Rearrange array to obtain maximum possible value of concatenation of prefix GCDs - GeeksforGeeks

28 Feb, 2022
Given an array arr[] consisting of N positive integers, the task is to rearrange the array elements such that the number formed by concatenating the GCD of elements of the array arr[] from index 0 to i for each index i is the maximum possible.
Examples:
Input: arr[] = {4, 2, 5}Output: 5 4 2Explanation:X = 511 is the maximum value of X that can be obtained among all the rearrangement of arr[].Possible arrangements of arr[] are:arr[] = [2, 4, 5] → X = 221arr[] = [2, 5, 4] → X = 211arr[] = [4, 2, 5] → X = 421arr[] = [4, 5, 2] → X = 411arr[] = [5, 4, 2] → X = 511arr[] = [5, 2, 4] → X = 511
Input: arr[] = {2, 4, 6, 8}Output: 8 4 6 2Explanation: X = 842 is the maximum value of X that can be obtained among all the rearrangement of arr[].Possible arrangements of arr[] are:arr[] = [4, 6, 8] → X = 422arr[] = [4, 8, 6] → X = 442arr[] = [6, 4, 8] → X = 622arr[] = [6, 8, 4] → X = 622arr[] = [8, 4, 6] → X = 842arr[] = [8, 6, 4] → X = 822
Approach: The GCD of a number alone is the number itself, thus the first digit of X i.e., X[0] would always be equal to arr[0]. Thus, to ensure that X is maximum among all obtainable numbers, arr[0] needs to be maximum. Then proceed by keeping track of the GCD of the longest prefix of arr[] that has been already arranged and find the values of the consecutive elements to be placed after this prefix. Follow the steps below to solve the above problem:
The largest element of the array is set as the first element, thus the first prefix correctly arranged in the array arr[].Now find the element consecutive to the last element of the prefix i.e., arr[1].Here the GCD of the longest prefix(say G) is equal to arr[0], thus traverse the remaining array to find the element that gives the greatest GCD with G.Now, swap the element arr[1] with the element that gives maximum GCD with value G, update the value of G to this maximum GCD obtained i.e., G = GCD(G, arr[1]).Now the longest fixed prefix becomes arr[0], arr[1], continue this process for finding arr[2], arr[3], ..., arr[N – 1], to obtain the required array.Print rearrange array after the above steps.
The largest element of the array is set as the first element, thus the first prefix correctly arranged in the array arr[].
Now find the element consecutive to the last element of the prefix i.e., arr[1].
Here the GCD of the longest prefix(say G) is equal to arr[0], thus traverse the remaining array to find the element that gives the greatest GCD with G.
Now, swap the element arr[1] with the element that gives maximum GCD with value G, update the value of G to this maximum GCD obtained i.e., G = GCD(G, arr[1]).
Now the longest fixed prefix becomes arr[0], arr[1], continue this process for finding arr[2], arr[3], ..., arr[N – 1], to obtain the required array.
Print rearrange array after the above steps.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Function to find the maximum number// obtainable from prefix GCDsvoid prefixGCD(int arr[], int N){    // Stores the GCD of the    // longest prefix    int gcc;     // Sort the array    sort(arr, arr + N);     // Reverse the array    reverse(arr, arr + N);     // GCD of a[0] is a[0]    gcc = arr[0];    int start = 0;     // Iterate to place the arr[start + 1]    // element at it's correct position    while (start < N - 1) {         int g = 0, s1;         for (int i = start + 1; i < N; i++) {             // Find the element with            // maximum GCD            int gc = __gcd(gcc, arr[i]);             // Update the value of g            if (gc > g) {                g = gc;                s1 = i;            }        }         // Update GCD of prefix        gcc = g;         // Place arr[s1] to it's        // correct position        swap(arr[s1], arr[start + 1]);         // Increment start for the        // remaining elements        start++;    }     // Print the rearranged array    for (int i = 0; i < N; i++) {        cout << arr[i] << " ";    }} // Driver Codeint main(){    // Given array arr[]    int arr[] = { 1, 2, 3, 4 };     int N = sizeof(arr) / sizeof(arr[0]);     // Function Call    prefixGCD(arr, N);     return 0;}
//Java program for// the above approachimport java.util.*;class GFG{ //Function to find the maximum number//obtainable from prefix GCDsstatic void prefixGCD(int arr[], int N){  // Stores the GCD of the  // longest prefix  int gcc;   // Sort the array  Arrays.sort(arr);   // Reverse the array  arr = reverse(arr);   // GCD of a[0] is a[0]  gcc = arr[0];  int start = 0;   // Iterate to place  // the arr[start + 1]  // element at it's  // correct position  while (start < N - 1)  {    int g = 0, s1 = 0;     for (int i = start + 1; i < N; i++)    {      // Find the element with      // maximum GCD      int gc = __gcd(gcc, arr[i]);       // Update the value of g      if (gc > g)      {        g = gc;        s1 = i;      }    }     // Update GCD of prefix    gcc = g;     // Place arr[s1] to it's    // correct position    arr = swap(arr, s1, start + 1);     // Increment start for the    // remaining elements    start++;  }   // Print the rearranged array  for (int i = 0; i < N; i++)  {    System.out.print(arr[i] + " ");  }}   static int __gcd(int a, int b) {   return b == 0 ? a : __gcd(b, a % b);    } static int[] reverse(int a[]){  int i, n = a.length, t;  for (i = 0; i < n / 2; i++)  {    t = a[i];    a[i] = a[n - i - 1];    a[n - i - 1] = t;  }  return a;} static int[] swap(int []arr,                  int i, int j){  int temp = arr[i];  arr[i] = arr[j];  arr[j] = temp;  return arr;}     //Driver Codepublic static void main(String[] args){  // Given array arr[]  int arr[] = {1, 2, 3, 4};   int N = arr.length;   // Function Call  prefixGCD(arr, N);}} //This code is contributed by 29AjayKumar
