Given an array arr[] and an integer K, the task is to merge K minimum elements of the array until there is only one element left in the array.
Note: If it is impossible to merge into only one element then print -1.
Input: arr[] = {3, 2, 4, 1}, K = 2
Output: 10
Explanation:
Merge K minimum elements of the Array({3, 2, 4, 1}) = 1 + 2 = 3
After Merging the Array will be {3, 3, 4}
Merge K minimum elements of the Array ({3, 3, 4}) = 3 + 3 = 6
After Merging the Array will be {4, 6}
Merge K minimum elements of the Array ({4, 6}) = 4 + 6 = 10
Input: arr[] = {3, 2, 4, 1}, K = 3
Output: -1
Explanation:
After merging there will be two elements left {6, 4} which cannot be merged further.
Approach: The idea is to sort the array and then merge the first K minimum elements of the array into one element, then insert the element into the array at its sorted position into the array with the help of the Binary Search. Similarly, repeat this step until there is only one element left in the array. If in the end there are less than K elements left then return -1.
Below is the implementation of the above approach:
C++
// C++ implementation to merge the K minimum elements of the // array until there is only one element is left in the // array #include <bits/stdc++.h> using namespace std; class GFG { public : // Utility function to insert the element at its correct // position in a sorted array (Binary search is used to // get the index of the correct position) void insertInSortedList(vector< int >& sortedList, int item, int start, int end) { int mid = start + (end - start) / 2; if (mid == 0 || mid == end - 1 || sortedList[mid] == item) { sortedList.insert(sortedList.begin() + mid, item); return ; } else if (sortedList[mid] < item) { insertInSortedList(sortedList, item, mid + 1, end); } else { insertInSortedList(sortedList, item, start, mid - 1); } } // Function to merge the element of the array until // there is only one element left in array int mergeStones(vector< int >& List, int k) { // Sort the array sort(List.begin(), List.end()); int cost = 0; // Loop to merge the elements until there are // greater than K elements in the array while (List.size() > k) { int Sum = 0; // Obtain the sum of K minimum elements for ( int i = 0; i < k; i++) { Sum += List[i]; } // Remove the K minimum elements from the array for ( int i = 0; i < k; i++) { List.erase(List.begin()); } // Insert the merged element into the sorted // array at its correct position insertInSortedList(List, Sum, 0, List.size()); cost += Sum; } // If there are only K elements left, // take the sum and return the sum, else return -1 if (List.size() == k) { int temp = 0; for ( int i = 0; i < k; i++) { temp += List[i]; } cost += temp; return cost; } else { return -1; } } }; // Driver code int main() { vector< int > stones; stones.push_back(3); stones.push_back(2); stones.push_back(4); stones.push_back(1); cout << GFG().mergeStones(stones, 3) << "\n" ; cout << GFG().mergeStones(stones, 2) << "\n" ; return 0; } // This code is contributed by akashish__ |
Java
// Java implementation to merge the // K minimum elements of the Array // until there is only one element // is left in the array // Imports import java.io.*; import java.lang.*; import java.util.*; class GFG { // Function to merge the element // of the array until there is // only one element left in array public static int mergeStones( List<Integer> list, final int k) { // Sorting the array Collections.sort(list); int cost = 0 ; // Loop to merge the elements // until there is element // greater than K elements while (list.size() > k) { int sum = 0 ; // Merging the K minimum // elements of the array for ( int i = 0 ; i < k; i++) { sum += list.get(i); } // Removing the K minimum // elements of the array list.subList( 0 , k).clear(); // Inserting the merged // element into the array insertInSortedList(list, sum); cost += sum; } // If there is only K element // left then return the element if (list.size() == k) { cost += list.stream().reduce( 0 , Integer::sum); return cost; } else { return - 1 ; } } // Function insert the element into // the sorted position in the array public static void insertInSortedList( List<Integer> sortedList, int item) { int len = sortedList.size(); insertInSortedList(sortedList, item, 0 , len - 1 ); } // Utility function to insert into the // array with the help of the position public static void insertInSortedList( List<Integer> sortedList, int item, int start, int end) { int mid = ( int ) ((end - start)/ 2.00 ); if (mid == 0 || (mid == sortedList.size() - 1 ) || sortedList.get(mid) == item) { sortedList.add(mid, item); return ; } else if (sortedList.get(mid) < item) { insertInSortedList(sortedList, item, mid + 1 , end); } else { insertInSortedList(sortedList, item, start, mid - 1 ); } } // Driver Code public static void main(String [] args) { List<Integer> stones = new ArrayList<>(); stones.add( 3 ); stones.add( 2 ); stones.add( 4 ); stones.add( 1 ); System.out.println(mergeStones(stones, 3 )); System.out.println(mergeStones(stones, 2 )); } } |
Python3
