Find GCD of Array after subtracting given value for M queries
Last Updated :
19 Jan, 2023
Given an array, arr[] of size N, Given M queries and each query consisting of a number X, the task is to subtract X from every element of arr[] for each query and print the Greatest Common Divisor of all elements of arr[].
Examples:
Input: arr[] = {9, 13, 17, 25}, Q[] = {1, 3, 5, 9}
Output: 4 2 4 4
Explanation: First Query: GCD(9 - 1, 13 -1, 17 - 1, 25 - 1) = GCD(8, 12, 16, 24) = 4
Second Query: GCD(9 - 3, 13 - 3, 17 - 3, 25 - 3) = GCD(6, 10, 14, 22) = 2
Third Query: GCD(9 - 5, 13 - 5, 17 - 5, 25 - 5) = GCD(4, 8, 12, 20) = 4
Fourth Query: GCD(9 - 9, 13 - 9, 17 - 9, 25 - 9) = GCD(0, 4, 8,, 16) = 4
Input: arr[] = {1 25 121 169}, Q[] = {1 2 7 23}
Output: 24 1 6 2
Naive approach: The basic way to solve the problem is as follows:
For each query Iterate through every element of arr[] and keep track of GCD.
Time Complexity: O(N * M * logD) where D is the maximum value of the array
Auxiliary Space: O(1)
Efficient Approach: The problem can be solved based on the following idea:
Property of Euclidean Algorithm for finding GCD can be used which is GCD(a, b) = GCD(a, b - a). For multiple numbers idea can be generalized as GCD(a, b, c, …) = GCD(a, b - a, c - a, …).
Follow the steps below to solve the problem:
- To calculate GCD(arr[0] - X, arr[1] - X, arr[2] - X, . . ., arr[N - 1] - X), subtract arr[0] - X from other Numbers.
- Then, GCD (arr[0] - X, arr[1] - X, arr[2] - X, . . ., arr[N - 1] - X) = GCD(arr[0] - X, arr[1] - arr[0], arr[2] - arr[0], . . ., arr[N - 1] - arr[0]).
- Find T = GCD( arr[1] - arr[0], arr[2] - arr[0], . . ., arr[N - 1] - arr[0]), then gcd for every query can be calculated to be GCD(arr[0] - X, T).
- For each query print the absolute of GCD(X, T) (print absolute since the answer can be negative after subtraction).
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to Calculate GCD for each query
vector<int> findGCDBySubtractingX(int arr[], int Q[],
int N, int M)
{
int T = 0;
vector<int> res;
// Find GCD of arr[1] - arr[0], arr[2] - arr[0],
// . . ., arr[N - 1] - arr[0]
for (int i = 1; i < N; i++) {
T = __gcd(T, arr[i] - arr[0]);
}
// Iterating for each of M Queries
for (int j = 0; j < M; j++) {
int X = Q[j];
// Finding GCD for each query with
// their absolute since subtraction
// can be negative
res.push_back(abs(__gcd(T, arr[0] - X)));
}
return res;
}
// Driver Code
int main()
{
// Input
int arr[] = { 9, 13, 17, 25 }, Q[] = { 1, 3, 5, 9 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(Q) / sizeof(Q[0]);
// Function Call
vector<int> ans = findGCDBySubtractingX(arr, Q, N, M);
for (int x : ans)
cout << x << " ";
return 0;
}
// Java code to implement the approach
import java.io.*;
import java.util.*;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to Calculate GCD for each query
public static List<Integer>
findGCDBySubtractingX(int[] arr, int[] Q, int N, int M)
{
int T = 0;
List<Integer> res = new ArrayList<>();
// Find GCD of arr[1] - arr[0], arr[2] - arr[0],
// . . ., arr[N - 1] - arr[0]
for (int i = 1; i < N; i++) {
T = gcd(T, arr[i] - arr[0]);
}
// Iterating for each of M Queries
for (int j = 0; j < M; j++) {
int X = Q[j];
// Finding GCD for each query with
// their absolute since subtraction
// can be negative
res.add(Math.abs(gcd(T, arr[0] - X)));
}
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input
int[] arr = { 9, 13, 17, 25 };
int[] Q = { 1, 3, 5, 9 };
int N = arr.length;
int M = Q.length;
// Function Call
List<Integer> ans
= findGCDBySubtractingX(arr, Q, N, M);
for (int x : ans)
System.out.print(x + " ");
}
}
// This code is contributed by lokeshmvs21.
import math
def findGCDBySubtractingX(arr, q, n, m):
t = 0
res = []
# find the gcd of arr[1]-arr[0]
# .... arr[n-1]-arr[0]
for i in range(1, n):
t = math.gcd(t, arr[i]-arr[0])
# Iterating for each of m queries
for j in range(m):
x = q[j]
# finding the gcd for each query with
# their absolute since subtraction
# can be negative
res.append(abs(math.gcd(t, arr[0]-x)))
return res
arr = [9, 13, 17, 25]
q = [1, 3, 5, 9]
n = len(arr)
m = len(q)
ans = findGCDBySubtractingX(arr, q, n, m)
for x in ans:
print(x, end=" ")
// C# code to implement the approach
using System;
using System.Collections.Generic;
public class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to Calculate GCD for each query
public static List<int>
findGCDBySubtractingX(int[] arr, int[] Q, int N, int M)
{
int T = 0;
List<int> res = new List<int>();
// Find GCD of arr[1] - arr[0], arr[2] - arr[0],
// . . ., arr[N - 1] - arr[0]
for (int i = 1; i < N; i++) {
T = gcd(T, arr[i] - arr[0]);
}
// Iterating for each of M Queries
for (int j = 0; j < M; j++) {
int X = Q[j];
// Finding GCD for each query with
// their absolute since subtraction
// can be negative
res.Add(Math.Abs(gcd(T, arr[0] - X)));
}
return res;
}
static public void Main()
{
// Input
int[] arr = { 9, 13, 17, 25 };
int[] Q = { 1, 3, 5, 9 };
int N = arr.Length;
int M = Q.Length;
// Function Call
List<int> ans = findGCDBySubtractingX(arr, Q, N, M);
foreach(int x in ans) Console.Write(x + " ");
}
}
// This code is contributed by lokesh.
// JavaScript code to implement the approach
// function to find gcd
function __gcd(a, b)
{
if(b==0)
return a;
else
return __gcd(b, a%b);
}
// Function to Calculate GCD for each query
function findGCDBySubtractingX(arr, Q, N, M)
{
let T = 0;
let res=[];
// Find GCD of arr[1] - arr[0], arr[2] - arr[0],
// . . ., arr[N - 1] - arr[0]
for (let i = 1; i < N; i++) {
T = __gcd(T, arr[i] - arr[0]);
}
// Iterating for each of M Queries
for (let j = 0; j < M; j++) {
let X = Q[j];
// Finding GCD for each query with
// their absolute since subtraction
// can be negative
res.push(Math.abs(__gcd(T, arr[0] - X)));
}
return res;
}
// Driver Code
// Input
let arr = [ 9, 13, 17, 25 ], Q = [ 1, 3, 5, 9 ];
let N = arr.length;
let M = Q.length;
// Function Call
let ans = findGCDBySubtractingX(arr, Q, N, M);
for (let x of ans)
console.log(x + " ");
// This code is contributed by poojaagarwal2.
Time Complexity: O(N * logD + M * logD) where D is the maximum element in the array
Auxiliary Space: O(1)
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