Given a binary array arr[] and q queries of following types:
- k: find the index of the kth set bit i.e. kth 1 in the array.
- (x, y): Update arr[x] = y where y can either be a 0 or 1.
Examples:
Input: arr[] = {1, 0, 1, 0, 0, 1, 1, 1}, q = 2
k = 4
(x, y) = (5, 1)
Output:
Index of 4th set bit: 6
Array after updation:
1 0 1 0 0 0 1 1
Approach: A simple solution is to run a loop from 0 to n – 1 and find the kth 1. To update a value, simply do arr[i] = x. The first operation takes O(n) time and second operation takes O(1) time.
Another solution is to create another array and store each 1’s index at the ith index in this array. Index of Kth 1 can now be calculated in O(1) time but update operation takes O(n) time now. This works well if the number of query operations are large and very few updates.
But, what if the number of query and updates are equal? We can perform both the operations in O(log n) time using a Segment Tree to do both operations in O(Logn) time.
Representation of Segment tree:
- Leaf Nodes are the elements of the input array.
- Each internal node represents sum merging of the leaf nodes.
An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at (i-1)/2.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <iostream> using namespace std;
// Function to build the tree void buildTree( int * tree, int * a, int s, int e, int idx)
{ // s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e) {
// store each element value at the
// leaf node
tree[idx] = a[s];
return ;
}
int mid = (s + e) / 2;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return ;
} // Function to return the index of the query int queryTree( int * tree, int s, int e, int k, int idx)
{ // s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1, e, k - tree[2 * idx], 2 * idx + 1);
} // Function to perform the update query void updateTree( int * tree, int s, int e, int i, int change, int idx)
{ // s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e) {
cout << "error" ;
return ;
}
// Leaf node of the required index i
if (s == e) {
// Replacing the node value with
// the new changed value
tree[idx] = change;
return ;
}
int mid = (s + e) / 2;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return ;
} // Function to perform queries void queries( int * tree, int * a, int q, int p, int k, int change, int n)
{ int s = 0, e = n - 1, idx = 1;
if (q == 1) {
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
cout << "Array after updation:\n" ;
for ( int i = 0; i < n; i++)
cout << a[i] << " " ;
cout << "\n" ;
}
else {
// q = 0, print kth bit
cout << "Index of " << k << "th set bit: "
<< queryTree(tree, s, e, k, idx) << "\n" ;
}
} // Driver code int main()
{ int a[] = { 1, 0, 1, 0, 0, 1, 1, 1 };
int n = sizeof (a) / sizeof ( int );
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int * tree = new int [4 * n + 1];
for ( int i = 0; i < 4 * n + 1; ++i) {
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
int s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1, p = 5, change = 0;
queries(tree, a, q, p, k, change, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
import java.util.*;
class GFG
{ // Function to build the tree
static void buildTree( int [] tree, int [] a,
int s, int e, int idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e)
{
// store each element value at the
// leaf node
tree[idx] = a[s];
return ;
}
int mid = (s + e) / 2 ;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1 , e, 2 * idx + 1 );
// Summing up the number of one's
tree[idx] = tree[ 2 * idx] + tree[ 2 * idx + 1 ];
return ;
}
// Function to return the index of the query
static int queryTree( int [] tree, int s,
int e, int k, int idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return - 1 ;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2 ;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[ 2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1 , e, k - tree[ 2 * idx],
2 * idx + 1 );
}
// Function to perform the update query
static void updateTree( int [] tree, int s, int e,
int i, int change, int idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e)
{
System.out.println( "Error" );
return ;
}
// Leaf node of the required index i
if (s == e)
{
// Replacing the node value with
// the new changed value
tree[idx] = change;
return ;
}
int mid = (s + e) / 2 ;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1 , e, i, change, 2 * idx + 1 );
