Number of elements less than or equal to a given number in a given subarray | Set 2 (Including Updates)

Last Updated : 10 Mar, 2023

Given an array 'a[]' and number of queries q there will be two type of queries

Query 0 update(i, v) : Two integers i and v which means set a[i] = v
Query 1 count(l, r, k): We need to print number of integers less than equal to k in the subarray l to r.

Given a[i], v <= 10000 Examples :

Input : arr[] = {5, 1, 2, 3, 4}
        q = 6
        1 1 3 1  // First value 1 means type of query is count()
        0 3 10   // First value 0 means type of query is update()
        1 3 3 4
        0 2 1
        0 0 2
        1 0 4 5
Output :
1
0
4
For first query number of values less than equal to 1 in 
arr[1..3] is 1(1 only), update a[3] = 10
There is no value less than equal to 4 in the a[3..3]
and similarly other queries are answered

We have discussed a solution that handles only count() queries in below post.Number of elements less than or equal to a given number in a given subarray Here update() query also needs to be handled. Naive Approach The naive approach is whenever there is update operation update the array and whenever type 2 query is there traverse the subarray and count the valid elements. Efficient Approach The idea is to use square root decomposition

Step 1 : Divide the array in sqrt(n) equal sized blocks. For each block keep a binary index tree of size equal to 1 more than the maximum possible element in the array of the elements in that block.
Step 2: For each element of the array find out the block to which it belongs and update the bit array of that block with the value 1 at arr[i].
Step 3: Whenever there is a update query, update the bit array of the corresponding block at the original value of the array at that index with value equal to -1 and update the bit array of the same block with value 1 at the new value of the array at that index.
Step 4: For type 2 query you can make a single query to the BIT (to count elements less than or equal to k) for each complete block in the range, and for the two partial blocks on the end, just loop through the elements.

C++

// Number of elements less than or equal to a given
// number in a given subarray and allowing update
// operations.
#include<bits/stdc++.h>
using namespace std;

const int MAX = 10001;

// updating the bit array of a valid block
void update(int idx, int blk, int val, int bit[][MAX])
{
    for (; idx<MAX; idx += (idx&-idx))
        bit[blk][idx] += val;
}

// answering the query
int query(int l, int r, int k, int arr[], int blk_sz,
                                      int bit[][MAX])
{
    // traversing the first block in range
    int sum = 0;
    while (l<r && l%blk_sz!=0 && l!=0)
    {
        if (arr[l] <= k)
            sum++;
        l++;
    }

    // Traversing completely overlapped blocks in
    // range for such blocks bit array of that block
    // is queried
    while (l + blk_sz <= r)
    {
        int idx = k;
        for (; idx > 0 ; idx -= idx&-idx)
            sum += bit[l/blk_sz][idx];
        l += blk_sz;
    }

    // Traversing the last block
    while (l <= r)
    {
        if (arr[l] <= k)
            sum++;
        l++;
    }
    return sum;
}

// Preprocessing the array
void preprocess(int arr[], int blk_sz, int n, int bit[][MAX])
{
    for (int i=0; i<n; i++)
        update(arr[i], i/blk_sz, 1, bit);
}

void preprocessUpdate(int i, int v, int blk_sz,
                      int arr[], int bit[][MAX])
{
    // updating the bit array at the original
    // and new value of array
    update(arr[i], i/blk_sz, -1, bit);
    update(v, i/blk_sz, 1, bit);
    arr[i] = v;
}

// driver function
int main()
{
    int arr[] = {5, 1, 2, 3, 4};
    int n = sizeof(arr)/sizeof(arr[0]);

    // size of block size will be equal to square root of n
    int blk_sz = sqrt(n);

    // initialising bit array of each block
    // as elements of array cannot exceed 10^4 so size
    // of bit array is accordingly
    int bit[blk_sz+1][MAX];
    memset(bit, 0, sizeof(bit));

    preprocess(arr, blk_sz, n, bit);
    cout << query  (1, 3, 1, arr, blk_sz, bit) << endl;

    preprocessUpdate(3, 10, blk_sz, arr, bit);
    cout << query(3, 3, 4, arr, blk_sz, bit) << endl;
    preprocessUpdate(2, 1, blk_sz, arr, bit);
    preprocessUpdate(0, 2, blk_sz, arr, bit);
    cout << query (0, 4, 5, arr, blk_sz, bit) << endl;
    return 0;
}

Java

// Number of elements less than or equal to a given
// number in a given subarray and allowing update
// operations.

class Test
{
    static final int MAX = 10001;
    
    // updating the bit array of a valid block
    static void update(int idx, int blk, int val, int bit[][])
    {
        for (; idx<MAX; idx += (idx&-idx))
            bit[blk][idx] += val;
    }
     
    // answering the query
    static int query(int l, int r, int k, int arr[], int blk_sz,
                                          int bit[][])
    {
        // traversing the first block in range
        int sum = 0;
        while (l<r && l%blk_sz!=0 && l!=0)
        {
            if (arr[l] <= k)
                sum++;
            l++;
        }
     
