Maximum sum of K-length subarray consisting of same number of distinct elements as the given array

Given an array arr[] consisting of N integers and an integer K, the task is to find a subarray of size K with maximum sum and count of distinct elements same as that of the original array.

Examples:

Input: arr[] = {7, 7, 2, 4, 2, 7, 4, 6, 6, 6}, K = 6
Output: 31
Explanation: The given array consists of 4 distinct elements, i.e. {2, 4, 6, 7}. The subarray of size K consisting of all these elements and maximum sum is {2, 7, 4, 6, 6, 6} which starts from 5^th index (1-based indexing) of the original array.
Therefore, the sum of the subarray = 2 + 7 + 4 + 6 + 6 + 6 = 31.

Input: arr[] = {1, 2, 5, 5, 19, 2, 1}, K = 4
Output: 27

Naive Approach: The simple approach is to generate all possible subarrays of size K and check if it has the same distinct elements as the original array. If yes then find the sum of this subarray. After checking all the subarrays print the maximum sum of all such subarrays.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// distinct elements present in the array
int distinct(int arr[], int n)
{
    map<int,int> mpp;

    // Insert all elements into the Set
    for (int i = 0; i < n; i++)
    {
        mpp[arr[i]] = 1;
    }

    // Return the size of set
    return mpp.size();
}

// Function that finds the maximum
// sum of K-length subarray having
// same unique elements as arr[]
int maxSubSum(int arr[], int n,int k, int totalDistinct)
{

    // Not possible to find a
    // subarray of size K
    if (k > n)
        return 0;
    int maxm = 0, sum = 0;
    for (int i = 0; i < n - k + 1; i++)
    {
        sum = 0;

        // Initialize Set
        set<int> st;

        // Calculate sum of the distinct elements
        for (int j = i; j < i + k; j++)
        {
            sum += arr[j];
            st.insert(arr[j]);
        }

        // If the set size is same as the
        // count of distinct elements
        if ((int) st.size() == totalDistinct)

            // Update the maximum value
            maxm = max(sum, maxm);
    }
    return maxm;
}

// Driver code
int main()
{
  int arr[] = { 7, 7, 2, 4, 2,
                7, 4, 6, 6, 6 };
  int K = 6;
  int N = sizeof(arr)/sizeof(arr[0]);

  // Stores the count of distinct elements
  int totalDistinct = distinct(arr, N);
  cout << (maxSubSum(arr, N, K, totalDistinct));

  return 0;
}

// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach
import java.util.*;

class GFG {

    // Function to count the number of
    // distinct elements present in the array
    static int distinct(int arr[], int n)
    {
        Set<Integer> set = new HashSet<>();

        // Insert all elements into the Set
        for (int i = 0; i < n; i++) {
            set.add(arr[i]);
        }

        // Return the size of set
        return set.size();
    }

    // Function that finds the maximum
    // sum of K-length subarray having
    // same unique elements as arr[]
    static int maxSubSum(int arr[], int n,
                         int k,
                         int totalDistinct)
    {
        // Not possible to find a
        // subarray of size K
        if (k > n)
            return 0;

        int max = 0, sum = 0;

        for (int i = 0; i < n - k + 1; i++) {
            sum = 0;

            // Initialize Set
            Set<Integer> set = new HashSet<>();

            // Calculate sum of the distinct elements
            for (int j = i; j < i + k; j++) {
                sum += arr[j];
                set.add(arr[j]);
            }

            // If the set size is same as the
            // count of distinct elements
            if (set.size() == totalDistinct)

                // Update the maximum value
                max = Math.max(sum, max);
        }
        return max;
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 7, 7, 2, 4, 2,
                      7, 4, 6, 6, 6 };
        int K = 6;
        int N = arr.length;

        // Stores the count of distinct elements
        int totalDistinct = distinct(arr, N);

        System.out.println(
            maxSubSum(arr, N, K, totalDistinct));
    }
}

Python3

# Python3 program for the above approach

# Function to count the number of
# distinct elements present in the array
def distinct(arr, n):

    mpp = {}

    # Insert all elements into the Set
    for i in range(n):
        mpp[arr[i]] = 1

    # Return the size of set
    return len(mpp)

# Function that finds the maximum
# sum of K-length subarray having
# same unique elements as arr[]
def maxSubSum(arr, n, k, totalDistinct):

    # Not possible to find a
    # subarray of size K
    if (k > n):
        return 0

    maxm = 0
    sum = 0

    for i in range(n - k + 1):
        sum = 0

        # Initialize Set
        st = set()

        # Calculate sum of the distinct elements
        for j in range(i, i + k, 1):
            sum += arr[j]
            st.add(arr[j])