// Merging both left and right sub-trees
tree[idx] = tree[ 2 * idx] + tree[ 2 * idx + 1 ];
return ;
}
// Function to perform queries
static void queries( int [] tree, int [] a, int q, int p,
int k, int change, int n)
{
int s = 0 , e = n - 1 , idx = 1 ;
if (q == 1 )
{
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
System.out.println( "Array after updation:" );
for ( int i = 0 ; i < n; i++)
System.out.print(a[i] + " " );
System.out.println();
}
else
{
// q = 0, print kth bit
System.out.printf( "Index of %dth set bit: %d\n" ,
k, queryTree(tree, s, e, k, idx));
}
}
// Driver Code
public static void main(String[] args)
{
int [] a = { 1 , 0 , 1 , 0 , 0 , 1 , 1 , 1 };
int n = a.length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int [] tree = new int [ 4 * n + 1 ];
for ( int i = 0 ; i < 4 * n + 1 ; ++i)
{
tree[i] = 0 ;
}
// s and e are the starting and ending
// indices respectively
int s = 0 , e = n - 1 , idx = 1 ;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0 , p = 0 , change = 0 , k = 4 ;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1 ;
p = 5 ;
change = 0 ;
queries(tree, a, q, p, k, change, n);
}
} // This code is contributed by // sanjeev2552 |
# Python3 implementation of the approach # Function to build the tree def buildTree(tree: list , a: list ,
s: int , e: int , idx: int ):
# s = starting index
# e = ending index
# a = array storing binary string of numbers
# idx = starting index = 1, for recurring the tree
if s = = e:
# store each element value at the
# leaf node
tree[idx] = a[s]
return
mid = (s + e) / / 2
# Recurring in two sub portions (left, right)
# to build the segment tree
# Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx)
# Calling for the right sub portion
buildTree(tree, a, mid + 1 , e, 2 * idx + 1 )
# Summing up the number of one's
tree[idx] = tree[ 2 * idx] + tree[ 2 * idx + 1 ]
return
# Function to return the index of the query def queryTree(tree: list , s: int , e: int ,
k: int , idx: int ) - > int :
# s = starting index
# e = ending index
# k = searching for kth bit
# idx = starting index = 1, for recurring the tree
# k > number of 1's in a binary string
if k > tree[idx]:
return - 1
# leaf node at which kth 1 is stored
if s = = e:
return s
mid = (s + e) / / 2
# If left sub-tree contains more or equal 1's
# than required kth 1
if tree[ 2 * idx] > = k:
return queryTree(tree, s, mid, k, 2 * idx)
# If left sub-tree contains less 1's than
# required kth 1 then recur in the right sub-tree
else :
return queryTree(tree, mid + 1 , e,
k - tree[ 2 * idx], 2 * idx + 1 )
# Function to perform the update query def updateTree(tree: list , s: int , e: int ,
i: int , change: int , idx: int ):
# s = starting index
# e = ending index
# i = index at which change is to be done
# change = new changed bit
# idx = starting index = 1, for recurring the tree
# Out of bounds request
if i < s or i > e:
print ( "error" )
return
# Leaf node of the required index i
if s = = e:
# Replacing the node value with
# the new changed value
tree[idx] = change
return
mid = (s + e) / / 2
# If the index i lies in the left sub-tree
if i > = s and i < = mid:
updateTree(tree, s, mid, i, change, 2 * idx)
# If the index i lies in the right sub-tree
else :
updateTree(tree, mid + 1 , e, i, change, 2 * idx + 1 )
# Merging both left and right sub-trees
tree[idx] = tree[ 2 * idx] + tree[ 2 * idx + 1 ]
return
# Function to perform queries def queries(tree: list , a: list , q: int , p: int ,
k: int , change: int , n: int ):
s = 0
e = n - 1
idx = 1
if q = = 1 :
# q = 1 update, p = index at which change
# is to be done, change = new bit
a[p] = change
updateTree(tree, s, e, p, change, idx)
print ( "Array after updation:" )
for i in range (n):
print (a[i], end = " " )
print ()
else :
# q = 0, print kth bit
print ( "Index of %dth set bit: %d" % (k,
queryTree(tree, s, e, k, idx)))
# Driver Code if __name__ = = "__main__" :
a = [ 1 , 0 , 1 , 0 , 0 , 1 , 1 , 1 ]
n = len (a)
# Declaring & initializing the tree with
# maximum possible size of the segment tree
# and each value initially as 0
tree = [ 0 ] * ( 4 * n + 1 )
# s and e are the starting and ending
# indices respectively
s = 0
e = n - 1
idx = 1
# Build the segment tree
buildTree(tree, a, s, e, idx)
# Find index of kth set bit
q = 0
p = 0
change = 0
k = 4
queries(tree, a, q, p, k, change, n)