# Python3 program for the above approachfrom math import gcd # Function to find the maximum number# obtainable from prefix GCDsdef prefixGCD(arr, N):         # Stores the GCD of the    # longest prefix    gcc = 0     # Sort the array    arr = sorted(arr)     # Reverse the array    arr = arr[::-1]     # GCD of a[0] is a[0]    gcc = arr[0]    start = 0     # Iterate to place the arr[start + 1]    # element at it's correct position    while (start < N - 1):        g = 0        s1 = 0         for i in range(start + 1, N):             # Find the element with            # maximum GCD            gc = gcd(gcc, arr[i])             # Update the value of g            if (gc > g):                g = gc                s1 = i         # Update GCD of prefix        gcc = g         # Place arr[s1] to it's        # correct position        arr[s1], arr[start + 1] = arr[start + 1], arr[s1]         # Increment start for the        # remaining elements        start += 1     # Print the rearranged array    for i in range(N):        print(arr[i], end = " ") # Driver Codeif __name__ == '__main__':         # Given array arr[]    arr = [ 1, 2, 3, 4 ]     N = len(arr)     # Function Call    prefixGCD(arr, N) # This code is contributed by mohit kumar 29
// C# program for the above approach using System;class GFG{ // Function to find the maximum number// obtainable from prefix GCDsstatic void prefixGCD(int[] arr, int N){       // Stores the GCD of the  // longest prefix  int gcc;   // Sort the array  Array.Sort(arr);   // Reverse the array  arr = reverse(arr);   // GCD of a[0] is a[0]  gcc = arr[0];  int start = 0;   // Iterate to place the  // arr[start + 1] element  // at it's correct position  while (start < N - 1)  {    int g = 0, s1 = 0;     for(int i = start + 1; i < N; i++)    {               // Find the element with      // maximum GCD      int gc = __gcd(gcc, arr[i]);       // Update the value of g      if (gc > g)      {        g = gc;        s1 = i;      }    }     // Update GCD of prefix    gcc = g;     // Place arr[s1] to it's    // correct position    arr = swap(arr, s1, start + 1);     // Increment start for the    // remaining elements    start++;  }   // Print the rearranged array  for(int i = 0; i < N; i++)  {    Console.Write(arr[i] + " ");  }}   static int __gcd(int a, int b) {   return b == 0 ? a : __gcd(b, a % b);    } static int[] reverse(int[] a){  int i, n = a.Length, t;     for(i = 0; i < n / 2; i++)  {    t = a[i];    a[i] = a[n - i - 1];    a[n - i - 1] = t;  }  return a;} static int[] swap(int []arr, int i,                             int j){  int temp = arr[i];  arr[i] = arr[j];  arr[j] = temp;  return arr;}     //Driver Codepublic static void Main(){       // Given array arr[]  int[] arr = { 1, 2, 3, 4 };   int N = arr.Length;   // Function call  prefixGCD(arr, N);}} // This code is contributed by sanjoy_62
<script> // Javascript program to implement// the above approach //Function to find the maximum number//obtainable from prefix GCDsfunction prefixGCD(arr, N){  // Stores the GCD of the  // longest prefix  let gcc;   // Sort the array  arr.sort();   // Reverse the array  arr = reverse(arr);   // GCD of a[0] is a[0]  gcc = arr[0];  let start = 0;   // Iterate to place  // the arr[start + 1]  // element at it's  // correct position  while (start < N - 1)  {    let g = 0, s1 = 0;     for (let i = start + 1; i < N; i++)    {      // Find the element with      // maximum GCD      let gc = __gcd(gcc, arr[i]);       // Update the value of g      if (gc > g)      {        g = gc;        s1 = i;      }    }     // Update GCD of prefix    gcc = g;     // Place arr[s1] to it's    // correct position    arr = swap(arr, s1, start + 1);     // Increment start for the    // remaining elements    start++;  }   // Print the rearranged array  for (let i = 0; i < N; i++)  {    document.write(arr[i] + " ");  }}   function __gcd(a, b) {   return b == 0 ? a : __gcd(b, a % b);    } function reverse(a){  let i, n = a.length, t;  for (i = 0; i < n / 2; i++)  {    t = a[i];    a[i] = a[n - i - 1];    a[n - i - 1] = t;  }  return a;} function swap(arr, i, j){  let temp = arr[i];  arr[i] = arr[j];  arr[j] = temp;  return arr;} // Driver Code     // Given array arr[]  let arr = [1, 2, 3, 4];   let N = arr.length;   // Function Call  prefixGCD(arr, N); </script>
4 2 3 1
 
Time Complexity: O(N2)Auxiliary Space: O(1)
mohit kumar 29
29AjayKumar
sanjoy_62
target_2
simranarora5sos
array-rearrange
HCF
Arrays
Mathematical
Arrays
Mathematical
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