        // Traversing completely overlapped blocks in
        // range for such blocks bit array of that block
        // is queried
        while (l + blk_sz <= r)
        {
            int idx = k;
            for (; idx > 0 ; idx -= idx&-idx)
                sum += bit[l/blk_sz][idx];
            l += blk_sz;
        }
     
        // Traversing the last block
        while (l <= r)
        {
            if (arr[l] <= k)
                sum++;
            l++;
        }
        return sum;
    }
     
    // Preprocessing the array
    static void preprocess(int arr[], int blk_sz, int n, int bit[][])
    {
        for (int i=0; i<n; i++)
            update(arr[i], i/blk_sz, 1, bit);
    }
     
    static void preprocessUpdate(int i, int v, int blk_sz,
                          int arr[], int bit[][])
    {
        // updating the bit array at the original
        // and new value of array
        update(arr[i], i/blk_sz, -1, bit);
        update(v, i/blk_sz, 1, bit);
        arr[i] = v;
    }
    
    // Driver method
    public static void main(String args[])
    {
        int arr[] = {5, 1, 2, 3, 4};
     
        // size of block size will be equal to square root of n
        int blk_sz = (int) Math.sqrt(arr.length);
     
        // initialising bit array of each block
        // as elements of array cannot exceed 10^4 so size
        // of bit array is accordingly
        int bit[][] = new int[blk_sz+1][MAX];
     
        preprocess(arr, blk_sz, arr.length, bit);
        System.out.println(query(1, 3, 1, arr, blk_sz, bit));
     
        preprocessUpdate(3, 10, blk_sz, arr, bit);
        System.out.println(query(3, 3, 4, arr, blk_sz, bit));
        preprocessUpdate(2, 1, blk_sz, arr, bit);
        preprocessUpdate(0, 2, blk_sz, arr, bit);
        System.out.println(query (0, 4, 5, arr, blk_sz, bit));
    }
}

Python3

# Number of elements less than or equal to a given
# number in a given subarray and allowing update
# operations.
MAX = 10001

# updating the bit array of a valid block
def update(idx, blk, val, bit):
    while idx < MAX:
        bit[blk][idx] += val

        idx += (idx & -idx)

# answering the query
def query(l, r, k, arr, blk_sz, bit):

    # traversing the first block in range
    summ = 0
    while l < r and l % blk_sz != 0 and l != 0:
        if arr[l] <= k:
            summ += 1
        l += 1

    # Traversing completely overlapped blocks in
    # range for such blocks bit array of that block
    # is queried
    while l + blk_sz <= r:
        idx = k
        while idx > 0:
            summ += bit[l // blk_sz][idx]
            idx -= (idx & -idx)

        l += blk_sz

    # Traversing the last block
    while l <= r:
        if arr[l] <= k:
            summ += 1
        l += 1

    return summ

# Preprocessing the array
def preprocess(arr, blk_sz, n, bit):
    for i in range(n):
        update(arr[i], i // blk_sz, 1, bit)

def preprocessUpdate(i, v, blk_sz, arr, bit):

    # updating the bit array at the original
    # and new value of array
    update(arr[i], i // blk_sz, -1, bit)
    update(v, i // blk_sz, 1, bit)
    arr[i] = v

# Driver Code
if __name__ == "__main__":
    arr = [5, 1, 2, 3, 4]
    n = len(arr)

    # size of block size will be equal 
    # to square root of n
    from math import sqrt
    blk_sz = int(sqrt(n))

    # initialising bit array of each block
    # as elements of array cannot exceed 10^4 
    # so size of bit array is accordingly
    bit = [[0 for i in range(MAX)] 
              for j in range(blk_sz + 1)]

    preprocess(arr, blk_sz, n, bit)
    print(query(1, 3, 1, arr, blk_sz, bit))

    preprocessUpdate(3, 10, blk_sz, arr, bit)
    print(query(3, 3, 4, arr, blk_sz, bit))
    preprocessUpdate(2, 1, blk_sz, arr, bit)
    preprocessUpdate(0, 2, blk_sz, arr, bit)
    print(query(0, 4, 5, arr, blk_sz, bit))

# This code is contributed by
# sanjeev2552

// Number of elements less than or equal
// to a given number in a given subarray
// and allowing update operations.
using System;
        
class GFG 
{
    static int MAX = 10001;
    
    // updating the bit array of a valid block
    static void update(int idx, int blk, 
                    int val, int [,]bit)
    {
        for (; idx < MAX; idx += (idx&-idx))
            bit[blk, idx] += val;
    }
    
    // answering the query
    static int query(int l, int r, int k, int []arr,
                             int blk_sz, int [,]bit)
    {
        // traversing the first block in range
        int sum = 0;
        while (l < r && l % blk_sz != 0 && l != 0)
        {
            if (arr[l] <= k)
                sum++;
            l++;
        }
    
