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Maximum number of chocolates to be distributed equally among k students

Given n boxes containing some chocolates arranged in a row. There are k number of students. The problem is to distribute maximum number of chocolates equally among k students by selecting a consecutive sequence of boxes from the given lot. Consider the boxes are arranged in a row with numbers from 1 to n from left to right. We have to select a group of boxes which are in consecutive order that could provide maximum number of chocolates equally to all the k students. An array arr[] is given representing the row arrangement of the boxes and arr[i] represents number of chocolates in that box at position ‘i’.

Examples:

Input : arr[] = {2, 7, 6, 1, 4, 5}, k = 3
Output : 6
The subarray is {7, 6, 1, 4} with sum 18.
Equal distribution of 18 chocolates among
3 students is 6.
Note that the selected boxes are in consecutive order
with indexes {1, 2, 3, 4}.

Source: Asked in Amazon.



The problem is to find maximum sum sub-array divisible by k and then return (sum / k).

Method 1 (Naive Approach): Consider the sum of all the sub-arrays. Select the maximum sum. Let it be maxSum. Return (maxSum / k). Time Complexity is of O(n2).

Method 2 (Efficient Approach): Create an array sum[] where sum[i] stores sum(arr[0]+..arr[i]). Create a hash table having tuple as (ele, idx), where ele represents an element of (sum[i] % k) and idx represents the element’s index of first occurrence when array sum[] is being traversed from left to right. Now traverse sum[] from i = 0 to n and follow the steps given below.

  1. Calculate current remainder as curr_rem = sum[i] % k.
  2. If curr_rem == 0, then check if maxSum < sum[i], update maxSum = sum[i].
  3. Else if curr_rem is not present in the hash table, then create tuple (curr_rem, i) in the hash table.
  4. Else, get the value associated with curr_rem in the hash table. Let this be idx. Now, if maxSum < (sum[i] – sum[idx]) then update maxSum = sum[i] – sum[idx].

Finally, return (maxSum / k).

Explanation:
If (sum[i] % k) == (sum[j] % k), where sum[i] = sum(arr[0]+..+arr[i]) and sum[j] = sum(arr[0]+..+arr[j]) and ‘i’ is less than ‘j’, then sum(arr[i+1]+..+arr[j]) must be divisible by ‘k’.

C++

// C++ implementation to find the maximum number
// of chocolates to be distributed equally among
// k students
#include <bits/stdc++.h>
using namespace std;
  
// function to find the maximum number of chocolates
// to be distributed equally among k students
int maxNumOfChocolates(int arr[], int n, int k)
{
    // unordered_map 'um' implemented as
    // hash table
    unordered_map<int, int> um;
  
    // 'sum[]' to store cumulative sum, where
    // sum[i] = sum(arr[0]+..arr[i])
    int sum[n], curr_rem;
  
    // to store sum of sub-array having maximum sum
    int maxSum = 0;
  
    // building up 'sum[]'
    sum[0] = arr[0];
    for (int i = 1; i < n; i++)
        sum[i] = sum[i - 1] + arr[i];
  
    // traversing 'sum[]'
    for (int i = 0; i < n; i++) {
  
        // finding current remainder
        curr_rem = sum[i] % k;
  
        // if true then sum(0..i) is divisible
        // by k
        if (curr_rem == 0) {
            // update 'maxSum'
            if (maxSum < sum[i])
                maxSum = sum[i];
        }
  
        // if value 'curr_rem' not present in 'um'
        // then store it in 'um' with index of its
        // first occurrence
        else if (um.find(curr_rem) == um.end())
            um[curr_rem] = i;
  
        else
            // if true, then update 'max'
            if (maxSum < (sum[i] - sum[um[curr_rem]]))
            maxSum = sum[i] - sum[um[curr_rem]];
    }
  
    // required maximum number of chocolates to be
    // distributed equally among 'k' students
    return (maxSum / k);
}
  
// Driver program to test above
int main()
{
    int arr[] = { 2, 7, 6, 1, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << "Maximum number of chocolates: "
         << maxNumOfChocolates(arr, n, k);
    return 0;
}

Java

// Java implementation to find the maximum number
// of chocolates to be distributed equally among
// k students
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum number of chocolates
// to be distributed equally among k students
static int maxNumOfChocolates(int arr[], int n, int k)
{
    // Hash table
    HashMap <Integer,Integer> um = new HashMap<Integer,Integer>();
  
    // 'sum[]' to store cumulative sum, where
    // sum[i] = sum(arr[0]+..arr[i])
    int[] sum=new int[n];
    int curr_rem;
  
    // To store sum of sub-array having maximum sum
    int maxSum = 0;
  
    // Building up 'sum[]'
    sum[0] = arr[0];
    for (int i = 1; i < n; i++)
        sum[i] = sum[i - 1] + arr[i];
  
