Given an array arr[] of length N ? 2. The task is to remove an element from the given array such that the LCM of the array after removing it is minimized.
Examples:
Input: arr[] = {18, 12, 24}
Output: 24
Remove 12: LCM(18, 24) = 72
Remove 18: LCM(12, 24) = 24
Remove 24: LCM(12, 18) = 36
Input: arr[] = {5, 15, 9, 36}
Output: 45
Approach:
- Idea is to find the LCM value of all the sub-sequences of length (N – 1) and removing the element which is not present in the sub-sequence with that LCM. The minimum LCM found would be the answer.
- To find the LCM of the sub-sequences optimally, maintain a prefixLCM[] and a suffixLCM[] array using single state dynamic programming.
- The minimum value of LCM(prefixLCM[i – 1], suffixLCM[i + 1]) is the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the LCM of two numbersint lcm(int a, int b){ int GCD = __gcd(a, b); return (a * b) / GCD;}// Function to return the minimum LCM// after removing a single element// from the given arrayint MinLCM(int a[], int n){ // Prefix and Suffix arrays int Prefix[n + 2]; int Suffix[n + 2]; // Single state dynamic programming relation // for storing LCM of first i elements // from the left in Prefix[i] Prefix[1] = a[0]; for (int i = 2; i <= n; i += 1) { Prefix[i] = lcm(Prefix[i - 1], a[i - 1]); } // Initializing Suffix array Suffix[n] = a[n - 1]; // Single state dynamic programming relation // for storing LCM of all the elements having // index greater than or equal to i in Suffix[i] for (int i = n - 1; i >= 1; i -= 1) { Suffix[i] = lcm(Suffix[i + 1], a[i - 1]); } // If first or last element of // the array has to be removed int ans = min(Suffix[2], Prefix[n - 1]); // If any other element is replaced for (int i = 2; i < n; i += 1) { ans = min(ans, lcm(Prefix[i - 1], Suffix[i + 1])); } // Return the minimum LCM return ans;}// Driver codeint main(){ int a[] = { 5, 15, 9, 36 }; int n = sizeof(a) / sizeof(a[0]); cout << MinLCM(a, n); return 0;} |
Java
// Java implementation of the above approachclass GFG{// Function to return the LCM of two numbersstatic int lcm(int a, int b){ int GCD = __gcd(a, b); return (a * b) / GCD;}// Function to return the minimum LCM// after removing a single element// from the given arraystatic int MinLCM(int a[], int n){ // Prefix and Suffix arrays int []Prefix = new int[n + 2]; int []Suffix = new int[n + 2]; // Single state dynamic programming relation // for storing LCM of first i elements // from the left in Prefix[i] Prefix[1] = a[0]; for (int i = 2; i <= n; i += 1) { Prefix[i] = lcm(Prefix[i - 1], a[i - 1]); } // Initializing Suffix array Suffix[n] = a[n - 1]; // Single state dynamic programming relation // for storing LCM of all the elements having // index greater than or equal to i in Suffix[i] for (int i = n - 1; i >= 1; i -= 1) { Suffix[i] = lcm(Suffix[i + 1], a[i - 1]); } // If first or last element of // the array has to be removed int ans = Math.min(Suffix[2], Prefix[n - 1]); // If any other element is replaced for (int i = 2; i < n; i += 1) { ans = Math.min(ans, lcm(Prefix[i - 1], Suffix[i + 1])); } // Return the minimum LCM return ans;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver codepublic static void main(String []args){ int a[] = { 5, 15, 9, 36 }; int n = a.length; System.out.println(MinLCM(a, n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of # the above approach from math import gcd# Function to return the LCM # of two numbers def lcm(a, b) : GCD = gcd(a, b); return (a * b) // GCD; # Function to return the minimum LCM # after removing a single element # from the given array def MinLCM(a, n) : # Prefix and Suffix arrays Prefix = [0] * (n + 2); Suffix = [0] * (n + 2); # Single state dynamic programming relation # for storing LCM of first i elements # from the left in Prefix[i] Prefix[1] = a[0]; for i in range(2, n + 1) : Prefix[i] = lcm(Prefix[i - 1], a[i - 1]); # Initializing Suffix array Suffix[n] = a[n - 1]; # Single state dynamic programming relation # for storing LCM of all the elements having # index greater than or equal to i in Suffix[i] for i in range(n - 1, 0, -1) : Suffix[i] = lcm(Suffix[i + 1], a[i - 1]); # If first or last element of # the array has to be removed ans = min(Suffix[2], Prefix[n - 1]); # If any other element is replaced for i in range(2, n) : ans = min(ans, lcm(Prefix[i - 1], Suffix[i + 1])); # Return the minimum LCM return ans; # Driver code if __name__ == "__main__" : a = [ 5, 15, 9, 36 ]; n = len(a); print(MinLCM(a, n));# This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approachusing System; class GFG{// Function to return the LCM of two numbersstatic int lcm(int a, int b){ int GCD = __gcd(a, b); return (a * b) / GCD;}// Function to return the minimum LCM// after removing a single element// from the given arraystatic int MinLCM(int []a, int n){ // Prefix and Suffix arrays int []Prefix = new int[n + 2]; int []Suffix = new int[n + 2]; // Single state dynamic programming relation // for storing LCM of first i elements // from the left in Prefix[i] Prefix[1] = a[0]; for (int i = 2; i <= n; i += 1) { Prefix[i] = lcm(Prefix[i - 1], a[i - 1]); } // Initializing Suffix array Suffix[n] = a[n - 1]; // Single state dynamic programming relation // for storing LCM of all the elements having // index greater than or equal to i in Suffix[i] for (int i = n - 1; i >= 1; i -= 1) { Suffix[i] = lcm(Suffix[i + 1], a[i - 1]); } // If first or last element of // the array has to be removed int ans = Math.Min(Suffix[2], Prefix[n - 1]); // If any other element is replaced for (int i = 2; i < n; i += 1) { ans = Math.Min(ans, lcm(Prefix[i - 1], Suffix[i + 1])); } // Return the minimum LCM return ans;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver codepublic static void Main(String []args){ int []a = { 5, 15, 9, 36 }; int n = a.Length; Console.WriteLine(MinLCM(a, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation of the above approach// Function to return the LCM of two numbersfunction lcm(a, b) { let GCD = __gcd(a, b); return Math.floor((a * b) / GCD);}function __gcd(a, b) { return b == 0 ? a : __gcd(b, a % b); }// Function to return the minimum LCM// after removing a single element// from the given arrayfunction MinLCM(a, n) { // Prefix and Suffix arrays let Prefix = new Array(n + 2); let Suffix = new Array(n + 2); // Single state dynamic programming relation // for storing LCM of first i elements // from the left in Prefix[i] Prefix[1] = a[0]; for (let i = 2; i <= n; i += 1) { Prefix[i] = lcm(Prefix[i - 1], a[i - 1]); } // Initializing Suffix array Suffix[n] = a[n - 1]; // Single state dynamic programming relation // for storing LCM of all the elements having // index greater than or equal to i in Suffix[i] for (let i = n - 1; i >= 1; i -= 1) { Suffix[i] = lcm(Suffix[i + 1], a[i - 1]); } // If first or last element of // the array has to be removed let ans = Math.min(Suffix[2], Prefix[n - 1]); // If any other element is replaced for (let i = 2; i < n; i += 1) { ans = Math.min(ans, lcm(Prefix[i - 1], Suffix[i + 1])); } // Return the minimum LCM return ans;}// Driver codelet a = [5, 15, 9, 36];let n = a.length;document.write(MinLCM(a, n));</script> |
Output:
45
Time Complexity: O(N * log(M)) where M is the maximum element from the array.
Auxiliary Space: O(N)
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