[i][x]=a[i][l]; 
21.	return; 
22.	} 
23.	int mid=(l+r)/2; 
24.	build(i,l,mid,x*2+1); 
25.	build(i,mid+1,r,x*2+2); 
26.	seg[i][x]=seg[i][x*2+1]+seg[i][x*2+2]; 
27.	 
28.	} 
29.	 
30.	void setbit(int i){ 
31.	 
32.	int x=a[24][i]; 
33.	int c=0; 
34.	while(x>0){ 
35.	a[c++][i]=x%2; 
36.	x/=2; 
37.	} 
38.	 
39.	} 
40.	 
41.	ll sum(int i,int ql,int qr,int l,int r,int x){ 
42.	 
43.	if(l>r||ql>r||qr<l) return 0; 
44.	 
45.	if(lazy[i][x]){ 
46.	 
47.	seg[i][x]=(r-l+1)-seg[i][x]; 
48.	 
49.	if(l!=r){ 
50.	lazy[i][2*x+1]^=lazy[i][x]; 
51.	lazy[i][2*x+2]^=lazy[i][x]; 
52.	} 
53.	 
54.	lazy[i][x]=0; 
55.	 
56.	} 
57.	 
58.	if(ql<=l&&qr>=r) return seg[i][x]; 
59.	 
60.	int mid=l+r>>1; 
61.	return (sum(i,ql,qr,l,mid,x*2+1)+sum(i,ql,qr,mid+1,r,2*x+2)); 
62.	 
63.	} 
64.	 
65.	void upd(int i,int ql,int qr,int l,int r,int x){ 
66.	 
67.	if(lazy[i][x]){ 
68.	 
69.	seg[i][x]=(r-l+1)-seg[i][x]; 
70.	 
71.	if(l!=r){ 
72.	lazy[i][2*x+1]^=lazy[i][x]; 
73.	lazy[i][2*x+2]^=lazy[i][x]; 
74.	} 
75.	 
76.	lazy[i][x]=0; 
77.	 
78.	} 
79.	 
80.	if(l>r||ql>r||l>qr) return; 
81.	 
82.	if(ql<=l&&r<=qr){ 
83.	 
84.	int temp=5123123; 
85.	temp=seg[i][x]; 
86.	temp=seg[i][x]=(r-l+1)-seg[i][x]; 
87.	temp=2; 
88.	if(l!=r){ 
89.	lazy[i][2*x+1]^=1; 
90.	lazy[i][2*x+2]^=1; 
91.	} 
92.	return; 
93.	} 
94.	int mid=l+r>>1; 
95.	upd(i,ql,qr,l,mid,x*2+1); 
96.	upd(i,ql,qr,mid+1,r,x*2+2); 
97.	seg[i][x]=seg[i][x*2+1]+seg[i][x*2+2]; 
98.	 
99.	} 
100.	 
101.	int main() 
102.	{ 
103.	fast; 
104.	//freopen("C:/Users/PRAJAL/Desktop/input.txt", "r+", stdin); 
105.	//freopen("C:/Users/PRAJAL/Desktop/input5.txt", "w+", stdout); 
106.	 
107.	cin>>n; 
108.	 
109.	for(int i=0;i<n;++i) cin>>a[24][i]; 
110.	for(int i=0;i<n;++i) setbit(i); 
111.	for(int i=0;i<20;++i) build(i,0,n-1,0); 
112.	 
113.	int m,l,r,x; 
114.	cin>>m; 
115.	 
116.	while(m--){ 
117.	 
118.	cin>>x>>l>>r; 
119.	l--; 
120.	r--; 
121.	if(x==1){ 
122.	ll ans=0; 
123.	for(ll i=0;i<20;++i) 
124.	ans+=sum(i,l,r,0,n-1,0)<<i; 
125.	cout<<ans<<endl; 
126.	} 
127.	else{ 
128.	cin>>x; 
129.	for(int i=0;i<20;++i,x/=2) if (x%2) upd(i,l,r,0,n-1,0); 
130.	} 
131.	 
132.	} 
133.	return 0; 
134.	}

Application and features:-

  • One of the important application for count and toggle query is digital counters in digital electronics and as Digital counter is the biggest application of flip-flop where we have to change the state of a flip from off i.e. 0 to on i.e. 1
  • And as we know this count and toggle query approach is based on a segment tree. It is used in Relational Database Management System for query improvisation. It is often used with Binary Index tree. A segment tree is also used in fragmentation of disk during an external sort.

Conclusion:
A simple solution for this problem is to traverse the complete range for “Toggle” query and when you get “Count” query then count all the 1’s for a given range. But the time complexity for this approach will be O(q*n) where q=total number of queries.

An efficient solution for a given problem is to use Segment Tree with Lazy Propagation. Here we collect the updates until we get a query for “Count”. When we get the query for “Count”, we make all the previously collected Toggle updates in an array and then count a number of 1’s within the given range.

To read more articles on Data Structures, click here.

By Yogesh Kumar