    // Traversing 'sum[]'
    for (int i = 0; i < n; i++) {
  
        // Finding current remainder
        curr_rem = sum[i] % k;
  
        // If true then sum(0..i) is divisible
        // by k
        if (curr_rem == 0) {
            // update 'maxSum'
            if (maxSum < sum[i])
                maxSum = sum[i];
        }
  
        // If value 'curr_rem' not present in 'um'
        // then store it in 'um' with index of its
        // first occurrence
        else if (!um.containsKey(curr_rem) )
            um.put(curr_rem , i);
  
        else
            // If true, then update 'max'
            if (maxSum < (sum[i] - sum[um.get(curr_rem)]))
            maxSum = sum[i] - sum[um.get(curr_rem)];
    }
  
    // Required maximum number of chocolates to be
    // distributed equally among 'k' students
    return (maxSum / k);
}
  
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 7, 6, 1, 4, 5 };
int n = arr.length;
int k = 3;
System.out.println("Maximum number of chocolates: "
                    + maxNumOfChocolates(arr, n, k));
}
    }
  
// This code is contributed by 'Gitanjali'.

Python3

# Python3 implementation to 
# find the maximum number
# of chocolates to be 
# distributed equally
# among k students
  
# function to find the
# maximum number of chocolates
# to be distributed equally
# among k students
def maxNumOfChocolates(arr, n, k):
      
    um, curr_rem, maxSum = {}, 0, 0
      
    # 'sm[]' to store cumulative sm,
    # where sm[i] = sm(arr[0]+..arr[i])
    sm = [0]*n
    sm[0] = arr[0]
      
    # building up 'sm[]'
    for i in range(1, n):
        sm[i] = sm[i - 1] + arr[i]
          
    # traversing 'sm[]'
    for i in range(n):
  
        # finding current remainder
        curr_rem = sm[i] % k
          
        if (not curr_rem and maxSum < sm[i]) : 
            maxSum = sm[i]
        elif (not curr_rem in um) :
            um[curr_rem] = i
        elif (maxSum < (sm[i] - sm[um[curr_rem]])):
              maxSum = sm[i] - sm[um[curr_rem]]
          
    return maxSum//k
      
# Driver program to test above
arr = [ 2, 7, 6, 1, 4, 5 ]
n, k = len(arr), 3
  
print("Maximum number of chocolates: " +
     str(maxNumOfChocolates(arr, n, k)))
  
# This code is contributed by Ansu Kumari

C#

// C# implementation to find 
// the maximum number of 
// chocolates to be distributed 
// equally among k students
using System;
using System.Collections.Generic;
  
class GFG 
{
    // Function to find the 
    // maximum number of 
    // chocolates to be distributed 
    // equally among k students
    static int maxNumOfChocolates(int []arr, 
                                  int n, int k)
    {
        // Hash table
        Dictionary <int, int> um = 
                    new Dictionary<int, int>();
      
        // 'sum[]' to store cumulative 
        // sum, where sum[i] = 
        // sum(arr[0]+..arr[i])
        int[] sum = new int[n];
        int curr_rem;
      
        // To store sum of sub-array
        // having maximum sum
        int maxSum = 0;
      
        // Building up 'sum[]'
        sum[0] = arr[0];
        for (int i = 1; i < n; i++)
            sum[i] = sum[i - 1] + arr[i];
      
        // Traversing 'sum[]'
        for (int i = 0; i < n; i++) 
        {
      
            // Finding current
            // remainder
            curr_rem = sum[i] % k;
      
            // If true then sum(0..i) 
            // is divisible by k
            if (curr_rem == 0) 
            {
                // update 'maxSum'
                if (maxSum < sum[i])
                    maxSum = sum[i];
            }
      
            // If value 'curr_rem' not
            // present in 'um' then store
            // it in 'um' with index of 
            // its first occurrence
            else if (!um.ContainsKey(curr_rem))
                um.Add(curr_rem , i);
      
            else
              
                // If true, then
                // update 'max'
                if (maxSum < (sum[i] - 
                    sum[um[curr_rem]]))
                maxSum = sum[i] - 
                         sum[um[curr_rem]];
        }
      
        // Required maximum number 
        // of chocolates to be 
        // distributed equally
        // among 'k' students
        return (maxSum / k);
    }
      
    // Driver Code
    static void Main()
    {
    int []arr = new int[]{ 2, 7, 6, 1, 4, 5 };
    int n = arr.Length;
    int k = 3;
    Console.Write("Maximum number of chocolates: "
                     maxNumOfChocolates(arr, n, k));
    }
}
  
// This code is contributed by
// Manish Shaw(manishshaw1)


Output :

Maximum number of chocolates: 6

Time Complexity: O(n).
Auxiliary Space: O(n).



This article is attributed to GeeksforGeeks.org

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