        # If the set size is same as the
        # count of distinct elements
        if (len(st) == totalDistinct):

            # Update the maximum value
            maxm = max(sum, maxm)

    return maxm

# Driver code
if __name__ == '__main__':

    arr = [ 7, 7, 2, 4, 2, 7, 4, 6, 6, 6 ]
    K = 6
    N = len(arr)

    # Stores the count of distinct elements
    totalDistinct = distinct(arr, N)
    print(maxSubSum(arr, N, K, totalDistinct))

# This code is contributed by ipg2016107

C#

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG
{


  // Function to count the number of
  // distinct elements present in the array
  static int distinct(int[] arr, int n)
  {
    HashSet<int> set = new HashSet<int>();

    // Insert all elements into the Set
    for (int i = 0; i < n; i++) {
      set.Add(arr[i]);
    }

    // Return the size of set
    return set.Count;
  }

  // Function that finds the maximum
  // sum of K-length subarray having
  // same unique elements as arr[]
  static int maxSubSum(int[] arr, int n,
                       int k,
                       int totalDistinct)
  {
    // Not possible to find a
    // subarray of size K
    if (k > n)
      return 0;

    int max = 0, sum = 0;

    for (int i = 0; i < n - k + 1; i++) {
      sum = 0;

      // Initialize Set
      HashSet<int> set = new HashSet<int>();

      // Calculate sum of the distinct elements
      for (int j = i; j < i + k; j++) {
        sum += arr[j];
        set.Add(arr[j]);
      }

      // If the set size is same as the
      // count of distinct elements
      if (set.Count == totalDistinct)

        // Update the maximum value
        max = Math.Max(sum, max);
    }
    return max;
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int[] arr = { 7, 7, 2, 4, 2,
                 7, 4, 6, 6, 6 };
    int K = 6;
    int N = arr.Length;

    // Stores the count of distinct elements
    int totalDistinct = distinct(arr, N);

    Console.WriteLine(
      maxSubSum(arr, N, K, totalDistinct));
  }
}

// This code is contributed by code_hunt.

Javascript

<script>

// Javascript program for the above approach

// Function to count the number of
// distinct elements present in the array
function distinct(arr, n)
{
    var mpp = new Map();

    // Insert all elements into the Set
    for (var i = 0; i < n; i++)
    {
        mpp.set(arr[i], 1);
    }

    // Return the size of set
    return mpp.size;
}

// Function that finds the maximum
// sum of K-length subarray having
// same unique elements as arr[]
function maxSubSum(arr, n,k, totalDistinct)
{

    // Not possible to find a
    // subarray of size K
    if (k > n)
        return 0;
    var maxm = 0, sum = 0;
    for (var i = 0; i < n - k + 1; i++)
    {
        sum = 0;

        // Initialize Set
        var st = new Set();

        // Calculate sum of the distinct elements
        for (var j = i; j < i + k; j++)
        {
            sum += arr[j];
            st.add(arr[j]);
        }

        // If the set size is same as the
        // count of distinct elements
        if ( st.size == totalDistinct)

            // Update the maximum value
            maxm = Math.max(sum, maxm);
    }
    return maxm;
}

// Driver code
var arr = [7, 7, 2, 4, 2,
              7, 4, 6, 6, 6];
var K = 6;
var N = arr.length;

// Stores the count of distinct elements
var totalDistinct = distinct(arr, N);
document.write(maxSubSum(arr, N, K, totalDistinct));

// This code is contributed by itsok.
</script>

Output:

31

Time Complexity: O(N²*log(N))
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to make use of Map. Follow the steps below to solve the problem:

1. Traverse the array once and keep updating the frequency of array elements in the Map.
2. Check if the size of the map is equal to the total number of distinct elements present in the original array or not. If found to be true, update the maximum sum.
3. While traversing the original array, if the i^th traversal crosses K elements in the array, update the Map by deleting an occurrence of (i – K)^th element.
4. After completing the above steps, print the maximum sum obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to count the number of
// distinct elements present in the array
int distinct(vector<int>arr, int N)
{
    set<int> st;

    // Insert array elements into set
    for(int i = 0; i < N; i++)
    {
        st.insert(arr[i]);
    }

    // Return the st size
    return st.size();
}

// Function to calculate maximum
// sum of K-length subarray having
// same unique elements as arr[]
int maxSubarraySumUtil(vector<int>arr, int N,
                       int K, int totalDistinct)
{

    // Not possible to find an
    // subarray of length K from
    // an N-sized array, if K > N
    if (K > N)
        return 0;

    int mx = 0;
    int sum = 0;

    map<int, int> mp;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Update the mp
        mp[arr[i]] += 1;
        sum += arr[i];

        // If i >= K, then decrement
        // arr[i-K] element's one
        // occurrence
        if (i >= K)
        {
            mp[arr[i - K]] -= 1;
            sum -= arr[i - K];

