# Find the number of different numbers in the array after applying the given operation q times

• Difficulty Level : Medium
• Last Updated : 25 Jun, 2022

Given an array of size N, initially consists of zeroes only. The task is to apply the given operation q times and find the number of different numbers in the array except for zeroes.
Operation Format: update(l, r, x):: update a[i] = x for all (l <= i <= r).

Examples:

Input : N = 5, Q = 3,
update(1, 3, 1)
update(0, 1, 2)
update(3, 3, 3)
Output :
Explanation : Initially array is {0, 0, 0, 0, 0}. After
applying the operation for the first time array becomes {0, 1, 1, 1, 0}.
After applying the operation for the second time the array becomes
{2, 2, 1, 1, 0}. After applying the operation for the third time the array
becomes {2, 2, 1, 3, 0}. So, a number of different numbers expect zero are 3.

Input : N = 5, Q = 3,
update(1, 1, 4)
update(0, 1, 2)
update(1, 4, 5)
Output :

Approach :
Each operation suggests a range update, hence try to update the array using lazy propagation. After applying the operation Q times using lazy propagation call a function that finds the number of different numbers in the array. This function uses a set to find the count of different numbers.
The update and query operations are similar to what they are in a segment tree with some changes. Whenever an update query gets executed in a segment tree, all the nodes associated with the current node also get updated whereas in lazy propagation those nodes will only get updated when required i.e. we create an array lazy[] of size equal to the given array all of whose elements will be initialized to 0 which means there are no updates for any node initially and any non-zero value at lazy[i] indicates that node i has an update pending which will only be updated while querying (when required).

Below is the implementation of the above approach :

