Given an array arr[] consisting of N integers and an integer K, the task is to split the given array into K non-overlapping subsets such that the maximum among the sum of all subsets is minimum.
Examples:
Input: arr[] = {1, 7, 9, 2, 12, 3, 3}, M = 3
Output: 13
Explanation:
One possible way to split the array into 3 non-overlapping subsets is {arr[4], arr[0]}, {arr[2], arr[6]}, and {arr[1], arr[5], arr[3]}.
The sum of each subset is 13, 12 and 12 respectively. Now, the maximum among all the sum of subsets is 13, which is the minimum possible sum.Input: arr[] = {1, 2, 3, 4, 5}, M = 2
Output: 8
Approach: The given problem can be solved by the Greedy Approach by using the priority queue and sorting the given array. Follow the steps below to solve the problem:
- Initialize a Min-Heap using priority_queue say PQ to store the sum of elements of each group.
- Iterate in the range [1, K] and push 0 into the PQ.
- Sort the array, arr[] in decreasing order.
- Traverse the array arr[] and for each array elements arr[i], add the element arr[i] to the smallest sum group which will be at the top of the priority queue PQ.
- After completing the above steps, print the last element of the priority queue PQ as the resultant minimized maximum sum among all the possible groups.
Below is the implementation of the above approach:
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// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to split the array into M // groups such that maximum of the sum // of all elements of all the groups // is minimized int findMinimumValue( int arr[], int N,
int M)
{ // Sort the array in decreasing order
sort(arr, arr + N, greater< int >());
// Initialize priority queue (Min heap)
priority_queue< int , vector< int >,
greater< int > >
pq;
// Push 0 for all the M groups
for ( int i = 1; i <= M; ++i) {
pq.push(0);
}
// Traverse the array, arr[]
for ( int i = 0; i < N; ++i) {
// Pop the group having the
// minimum sum
int val = pq.top();
pq.pop();
// Increment val by arr[i]
val += arr[i];
// Push the new sum of the
// group into the pq
pq.push(val);
}
// Iterate while size of the pq
// is greater than 1
while (pq.size() > 1) {
pq.pop();
}
// Return result
return pq.top();
} // Driver Code int main()
{ int arr[] = { 1, 7, 9, 2, 12, 3, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
int K = 3;
cout << findMinimumValue(arr, N, K);
return 0;
} |
// Java program for the above approach import java.util.*;
public class Main
{ // Function to split the array into M
// groups such that maximum of the sum
// of all elements of all the groups
// is minimized
static int findMinimumValue(Vector<Integer> arr, int N, int M)
{
// Sort the array in decreasing order
Collections.sort(arr);
Collections.reverse(arr);
// Initialize priority queue (Min heap)
Vector<Integer> pq = new Vector<Integer>();
// Push 0 for all the M groups
for ( int i = 1 ; i <= M; ++i) {
pq.add( 0 );
}
Collections.sort(pq);
// Traverse the array, arr[]
for ( int i = 0 ; i < N; ++i) {
// Pop the group having the
// minimum sum
int val = pq.get( 0 );
pq.remove( 0 );
// Increment val by arr[i]
val += arr.get(i);
// Push the new sum of the
// group into the pq
pq.add(val);
Collections.sort(pq);
}
// Iterate while size of the pq
// is greater than 1
while (pq.size() > 1 ) {
pq.remove( 0 );
}
// Return result
return pq.get( 0 );
}
public static void main(String[] args) {
Integer[] arr = { 1 , 7 , 9 , 2 , 12 , 3 , 3 };
Vector<Integer> Arr = new Vector<Integer>();
Collections.addAll(Arr, arr);
int N = Arr.size();
int K = 3 ;
System.out.println(findMinimumValue(Arr, N, K));
}
} // This code is contributed by divyesh072019. |
# Python3 program for the above approach # Function to split the array into M # groups such that maximum of the sum # of all elements of all the groups # is minimized def findMinimumValue(arr, N, M):
# Sort the array in decreasing order
arr.sort()
arr.reverse()
# Initialize priority queue (Min heap)
pq = []
# Push 0 for all the M groups
for i in range ( 1 , M + 1 ):
pq.append( 0 )
pq.sort()
# Traverse the array, arr[]
for i in range (N):
# Pop the group having the
# minimum sum
val = pq[ 0 ]
del pq[ 0 ]
# Increment val by arr[i]
val + = arr[i]
# Push the new sum of the
# group into the pq
pq.append(val)
pq.sort()
# Iterate while size of the pq
# is greater than 1
while ( len (pq) > 1 ) :
del pq[ 0 ]
# Return result
return pq[ 0 ]
arr = [ 1 , 7 , 9 , 2 , 12 , 3 , 3 ]
N = len (arr)
K = 3
print (findMinimumValue(arr, N, K))
# This code is contributed by suresh07. |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG {
// Function to split the array into M
// groups such that maximum of the sum
// of all elements of all the groups
// is minimized
static int findMinimumValue( int [] arr, int N, int M)
{
// Sort the array in decreasing order
Array.Sort(arr);
Array.Reverse(arr);
// Initialize priority queue (Min heap)
List< int > pq = new List< int >();
// Push 0 for all the M groups
for ( int i = 1; i <= M; ++i) {
pq.Add(0);
}
pq.Sort();
// Traverse the array, arr[]
for ( int i = 0; i < N; ++i) {
// Pop the group having the
// minimum sum
int val = pq[0];
pq.RemoveAt(0);
// Increment val by arr[i]
val += arr[i];
// Push the new sum of the
// group into the pq
pq.Add(val);
pq.Sort();
}
// Iterate while size of the pq
// is greater than 1
while (pq.Count > 1) {
pq.RemoveAt(0);
}
// Return result
return pq[0];
}
static void Main() {
int [] arr = { 1, 7, 9, 2, 12, 3, 3 };
int N = arr.Length;
int K = 3;
Console.Write(findMinimumValue(arr, N, K));
}
} // This code is contributed by divyeshrabadiya07. |
<script> // Javascript program for the above approach
// Function to split the array into M
// groups such that maximum of the sum
// of all elements of all the groups
// is minimized
function findMinimumValue(arr, N, M)
{
// Sort the array in decreasing order
arr.sort( function (a, b){ return a - b});
arr.reverse();
// Initialize priority queue (Min heap)
let pq = [];
// Push 0 for all the M groups
for (let i = 1; i <= M; ++i) {
pq.push(0);
}
pq.sort( function (a, b){ return a - b});
// Traverse the array, arr[]
for (let i = 0; i < N; ++i) {
// Pop the group having the
// minimum sum
let val = pq[0];
pq.shift();
// Increment val by arr[i]
val += arr[i];
// Push the new sum of the
// group into the pq
pq.push(val);
pq.sort( function (a, b){ return a - b});
}
// Iterate while size of the pq
// is greater than 1
while (pq.length > 1) {
pq.shift();
}
// Return result
return pq[0];
}
let arr = [ 1, 7, 9, 2, 12, 3, 3 ];
let N = arr.length;
let K = 3;
document.write(findMinimumValue(arr, N, K));
// This code is contributed by decode2207.
</script> |
13
Time Complexity: O(N*log K)
Auxiliary Space: O(M)