# Update query
q = 1
p = 5
change = 0
queries(tree, a, q, p, k, change, n)
# This code is contributed by # sanjeev2552 |
// C# implementation of the approach using System;
class GFG
{ // Function to build the tree
static void buildTree( int [] tree, int [] a,
int s, int e, int idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e)
{
// store each element value at the
// leaf node
tree[idx] = a[s];
return ;
}
int mid = (s + e) / 2;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return ;
}
// Function to return the index of the query
static int queryTree( int [] tree, int s,
int e, int k, int idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1,
e, k - tree[2 * idx],
2 * idx + 1);
}
// Function to perform the update query
static void updateTree( int [] tree, int s, int e,
int i, int change, int idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e)
{
Console.WriteLine( "Error" );
return ;
}
// Leaf node of the required index i
if (s == e)
{
// Replacing the node value with
// the new changed value
tree[idx] = change;
return ;
}
int mid = (s + e) / 2;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i,
change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i,
change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] +
tree[2 * idx + 1];
return ;
}
// Function to perform queries
static void queries( int [] tree, int [] a, int q, int p,
int k, int change, int n)
{
int s = 0, e = n - 1, idx = 1;
if (q == 1)
{
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
Console.WriteLine( "Array after updation:" );
for ( int i = 0; i < n; i++)
Console.Write(a[i] + " " );
Console.WriteLine();
}
else
{
// q = 0, print kth bit
Console.Write( "Index of {0}th set bit: {1}\n" ,
k, queryTree(tree, s, e, k, idx));
}
}
// Driver Code
public static void Main(String[] args)
{
int [] a = { 1, 0, 1, 0, 0, 1, 1, 1 };
int n = a.Length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int [] tree = new int [4 * n + 1];
for ( int i = 0; i < 4 * n + 1; ++i)
{
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
int s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1;
p = 5;
change = 0;
queries(tree, a, q, p, k, change, n);
}
} // This code is contributed by Rajput-Ji |
<script> // JavaScript implementation of the approach // Function to build the tree function buildTree(tree, a, s, e, idx)
{ // s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e) {
// store each element value at the
// leaf node
tree[idx] = a[s];
return ;
}
let mid = Math.floor((s + e) / 2);
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return ;
} // Function to return the index of the query function queryTree(tree, s, e, k, idx)
{ // s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
let mid = Math.floor((s + e) / 2);
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1, e, k -
tree[2 * idx], 2 * idx + 1);
} // Function to perform the update query function updateTree(tree, s, e, i, change, idx)
{ // s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e) {
document.write( "error" );
return ;
}
// Leaf node of the required index i
if (s == e) {
// Replacing the node value with
// the new changed value
tree[idx] = change;
return ;
}
let mid = Math.floor((s + e) / 2);
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return ;
} // Function to perform queries function queries(tree, a, q, p, k, change, n)
{ let s = 0, e = n - 1, idx = 1;
if (q == 1) {
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
document.write( "Array after updation:<br>" );
for (let i = 0; i < n; i++)
document.write(a[i] + " " );
document.write( "<br>" );
}
else {
// q = 0, print kth bit
document.write( "Index of " + k +
"th set bit: " + queryTree(tree, s, e, k, idx) + "<br>" );
}
} // Driver code let a = [ 1, 0, 1, 0, 0, 1, 1, 1 ];
let n = a.length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
let tree = new Array(4 * n + 1);
for (let i = 0; i < 4 * n + 1; ++i) {
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
let s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
let q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1, p = 5, change = 0;
queries(tree, a, q, p, k, change, n);
// This code is contributed by _saurabh_jaiswal </script> |
Index of 4th set bit: 6 Array after updation: 1 0 1 0 0 0 1 1
Time Complexity: O(logN) per query and O(NlogN) for building segment tree.
Auxiliary Space: O(N)