## C++

 `// CPP implementation for above approach``#include ``using` `namespace` `std;` `#define N 100005` `// To store the tree in lazy propagation``int` `lazy[4 * N];` `// To store the different numbers``set<``int``> se;` `// Function to update in the range [x, y) with given value``void` `update(``int` `x, ``int` `y, ``int` `value, ``int` `id, ``int` `l, ``int` `r)``{``    ``// check out of bound``    ``if` `(x >= r or l >= y)``        ``return``;` `    ``// check for complete overlap``    ``if` `(x <= l && r <= y) {``        ``lazy[id] = value;``        ``return``;``    ``}` `    ``// find the mid number``    ``int` `mid = (l + r) / 2;` `    ``// check for pending updates``    ``if` `(lazy[id])``        ``lazy[2 * id] = lazy[2 * id + 1] = lazy[id];` `    ``// make lazy[id] = 0, so that it has no pending updates``    ``lazy[id] = 0;` `    ``// call for two child nodes``    ``update(x, y, value, 2 * id, l, mid);``    ``update(x, y, value, 2 * id + 1, mid, r);``}` `// Function to find non-zero integers in the range [l, r)``void` `query(``int` `id, ``int` `l, ``int` `r)``{``    ``// if id contains positive number``    ``if` `(lazy[id]) {``        ``se.insert(lazy[id]);``        ``// There is no need to see the children,``        ``// because all the interval have same number``        ``return``;``    ``}` `    ``// check for out of bound``    ``if` `(r - l < 2)``        ``return``;` `    ``// find the middle number``    ``int` `mid = (l + r) / 2;` `    ``// call for two child nodes``    ``query(2 * id, l, mid);``    ``query(2 * id + 1, mid, r);``}` `// Driver code``int` `main()``{``    ``// size of the array and number of queries``    ``int` `n = 5, q = 3;` `    ``// Update operation for l, r, x, id, 0, n``    ``update(1, 4, 1, 1, 0, n);``    ``update(0, 2, 2, 1, 0, n);``    ``update(3, 4, 3, 1, 0, n);` `    ``// Query operation to get answer in the range [0, n-1]``    ``query(1, 0, n);` `    ``// Print the count of non-zero elements``    ``cout << se.size() << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation for above approach``import` `java.util.*;` `class` `geeks``{``    ` `    ``static` `int` `N = ``100005``;` `    ``// To store the tree in lazy propagation``    ``static` `int``[] lazy = ``new` `int``[``4``*N];` `    ``// To store the different numbers``    ``static` `Set se = ``new` `HashSet();` `    ``// Function to update in the range [x, y) with given value``    ``public` `static` `void` `update(``int` `x, ``int` `y, ``int` `value,``                            ``int` `id, ``int` `l, ``int` `r)``    ``{` `        ``// check out of bound``        ``if` `(x >= r || l >= y)``            ``return``;``    ` `        ``// check for complete overlap``        ``if` `(x <= l && r <= y)``        ``{``            ``lazy[id] = value;``            ``return``;``        ``}``    ` `        ``// find the mid number``        ``int` `mid = (l + r) / ``2``;``    ` `        ``// check for pending updates``        ``if` `(lazy[id] != ``0``)``            ``lazy[``2` `* id] = lazy[``2` `* id + ``1``] = lazy[id];``    ` `        ``// make lazy[id] = 0, so that it has no pending updates``        ``lazy[id] = ``0``;``    ` `        ``// call for two child nodes``        ``update(x, y, value, ``2` `* id, l, mid);``        ``update(x, y, value, ``2` `* id + ``1``, mid, r);``    ``}` `    ``// Function to find non-zero integers in the range [l, r)``    ``public` `static` `void` `query(``int` `id, ``int` `l, ``int` `r)``    ``{``        ` `        ``// if id contains positive number``        ``if` `(lazy[id] != ``0``)``        ``{``            ``se.add(lazy[id]);``            ` `            ``// There is no need to see the children,``            ``// because all the interval have same number``            ``return``;``        ``}` `        ``// check for out of bound``        ``if` `(r - l < ``2``)``            ``return``;` `        ``// find the middle number``        ``int` `mid = (l + r) / ``2``;` `        ``// call for two child nodes``        ``query(``2` `* id, l, mid);``        ``query(``2` `* id + ``1``, mid, r);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ` `        ``// size of the array and number of queries``        ``int` `n = ``5``, q = ``3``;` `        ``// Update operation for l, r, x, id, 0, n``        ``update(``1``, ``4``, ``1``, ``1``, ``0``, n);``        ``update(``0``, ``2``, ``2``, ``1``, ``0``, n);``        ``update(``3``, ``4``, ``3``, ``1``, ``0``, n);` `        ``// Query operation to get answer in the range [0, n-1]``        ``query(``1``, ``0``, n);` `        ``// Print the count of non-zero elements``        ``System.out.println(se.size());``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 implementation for above approach``N ``=` `100005` `# To store the tree in lazy propagation``lazy ``=` `[``0``] ``*` `(``4` `*` `N);` `# To store the different numbers``se ``=` `set``()` `# Function