# Count and Toggle Queries on a Binary Array

Given a size n in which initially all elements are 0. The task is to perform multiple multiple queries of following two types. The queries can appear in any order.

1. toggle(start, end) : Toggle (0 into 1 or 1 into 0) the values from range ‘start’ to ‘end’.
2. count(start, end) : Count the number of 1’s within given range from ‘start’ to ‘end’.
```Input : n = 5       // we have n = 5 blocks
toggle 1 2  // change 1 into 0 or 0 into 1
Toggle 2 4
Count  2 3  // count all 1's within the range
Toggle 2 4
Count  1 4  // count all 1's within the range
Output : Total number of 1's in range 2 to 3 is = 1
Total number of 1's in range 1 to 4 is = 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

An efficient solution for this 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 array and then count number of 1’s with in the given range.

```// C++ program to implement toggle and count
// queries on a binary array.
#include<bits/stdc++.h>
using namespace std;
const int MAX = 100000;

// segment tree to store count of 1's within range
int tree[MAX] = {0};

// bool type tree to collect the updates for toggling
// the values of 1 and 0 in given range
bool lazy[MAX] = {false};

// function for collecting updates of toggling
// node --> index of current node in segment tree
// st --> starting index of current node
// en --> ending index of current node
// us --> starting index of range update query
// ue --> ending index of range update query
void toggle(int node, int st, int en, int us, int ue)
{
// If lazy value is non-zero for current node of segment
// tree, then there are some pending updates. So we need
// to make sure that the pending updates are done before
// making new updates. Because this value may be used by
// parent after recursive calls (See last line of this
// function)
if (lazy[node])
{
// Make pending updates using value stored in lazy nodes
lazy[node] = false;
tree[node] = en - st + 1 - tree[node];

// checking if it is not leaf node because if
// it is leaf node then we cannot go further
if (st < en)
{
// We can postpone updating children we don't
// need their new values now.
// Since we are not yet updating children of 'node',
// we need to set lazy flags for the children
lazy[node<<1] = !lazy[node<<1];
lazy[1+(node<<1)] = !lazy[1+(node<<1)];
}
}

// out of range
if (st>en || us > en || ue < st)
return ;

// Current segment is fully in range
if (us<=st && en<=ue)
{
// Add the difference to current node
tree[node] = en-st+1 - tree[node];

// same logic for checking leaf node or not
if (st < en)
{
// This is where we store values in lazy nodes,
// rather than updating the segment tree itelf
// Since we don't need these updated values now
// we postpone updates by storing values in lazy[]
lazy[node<<1] = !lazy[node<<1];
lazy[1+(node<<1)] = !lazy[1+(node<<1)];
}
return;
}

// If not completely in rang, but overlaps, recur for
// children,
int mid = (st+en)/2;
toggle((node<<1), st, mid, us, ue);
toggle((node<<1)+1, mid+1,en, us, ue);

// And use the result of children calls to update this node
if (st < en)
tree[node] = tree[node<<1] + tree[(node<<1)+1];
}

/* node --> Index of current node in the segment tree.
Initially 0 is passed as root is always at'
index 0
st & en  --> Starting and ending indexes of the
segment represented by current node,
i.e., tree[node]
qs & qe  --> Starting and ending indexes of query
range */
// function to count number of 1's within given range
int countQuery(int node, int st, int en, int qs, int qe)
{
// current node is out of range
if (st>en || qs > en || qe < st)
return 0;

// If lazy flag is set for current node of segment tree,
// then there are some pending updates. So we need to
// make sure that the pending updates are done before
// processing the sub sum query
if (lazy[node])
{
// Make pending updates to this node. Note that this
// node represents sum of elements in arr[st..en] and
// all these elements must be increased by lazy[node]
lazy[node] = false;
tree[node] = en-st+1-tree[node];

// checking if it is not leaf node because if
// it is leaf node then we cannot go further
if (st<en)
{
// Since we are not yet updating children os si,
// we need to set lazy values for the children
lazy[node<<1] = !lazy[node<<1];
lazy[(node<<1)+1] = !lazy[(node<<1)+1];
}
}

// At this point we are sure that pending lazy updates
// are done for current node. So we can return value
// If this segment lies in range
if (qs<=st && en<=qe)
return tree[node];

// If a part of this segment overlaps with the given range
int mid = (st+en)/2;
return countQuery((node<<1), st, mid, qs, qe) +
countQuery((node<<1)+1, mid+1, en, qs, qe);
}

// Driver program to run the case
int main()
{
int n = 5;
toggle(1, 0, n-1, 1, 2);  //  Toggle 1 2
toggle(1, 0, n-1, 2, 4);  //  Toggle 2 4

cout << countQuery(1, 0, n-1, 2, 3) << endl;  //  Count 2 3

toggle(1, 0, n-1, 2, 4);  //  Toggle 2 4

cout << countQuery(1, 0, n-1, 1, 4) << endl;  //  Count 1 4

return 0;
}
```

Output:

```1
2
```

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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