Check if all array elements can be reduced to less than X
Given an array A[] consisting of N positive integers and an integer X, the task is to determine if it is possible to convert all array elements to less than X by performing the following operations:
- Select 2 distinct indices j and k.
- Select an index i, where A[i] > X.
- Replace A[i] = gcd(A[j], A[k]) if and only if gcd(A[j], A[k]) ≠ 1.
Examples:
Input: A[] = {2, 1, 5, 3, 6}, X = 4
Output: Yes
Explanation:
- Selecting i = 3, j = 4, k = 5, set A[i] = gcd(A[j], A[k]) = 3. Therefore, A[] modifies to {2, 1, 3, 3, 6}.
- Selecting i = 5, j = 4, k = 5, set A[i] = gcd(A[j], A[k]) = 3. Therefore, A[] modifies to {2, 1, 3, 3, 3}.
Input: A[] = {2, 3, 2, 5, 4}, X = 3
Output: Yes
Approach: Follow the steps below to solve the problem:
- Find the two numbers having gcd ≠ 1 as well as gcd ≤ X, then, by using these two numbers, the required number A[i] can be replaced with gcd(A[j], A[k]).
- Using the fact that gcd(x, y) ≤ min(x, y), the array elements can be reduced to ≤ X.
- This way, the rest of the array can be converted to ≤ X using step 2.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if all array // elements are≤ Xbool check(int A[], int X, int N){ for(int i = 0; i < N; i++) { if (A[i] > X) { return false; } } return true;}// Function to check if all array elements// can be reduced to less than X or notbool findAns(int A[], int N, int X){ // Checks if all array elements // are already ≤ X or not if (check(A, X, N)) { return true; } // Traverse every possible pair for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { // Calculate GCD of two // array elements int g = __gcd(A[i], A[j]); // If gcd is ≠ 1 if (g != 1) { // If gcd is ≤ X, then a pair // is present to reduce all // array elements to ≤ X if (g <= X) { return true; } } } } // If no pair is present // with gcd is ≤ X return false;}// Driver Codeint main(){ int X = 4; int A[] = { 2, 1, 5, 3, 6 }; int N = 5; if (findAns(A, N, X)) { cout << "true"; } else { cout << "false"; }}// This code is contributed by mohit kumar 29 |
chevron_right
filter_none
Java
// Java Program to implement// the above approachimport java.io.*;import java.util.Arrays;class GFG { // Function to check if all array elements // can be reduced to less than X or not public static boolean findAns( int[] A, int N, int X) { // Checks if all array elements // are already ≤ X or not if (check(A, X)) { return true; } // Traverse every possible pair for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Calculate GCD of two // array elements int gcd = gcd(A[i], A[j]); // If gcd is ≠ 1 if (gcd != 1) { // If gcd is ≤ X, then a pair // is present to reduce all // array elements to ≤ X if (gcd <= X) { return true; } } } } // If no pair is present // with gcd is ≤ X return false; } // Function to check if all array elements are≤ X public static boolean check(int[] A, int X) { for (int i = 0; i < A.length; i++) { if (A[i] > X) { return false; } } return true; } // Function to calculate gcd of two numbers public static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } // Driver Code public static void main(String[] args) { int X = 4; int[] A = { 2, 1, 5, 3, 6 }; int N = 5; System.out.println(findAns(A, N, X)); }} |
chevron_right
filter_none
Python3
# Python3 program to implement# the above approach# Function to check if all array elements# can be reduced to less than X or notdef findAns(A, N, X): # Checks if all array elements # are already ≤ X or not if (check(A, X)): return True # Traverse every possible pair for i in range(N): for j in range(i + 1, N): # Calculate GCD of two # array elements gcd = GCD(A[i], A[j]) # If gcd is ≠ 1 if (gcd != 1): # If gcd is ≤ X, then a pair # is present to reduce all # array elements to ≤ X if (gcd <= X): return True # If no pair is present # with gcd is ≤ X return False# Function to check if all array elements are≤ Xdef check(A, X): for i in range(len(A)): if (A[i] > X): return False return True# Function to calculate gcd of two numbersdef GCD(a, b): if (b == 0): return a return GCD(b, a % b)# Driver CodeX = 4A = [ 2, 1, 5, 3, 6 ]N = 5print(findAns(A, N, X))# This code is contributed by rohitsingh07052 |
chevron_right
filter_none
C#
// C# Program to implement// the above approachusing System;class GFG { // Function to check if all array elements // can be reduced to less than X or not public static bool findAns( int[] A, int N, int X) { // Checks if all array elements // are already ≤ X or not if (check(A, X)) { return true; } // Traverse every possible pair for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Calculate GCD of two // array elements int gcd = gcdFoo(A[i], A[j]); // If gcd is ≠ 1 if (gcd != 1) { // If gcd is ≤ X, then a pair // is present to reduce all // array elements to ≤ X if (gcd <= X) { return true; } } } } // If no pair is present // with gcd is ≤ X return false; } // Function to check if all array elements are≤ X public static bool check(int[] A, int X) { for (int i = 0; i < A.Length; i++) { if (A[i] > X) { return false; } } return true; } // Function to calculate gcd of two numbers static int gcdFoo(int a, int b) { if (b == 0) return a; return gcdFoo(b, a % b); } // Driver Code public static void Main(String[] args) { int X = 4; int[] A = { 2, 1, 5, 3, 6 }; int N = 5; Console.WriteLine(findAns(A, N, X)); }}// This code is contributed by 29AjayKumar |
chevron_right
filter_none
Output
true
Time Complexity: O(N2)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
