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# Queries to calculate GCD of an array after multiplying first or last K elements by X

Given an array arr[] consisting of N positive integers and a 2D array queries[][] of the type {a, K, X} such that if the value of a is 1, then multiply first K array elements by X. Otherwise, multiply last K array elements by X. The task is to calculate GCD of the array after performing each query on the original array.

Examples:

Input: arr[] = {2, 3, 4, 8}, Queries[][3] = {{1, 2, 2}, {2, 4, 5}}
Output: 2 5
Explanation:
Query 1: The given query is {1, 2, 2}. After multiplying the first 2 array elements by 2, arr[] modifies to {4, 6, 4, 8}. GCD of the modified array is 2.
Query 2: The given query is {2, 4, 5}. After multiplying the last 4 elements array elements by 5, arr[] modifies to {10, 15, 20, 40}. GCD of the updated array is 5.

Input: arr[] = {4, 12, 4, 9}, Queries[][3] = {{1, 3, 3}, {2, 4, 1}}
Output: 3 1

Naive Approach: The simplest approach is to update the given array by performing each query and then find the GCD of the updated array.

Time Complexity: O(N * Q * log(M)), where M is the maximum element present in the array.
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by storing the prefix and suffix GCD arrays of the given array and solving each query in O(1) time by following the below steps:

• Initialize an array prefix[] and suffix[] of size N to store the prefix and suffix GCD arrays of the given array.
• Traverse the array from the front and the back and find the prefix and suffix GCD at each index and store it in prefix[] and suffix[] respectively.
• Now, traverse the array queries[] and for each query {a, K, X} perform the following:
• If the value of K is N then print the value of prefix[N – 1] * X as the result.
• If the value of a is 1, then find the GCD of prefix[K – 1] * X and suffix[K] as the result as the prefix[K – 1] * X is the new GCD of first K numbers and suffix[K + 1] is the GCD of the remaining array elements.
• If the value of a is 2, then find the GCD of prefix[N – K – 1] and suffix[N – K] * X as the result as the prefix[N – K – 1] * X is the new GCD of first K numbers and suffix[N – K] is the GCD of the remaining array elements.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the GCD after performing``// each query on array elements``void` `findGCDQueries(``int` `arr[], ``int` `N,``                    ``int` `Queries[][3],``                    ``int` `Q)``{``    ``// Stores prefix array and suffix``    ``// array``    ``int` `prefix[N], suffix[N];` `    ``prefix[0] = arr[0];``    ``suffix[N - 1] = arr[N - 1];` `    ``// Build prefix array``    ``for` `(``int` `i = 1; i < N; i++) {``        ``prefix[i] = __gcd(prefix[i - 1],``                          ``arr[i]);``    ``}` `    ``// Build suffix array``    ``for` `(``int` `i = N - 2; i >= 0; i--) {``        ``suffix[i] = __gcd(suffix[i + 1],``                          ``arr[i]);``    ``}` `    ``// Traverse queries array``    ``for` `(``int` `i = 0; i < Q; i++) {` `        ``int` `a = Queries[i][0];``        ``int` `K = Queries[i][1];``        ``int` `X = Queries[i][2];` `        ``// Edge Case when update is``        ``// is required till the end``        ``if` `(K == N) {``            ``cout << prefix[N - 1] * X;``            ``continue``;``        ``}` `        ``// Edge Case when update is``        ``// is required till the front``        ``if` `(a == 1) {``            ``cout << __gcd(prefix[K - 1] * X,``                          ``suffix[K]);``        ``}` `        ``// Find the resultant operation``        ``// for each query``        ``else` `{``            ``cout << __gcd(suffix[N - K] * X,``                          ``prefix[N - K - 1]);``        ``}` `        ``cout << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 4, 8 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `Queries[][3] = {``        ``{ 1, 2, 2 },``        ``{ 2, 4, 5 }``    ``};``    ``int` `Q = ``sizeof``(Queries)``            ``/ ``sizeof``(Queries[0]);` `    ``findGCDQueries(arr, N, Queries, Q);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{` `  ``// Recursive function to return gcd of a and b``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{` `    ``// Everything divides 0``    ``if` `(a == ``0``)``      ``return` `b;``    ``if` `(b == ``0``)``      ``return` `a;` `    ``// base case``    ``if` `(a == b)``      ``return` `a;` `    ``// a is greater``    ``if` `(a > b)``      ``return` `gcd(a - b, b);``    ``return` `gcd(a, b - a);``  ``}`  `  ``// Function to find the GCD after performing``  ``// each query on array elements``  ``static` `void` `findGCDQueries(``int` `arr[], ``int` `N,``                             ``int` `Queries[][],``                             ``int` `Q)``  ``{` `    ``// Stores prefix array and suffix``    ``// array``    ``int` `prefix[] = ``new` `int``[N], suffix[] = ``new` `int``[N];` `    ``prefix[``0``] = arr[``0``];``    ``suffix[N - ``1``] = arr[N - ``1``];` `    ``// Build prefix array``    ``for` `(``int` `i = ``1``; i < N; i++) {``      ``prefix[i] = gcd(prefix[i - ``1``],``                      ``arr[i]);``    ``}` `    ``// Build suffix array``    ``for` `(``int` `i = N - ``2``; i >= ``0``; i--) {``      ``suffix[i] = gcd(suffix[i + ``1``],``                      ``arr[i]);``    ``}` `    ``// Traverse queries array``    ``for` `(``int` `i = ``0``; i < Q; i++) {` `      ``int` `a = Queries[i][``0``];``      ``int` `K = Queries[i][``1``];``      ``int` `X = Queries[i][``2``];` `      ``// Edge Case when update is``      ``// is required till the end``      ``if` `(K == N) {``        ``System.out.print(prefix[N - ``1``] * X);``        ``continue``;``      ``}` `      ``// Edge Case when update is``      ``// is required till the front``      ``if` `(a == ``1``) {``        ``System.out.print(gcd(prefix[K - ``1``] * X,``                             ``suffix[K]));``      ``}` `      ``// Find the resultant operation``      ``// for each query``      ``else` `{``        ``System.out.print(gcd(suffix[N - K] * X,``                             ``prefix[N - K - ``1``]));``      ``}` `      ``System.out.print(``" "``);``    ``}``  ``}``  ` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{` `    ``int` `arr[] = { ``2``, ``3``, ``4``, ``8` `};``    ``int` `N = arr.length;``    ``int` `Queries[][] = {``      ``{ ``1``, ``2``, ``2` `},``      ``{ ``2``, ``4``, ``5` `}``    ``};``    ``int` `Q = Queries.length;` `    ``findGCDQueries(arr, N, Queries, Q);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Python3