            // If frequency of any
            // element is 0 then
            // remove the element
            if (mp[arr[i - K]] == 0)
                mp.erase(arr[i - K]);
        }

        // If mp size is same as the
        // count of distinct elements
        // of array arr[] then update
        // maximum sum
        if (mp.size() == totalDistinct)
            mx = max(mx, sum);
    }
    return mx;
}

// Function that finds the maximum
// sum of K-length subarray having
// same number of distinct elements
// as the original array
void maxSubarraySum(vector<int>arr,
                           int K)
{
    // Size of array
    int N = arr.size();

    // Stores count of distinct elements
    int totalDistinct = distinct(arr, N);

    // Print maximum subarray sum
    cout<<maxSubarraySumUtil(arr, N, K, totalDistinct);
}

// Driver Code
int main()
{
    vector<int>arr { 7, 7, 2, 4, 2,
                     7, 4, 6, 6, 6 };
    int K = 6;

    // Function Call
    maxSubarraySum(arr, K);
}

// This code is contributed by ipg2016107

Java

// Java program for the above approach

import java.util.*;
class GFG {

    // Function to count the number of
    // distinct elements present in the array
    static int distinct(int arr[], int N)
    {
        Set<Integer> set = new HashSet<>();

        // Insert array elements into Set
        for (int i = 0; i < N; i++) {
            set.add(arr[i]);
        }

        // Return the Set size
        return set.size();
    }

    // Function to calculate maximum
    // sum of K-length subarray having
    // same unique elements as arr[]
    static int maxSubarraySumUtil(
        int arr[], int N, int K,
        int totalDistinct)
    {
        // Not possible to find an
        // subarray of length K from
        // an N-sized array, if K > N
        if (K > N)
            return 0;

        int max = 0;
        int sum = 0;

        Map<Integer, Integer> map
            = new HashMap<>();

        // Traverse the array
        for (int i = 0; i < N; i++) {

            // Update the map
            map.put(arr[i],
                    map.getOrDefault(arr[i], 0) + 1);
            sum += arr[i];

            // If i >= K, then decrement
            // arr[i-K] element's one
            // occurrence
            if (i >= K) {
                map.put(arr[i - K],
                        map.get(arr[i - K]) - 1);
                sum -= arr[i - K];

                // If frequency of any
                // element is 0 then
                // remove the element
                if (map.get(arr[i - K]) == 0)
                    map.remove(arr[i - K]);
            }

            // If map size is same as the
            // count of distinct elements
            // of array arr[] then update
            // maximum sum
            if (map.size() == totalDistinct)
                max = Math.max(max, sum);
        }
        return max;
    }

    // Function that finds the maximum
    // sum of K-length subarray having
    // same number of distinct elements
    // as the original array
    static void maxSubarraySum(int arr[],
                               int K)
    {
        // Size of array
        int N = arr.length;

        // Stores count of distinct elements
        int totalDistinct = distinct(arr, N);

        // Print maximum subarray sum
        System.out.println(
            maxSubarraySumUtil(arr, N, K,
                               totalDistinct));
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 7, 7, 2, 4, 2,
                      7, 4, 6, 6, 6 };
        int K = 6;

        // Function Call
        maxSubarraySum(arr, K);
    }
}

Python3

# Python 3 program for the above approach

# Function to count the number of
# distinct elements present in the array
def distinct(arr, N):
    st = set()

    # Insert array elements into set
    for i in range(N):
        st.add(arr[i])

    # Return the st size
    return len(st)

# Function to calculate maximum
# sum of K-length subarray having
# same unique elements as arr[]
def maxSubarraySumUtil(arr, N, K, totalDistinct):
    # Not possible to find an
    # subarray of length K from
    # an N-sized array, if K > N
    if (K > N):
        return 0

    mx = 0
    sum = 0

    mp = {}

    # Traverse the array
    for i in range(N):
        # Update the mp
        if(arr[i] in mp):
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
        sum += arr[i]

        # If i >= K, then decrement
        # arr[i-K] element's one
        # occurrence
        if (i >= K):
            if(arr[i-K] in mp):
                mp[arr[i - K]] -= 1
                sum -= arr[i - K]

            # If frequency of any
            # element is 0 then
            # remove the element
            if (arr[i-K] in mp and mp[arr[i - K]] == 0):
                mp.remove(arr[i - K])

        # If mp size is same as the
        # count of distinct elements
        # of array arr[] then update
        # maximum sum
        if (len(mp) == totalDistinct):
            mx = max(mx, sum)
    return mx

# Function that finds the maximum
# sum of K-length subarray having
# same number of distinct elements
# as the original array
def maxSubarraySum(arr, K):