# Python implementation to merge the K minimum elements of the array # until there is only one element is left in the array class GFG: # Utility function to insert the element at its correct position in a sorted array # (Binary search is used to get the index of the correct position) def insertInSortedList( self , sortedList, item, start, end): mid = start + (end - start) / / 2 if mid = = 0 or mid = = end - 1 or sortedList[mid] = = item: sortedList.insert(mid, item) return elif sortedList[mid] < item: self .insertInSortedList(sortedList, item, mid + 1 , end) else : self .insertInSortedList(sortedList, item, start, mid - 1 ) # Function to merge the element of the array until there is # only one element left in array def mergeStones( self , List , k): # Sort the array List .sort() cost = 0 # Loop to merge the elements until there are # greater than K elements in the array while len ( List ) > k: Sum = 0 # Obtain the sum of K minimum elements for i in range (k): Sum + = List [i] # Remove the K minimum elements from the array for _ in range (k): List .pop( 0 ) # Insert the merged element into the sorted array # at its correct position self .insertInSortedList( List , Sum , 0 , len ( List )) cost + = Sum # If there are only K elements left, # take the sum and return the sum, else return -1 if len ( List ) = = k: cost + = sum ( List [:]) return cost else : return - 1 # Driver code if __name__ = = '__main__' : stones = [] stones.append( 3 ) stones.append( 2 ) stones.append( 4 ) stones.append( 1 ) print (GFG().mergeStones(stones, 3 )) print (GFG().mergeStones(stones, 2 )) # This code is contributed by keshavrathi |
C#
// Include namespace system using System; using System.Collections.Generic; using System.Linq; using System.Collections; public class GFG { // Function to merge the element // of the array until there is // only one element left in array public static int mergeStones(List< int > list, int k) { // Sorting the array list.Sort(); var cost = 0; // Loop to merge the elements // until there is element // greater than K elements while (list.Count > k) { var sum = 0; // Merging the K minimum // elements of the array for ( int i = 0; i < k; i++) { sum += list[i]; } // Removing the K minimum // elements of the array for ( int t = 0; t < k; t++){ list.RemoveAt(0); } // Inserting the merged // element into the array GFG.insertInSortedList(list, sum); cost += sum; } // If there is only K element // left then return the element if (list.Count == k) { cost += list.AsQueryable().Sum(); return cost; } else { return -1; } } // Function insert the element into // the sorted position in the array public static void insertInSortedList(List< int > sortedList, int item) { var len = sortedList.Count; GFG.insertInSortedList(sortedList, item, 0, len - 1); } // Utility function to insert into the // array with the help of the position public static void insertInSortedList(List< int > sortedList, int item, int start, int end) { var mid = ( int )((end - start) / 2.0); if (mid == 0 || (mid == sortedList.Count - 1) || sortedList[mid] == item) { sortedList.Insert(mid,item); return ; } else if (sortedList[mid] < item) { GFG.insertInSortedList(sortedList, item, mid + 1, end); } else { GFG.insertInSortedList(sortedList, item, start, mid - 1); } } // Driver Code public static void Main(String[] args) { var stones = new List< int >(); stones.Add(3); stones.Add(2); stones.Add(4); stones.Add(1); Console.WriteLine(GFG.mergeStones(stones, 3)); Console.WriteLine(GFG.mergeStones(stones, 2)); } } // This code is contributed by aadityaburujwale. |
Javascript
// JavaScript implementation to merge the K minimum elements of the array // until there is only one element is left in the array class GFG { // Utility function to insert the element at its correct position in a sorted array // (Binary search is used to get the index of the correct position) insertInSortedList(sortedList, item, start, end) { let mid = start + Math.floor((end - start) / 2); if (mid === 0 || mid === end - 1 || sortedList[mid] === item) { sortedList.splice(mid, 0, item); return ; } else if (sortedList[mid] < item) { this .insertInSortedList(sortedList, item, mid + 1, end); } else { this .insertInSortedList(sortedList, item, start, mid - 1); } } // Function to merge the element of the array until there is // only one element left in array mergeStones(List, k) { // Sort the array List.sort((a, b) => a - b); let cost = 0; // Loop to merge the elements until there are // greater than K elements in the array while (List.length > k) { let Sum = 0; // Obtain the sum of K minimum elements for (let i = 0; i < k; i++) { Sum += List[i]; } // Remove the K minimum elements from the array for (let i = 0; i < k; i++) { List.shift(); } // Insert the merged element into the sorted array // at its correct position this .insertInSortedList(List, Sum, 0, List.length); cost += Sum; } // If there are only K elements left, // take the sum and return the sum, else return -1 if (List.length === k) { cost += List.reduce((a, b) => a + b); return cost; } else { return -1; } } } // Driver code let stones = []; stones.push(3); stones.push(2); stones.push(4); stones.push(1); console.log( new GFG().mergeStones(stones, 3)); console.log( new GFG().mergeStones(stones, 2)); // This code is contributed by akashish__ |
-1 10
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