to update in the range [x, y)``# with given value``def` `update(x, y, value, ``id``, l, r) :``    ` `    ``# check out of bound``    ``if` `(x >``=` `r ``or` `l >``=` `y):``        ``return``;` `    ``# check for complete overlap``    ``if` `(x <``=` `l ``and` `r <``=` `y) :``        ``lazy[``id``] ``=` `value;``        ``return``;` `    ``# find the mid number``    ``mid ``=` `(l ``+` `r) ``/``/` `2``;` `    ``# check for pending updates``    ``if` `(lazy[``id``]) :``        ``lazy[``2` `*` `id``] ``=` `lazy[``2` `*` `id` `+` `1``] ``=` `lazy[``id``];` `    ``# make lazy[id] = 0,``    ``# so that it has no pending updates``    ``lazy[``id``] ``=` `0``;` `    ``# call for two child nodes``    ``update(x, y, value, ``2` `*` `id``, l, mid);``    ``update(x, y, value, ``2` `*` `id` `+` `1``, mid, r);` `# Function to find non-zero integers``# in the range [l, r)``def` `query(``id``, l, r) :``    ` `    ``# if id contains positive number``    ``if` `(lazy[``id``]) :``        ` `        ``se.add(lazy[``id``]);``        ` `        ``# There is no need to see the children,``        ``# because all the interval have same number``        ``return``;``    ` `    ``# check for out of bound``    ``if` `(r ``-` `l < ``2``) :``        ``return``;` `    ``# find the middle number``    ``mid ``=` `(l ``+` `r) ``/``/` `2``;` `    ``# call for two child nodes``    ``query(``2` `*` `id``, l, mid);``    ``query(``2` `*` `id` `+` `1``, mid, r);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``# size of the array and number of queries``    ``n ``=` `5``; q ``=` `3``;` `    ``# Update operation for l, r, x, id, 0, n``    ``update(``1``, ``4``, ``1``, ``1``, ``0``, n);``    ``update(``0``, ``2``, ``2``, ``1``, ``0``, n);``    ``update(``3``, ``4``, ``3``, ``1``, ``0``, n);` `    ``# Query operation to get answer``    ``# in the range [0, n-1]``    ``query(``1``, ``0``, n);` `    ``# Print the count of non-zero elements``    ``print``(``len``(se));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation for above approach``using` `System;``using` `System.Collections.Generic;``    ` `public` `class` `geeks``{``    ` `    ``static` `int` `N = 100005;` `    ``// To store the tree in lazy propagation``    ``static` `int``[] lazy = ``new` `int``[4*N];` `    ``// To store the different numbers``    ``static` `HashSet<``int``> se = ``new` `HashSet<``int``>();` `    ``// Function to update in the range [x, y) with given value``    ``public` `static` `void` `update(``int` `x, ``int` `y, ``int` `value,``                            ``int` `id, ``int` `l, ``int` `r)``    ``{` `        ``// check out of bound``        ``if` `(x >= r || l >= y)``            ``return``;``    ` `        ``// check for complete overlap``        ``if` `(x <= l && r <= y)``        ``{``            ``lazy[id] = value;``            ``return``;``        ``}``    ` `        ``// find the mid number``        ``int` `mid = (l + r) / 2;``    ` `        ``// check for pending updates``        ``if` `(lazy[id] != 0)``            ``lazy[2 * id] = lazy[2 * id + 1] = lazy[id];``    ` `        ``// make lazy[id] = 0, so that it has no pending updates``        ``lazy[id] = 0;``    ` `        ``// call for two child nodes``        ``update(x, y, value, 2 * id, l, mid);``        ``update(x, y, value, 2 * id + 1, mid, r);``    ``}` `    ``// Function to find non-zero integers in the range [l, r)``    ``public` `static` `void` `query(``int` `id, ``int` `l, ``int` `r)``    ``{``        ` `        ``// if id contains positive number``        ``if` `(lazy[id] != 0)``        ``{``            ``se.Add(lazy[id]);``            ` `            ``// There is no need to see the children,``            ``// because all the interval have same number``            ``return``;``        ``}` `        ``// check for out of bound``        ``if` `(r - l < 2)``            ``return``;` `        ``// find the middle number``        ``int` `mid = (l + r) / 2;` `        ``// call for two child nodes``        ``query(2 * id, l, mid);``        ``query(2 * id + 1, mid, r);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ` `        ``// size of the array and number of queries``        ``int` `n = 5, q = 3;` `        ``// Update operation for l, r, x, id, 0, n``        ``update(1, 4, 1, 1, 0, n);``        ``update(0, 2, 2, 1, 0, n);``        ``update(3, 4, 3, 1, 0, n);` `        ``// Query operation to get answer in the range [0, n-1]``        ``query(1, 0, n);` `        ``// Print the count of non-zero elements``        ``Console.WriteLine(se.Count);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N*logN), as we are using two recursive calls and in each recursive call, we are decrementing mid by floor division of 2.

Auxiliary Space: O(N), as we are using the implicit extra space for the recursive stack for the recursive calls.

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