 `# Python 3 program for the above approach``from` `math ``import` `gcd` `# Function to find the GCD after performing``# each query on array elements``def` `findGCDQueries(arr, N, Queries, Q):``  ` `    ``# Stores prefix array and suffix``    ``# array``    ``prefix ``=` `[``0` `for` `i ``in` `range``(N)]``    ``suffix ``=` `[``0` `for` `i ``in` `range``(N)]` `    ``prefix[``0``] ``=` `arr[``0``]``    ``suffix[N ``-` `1``] ``=` `arr[N ``-` `1``]` `    ``# Build prefix array``    ``for` `i ``in` `range``(``1``,N,``1``):``        ``prefix[i] ``=` `gcd(prefix[i ``-` `1``], arr[i])` `    ``# Build suffix array``    ``i ``=` `N ``-` `2``    ``while``(i>``=` `0``):``        ``suffix[i] ``=` `gcd(suffix[i ``+` `1``], arr[i])``        ``i ``-``=` `1` `    ``# Traverse queries array``    ``for` `i ``in` `range``(Q):``        ``a ``=` `Queries[i][``0``]``        ``K ``=` `Queries[i][``1``]``        ``X ``=` `Queries[i][``2``]` `        ``# Edge Case when update is``        ``# is required till the end``        ``if` `(K ``=``=` `N):``            ``print``(prefix[N ``-` `1``] ``*` `X,end ``=` `" "``)``            ``continue` `        ``# Edge Case when update is``        ``# is required till the front``        ``if` `(a ``=``=` `1``):``            ``print``(gcd(prefix[K ``-` `1``] ``*` `X,suffix[K]),end ``=` `" "``)` `        ``# Find the resultant operation``        ``# for each query``        ``else``:``            ``print``(gcd(suffix[N ``-` `K] ``*` `X, prefix[N ``-` `K ``-` `1``]),end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=`  `[``2``, ``3``, ``4``, ``8``]``    ``N ``=` `len``(arr)``    ``Queries ``=` `[[``1``, ``2``, ``2``], [``2``, ``4``, ``5``]]``    ``Q ``=` `len``(Queries)``    ``findGCDQueries(arr, N, Queries, Q)``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program to implement``// the above approach``using` `System;``public` `class` `GFG``{` `  ``// Recursive function to return gcd of a and b``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{` `    ``// Everything divides 0``    ``if` `(a == 0)``      ``return` `b;``    ``if` `(b == 0)``      ``return` `a;` `    ``// base case``    ``if` `(a == b)``      ``return` `a;` `    ``// a is greater``    ``if` `(a > b)``      ``return` `gcd(a - b, b);``    ``return` `gcd(a, b - a);``  ``}` `  ``// Function to find the GCD after performing``  ``// each query on array elements``  ``static` `void` `findGCDQueries(``int` `[]arr, ``int` `N,``                             ``int` `[,]Queries,``                             ``int` `Q)``  ``{` `    ``// Stores prefix array and suffix``    ``// array``    ``int` `[]prefix = ``new` `int``[N];``    ``int` `[]suffix = ``new` `int``[N];` `    ``prefix[0] = arr[0];``    ``suffix[N - 1] = arr[N - 1];` `    ``// Build prefix array``    ``for` `(``int` `i = 1; i < N; i++) {``      ``prefix[i] = gcd(prefix[i - 1],``                      ``arr[i]);``    ``}` `    ``// Build suffix array``    ``for` `(``int` `i = N - 2; i >= 0; i--) {``      ``suffix[i] = gcd(suffix[i + 1],``                      ``arr[i]);``    ``}` `    ``// Traverse queries array``    ``for` `(``int` `i = 0; i < Q; i++) {` `      ``int` `a = Queries[i,0];``      ``int` `K = Queries[i,1];``      ``int` `X = Queries[i,2];` `      ``// Edge Case when update is``      ``// is required till the end``      ``if` `(K == N) {``        ``Console.Write(prefix[N - 1] * X);``        ``continue``;``      ``}` `      ``// Edge Case when update is``      ``// is required till the front``      ``if` `(a == 1) {``        ``Console.Write(gcd(prefix[K - 1] * X,``                          ``suffix[K]));``      ``}` `      ``// Find the resultant operation``      ``// for each query``      ``else` `{``        ``Console.Write(gcd(suffix[N - K] * X,``                          ``prefix[N - K - 1]));``      ``}` `      ``Console.Write(``" "``);``    ``}``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{` `    ``int` `[]arr = { 2, 3, 4, 8 };``    ``int` `N = arr.Length;``    ``int` `[,]Queries = {``      ``{ 1, 2, 2 },``      ``{ 2, 4, 5 }``    ``};``    ``int` `Q = Queries.GetLength(0);` `    ``findGCDQueries(arr, N, Queries, Q);``  ``}``}` `// This code is contributed by AnkThon.`

## Javascript

 ``

Output

`2 5`

Time Complexity: O((N + Q)* log M), where M is the maximum element of the array.
Auxiliary Space: O(N)

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