        // Traversing completely overlapped blocks in
        // range for such blocks bit array of that block
        // is queried
        while (l + blk_sz <= r)
        {
            int idx = k;
            for (; idx > 0 ; idx -= idx&-idx)
                sum += bit[l/blk_sz,idx];
            l += blk_sz;
        }
    
        // Traversing the last block
        while (l <= r)
        {
            if (arr[l] <= k)
                sum++;
            l++;
        }
        return sum;
    }
    
    // Preprocessing the array
    static void preprocess(int []arr, int blk_sz,
                               int n, int [,]bit)
    {
        for (int i=0; i<n; i++)
            update(arr[i], i / blk_sz, 1, bit);
    }
    
    static void preprocessUpdate(int i, int v, int blk_sz,
                                    int []arr, int [,]bit)
    {
        // updating the bit array at the original
        // and new value of array
        update(arr[i], i/blk_sz, -1, bit);
        update(v, i/blk_sz, 1, bit);
        arr[i] = v;
    }
    
    // Driver method
    public static void Main()
    {
        int []arr = {5, 1, 2, 3, 4};
    
        // size of block size will be 
        // equal to square root of n
        int blk_sz = (int) Math.Sqrt(arr.Length);
    
        // initialising bit array of each block
        // as elements of array cannot exceed 10^4 so size
        // of bit array is accordingly
        int [,]bit = new int[blk_sz+1,MAX];
    
        preprocess(arr, blk_sz, arr.Length, bit);
        Console.WriteLine(query(1, 3, 1, arr, blk_sz, bit));
    
        preprocessUpdate(3, 10, blk_sz, arr, bit);
        Console.WriteLine(query(3, 3, 4, arr, blk_sz, bit));
        preprocessUpdate(2, 1, blk_sz, arr, bit);
        preprocessUpdate(0, 2, blk_sz, arr, bit);
        Console.WriteLine(query (0, 4, 5, arr, blk_sz, bit));
    }
}

// This code is contributed by Sam007

JavaScript

<script>

// Number of elements less than or equal to a given
// number in a given subarray and allowing update
// operations.

const MAX = 10001;

// updating the bit array of a valid block
function update(idx, blk, val, bit)
{
    while(idx<MAX)
    {
        bit[blk][idx] += val;
        idx += (idx&-idx);
    }
}
  
// answering the query
function query(l, r,  k,  arr,  blk_sz, bit)
{
    // traversing the first block in range
    var sum = 0;
    while (l<r && l%blk_sz!=0 && l!=0)
    {
        if (arr[l] <= k)
            sum++;
        l++;
    }
  
    // Traversing completely overlapped blocks in
    // range for such blocks bit array of that block
    // is queried
    while (l + blk_sz <= r)
    {
        var idx = k;
        for (; idx > 0 ; idx -= idx&-idx)
            sum += bit[parseInt(l/blk_sz)][idx];
        l += blk_sz;
    }
  
    // Traversing the last block
    while (l <= r)
    {
        if (arr[l] <= k)
            sum++;
        l++;
    }
    return sum;
}
  
// Preprocessing the array
function preprocess(arr,  blk_sz,  n,  bit)
{
    for (var i=0; i<n; i++)
        update(arr[i], parseInt(i/blk_sz), 1, bit);
}
  
function preprocessUpdate( i,  v,  blk_sz,
                       arr,  bit)
{
    // updating the bit array at the original
    // and new value of array
    update(arr[i], parseInt(i/blk_sz), -1, bit);
    update(v, parseInt(i/blk_sz), 1, bit);
    arr[i] = v;
}

// Driver code

// Test Case 1
let arr = [ 5, 1, 2, 3, 4 ];
var n = arr.length;
// size of block size will be equal to square root of n
var blk_sz = parseInt(Math.sqrt(n));

// initialising bit array of each block
// as elements of array cannot exceed 10^4 so size
// of bit array is accordingly
var bit = new Array(blk_sz+1);

for (var i = 0; i < bit.length; i++) 
{
  bit[i] = new Array(MAX);
}
for (let i = 0; i < bit.length; i++) 
{
  for (let j = 0; j < MAX; j++) 
  {
    bit[i][j] = 0;
    }
}

preprocess(arr, blk_sz, n, bit);
document.write(query  (1, 3, 1, arr, blk_sz, bit));
document.write("<br>");

preprocessUpdate(3, 10, blk_sz, arr, bit);
document.write(query(3, 3, 4, arr, blk_sz, bit));
document.write("<br>");
preprocessUpdate(2, 1, blk_sz, arr, bit);
preprocessUpdate(0, 2, blk_sz, arr, bit);
document.write(query (0, 4, 5, arr, blk_sz, bit));
document.write("<br>");        

// This code is contributed by Aarti_Rathi

</script>

Output:

1
0
4

Time Complexity: O(n log(max_element))

Space Complexity: O(sqrt(n) log(max_element))

The question is know why not to use this method when there were no update operations the answer lies in space complexity in this method 2-d bit array is used as well as its size depends upon the maximum possible value of the array but when there was no update operation our bit array was only dependent on the size of array.