    # Size of array
    N = len(arr)

    # Stores count of distinct elements
    totalDistinct = distinct(arr, N)

    # Print maximum subarray sum
    print(maxSubarraySumUtil(arr, N, K, totalDistinct))

# Driver Code
if __name__ == '__main__':
    arr  =  [7, 7, 2, 4, 2,7, 4, 6, 6, 6]
    K = 6

    # Function Call
    maxSubarraySum(arr, K)

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to count the number of
// distinct elements present in the array
static int distinct(List<int>arr, int N)
{
    HashSet<int> st = new HashSet<int>();

    // Insert array elements into set
    for(int i = 0; i < N; i++)
    {
        st.Add(arr[i]);
    }

    // Return the st size
    return st.Count;
}

// Function to calculate maximum
// sum of K-length subarray having
// same unique elements as arr[]
static int maxSubarraySumUtil(List<int>arr, int N,
                       int K, int totalDistinct)
{

    // Not possible to find an
    // subarray of length K from
    // an N-sized array, if K > N
    if (K > N)
        return 0;
    int mx = 0;
    int sum = 0;
    Dictionary<int,int> mp = new Dictionary<int,int>();

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Update the mp
        if(mp.ContainsKey(arr[i]))
            mp[arr[i]] += 1;
        else
            mp[arr[i]] = 1;
        sum += arr[i];

        // If i >= K, then decrement
        // arr[i-K] element's one
        // occurrence
        if (i >= K)
        {
           if(mp.ContainsKey(arr[i - K]))
              mp[arr[i - K]] -= 1;
           else
              mp[arr[i - K]] = 1;
            sum -= arr[i - K];

            // If frequency of any
            // element is 0 then
            // remove the element
            if (mp[arr[i - K]] == 0)
                mp.Remove(arr[i - K]);
        }

        // If mp size is same as the
        // count of distinct elements
        // of array arr[] then update
        // maximum sum
        if (mp.Count == totalDistinct)
            mx = Math.Max(mx, sum);
    }
    return mx;
}

// Function that finds the maximum
// sum of K-length subarray having
// same number of distinct elements
// as the original array
static void maxSubarraySum(List<int>arr,
                           int K)
{

    // Size of array
    int N = arr.Count;

    // Stores count of distinct elements
    int totalDistinct = distinct(arr, N);

    // Print maximum subarray sum
    Console.WriteLine(maxSubarraySumUtil(arr, N, K, totalDistinct));
}

// Driver Code
public static void Main()
{
    List<int>arr = new List<int>{ 7, 7, 2, 4, 2,
                     7, 4, 6, 6, 6 };
    int K = 6;

    // Function Call
    maxSubarraySum(arr, K);
}
}

// This code is contributed by bgangwar59.

Javascript

<script>

// JavaScript program for the above approach

// Function to count the number of
// distinct elements present in the array
function distinct(arr, N)
{
    var st = new Set();

    // Insert array elements into set
    for(var i = 0; i < N; i++)
    {
        st.add(arr[i]);
    }

    // Return the st size
    return st.size;
}

// Function to calculate maximum
// sum of K-length subarray having
// same unique elements as arr[]
function maxSubarraySumUtil(arr, N, K, totalDistinct)
{

    // Not possible to find an
    // subarray of length K from
    // an N-sized array, if K > N
    if (K > N)
        return 0;


    var mx = 0;
    var sum = 0;

    var mp = new Map();

    // Traverse the array
    for(var i=0; i<N; i++)
    {

        // Update the mp
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1)

        sum += arr[i];

        // If i >= K, then decrement
        // arr[i-K] element's one
        // occurrence
        if (i >= K)
        {
            if(mp.has(arr[i-K]))
                mp.set(arr[i-K], mp.get(arr[i-K])-1)

            sum -= arr[i - K];

            // If frequency of any
            // element is 0 then
            // remove the element
            if (mp.has(arr[i - K]) &&  mp.get(arr[i - K])== 0)
                mp.delete(arr[i - K]);
        }

        // If mp size is same as the
        // count of distinct elements
        // of array arr[] then update
        // maximum sum
        if (mp.size == totalDistinct)
            mx = Math.max(mx, sum);
    }
    return mx;
}

// Function that finds the maximum
// sum of K-length subarray having
// same number of distinct elements
// as the original array
function maxSubarraySum(arr, K)
{
    // Size of array
    var N = arr.length;

    // Stores count of distinct elements
    var totalDistinct = distinct(arr, N);

    // Print maximum subarray sum
    document.write( maxSubarraySumUtil(arr, N, K, totalDistinct));
}

// Driver Code

var arr = [7, 7, 2, 4, 2,
           7, 4, 6, 6, 6 ];
var K = 6;

// Function Call
maxSubarraySum(arr, K);


</script>

Output:

31

Time Complexity: O(N*log (N))
Auxiliary Space: O(N)