# Number of ways to choose elements from the array such that their average is K

Given an array arr[] of N integers and an integer K. The task is to find the number of ways to select one or more elements from the array such that the average of the selected integers is equal to given number K.

Examples:

Input: arr[] = {7, 9, 8, 9}, K = 8
Output: 5
{8}, {7, 9}, {7, 9}, {7, 8, 9} and {7, 8, 9}

Input: arr[] = {3, 6, 2, 8, 7, 6, 5, 9}, K = 5
Output: 19

Input: arr[] = {6, 6, 9}, K = 8
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: A simple solution would be to try for all possibilities since N can be large. Time complexity can be 2N.

Efficient Approach: The above approach can be optimized by using dynamic programming to solve this problem. Suppose we are at i_th index, and let val be the current value of that index. We have two possibilities either to choose that element in the answer or discard the element. Hence we are done now. We will also keep track of the number of elements in our current set of chosen elements.

Following is the recursive formula.

ways(index, sum, count)
= ways(index - 1, sum, count)
+ ways(index - 1, sum + arr[index], count + 1)

Below is the implementation of the above approach:

## C++

 #include using namespace std;    #define MAX_INDEX 51 #define MAX_SUM 2505    // This dp array is used to store our values // so that we don't have to calculate same // values again and again int dp[MAX_INDEX][MAX_SUM][MAX_INDEX];    int waysutil(int index, int sum, int count,              vector& arr, int K) {     // Base cases     // Index can't be less than 0     if (index < 0)         return 0;        if (index == 0) {            // No element is picked hence         // average cannot be calculated         if (count == 0)             return 0;         int remainder = sum % count;            // If remainder is non zero, we cannot         // divide the sum by count i.e. the average         // will not be an integer         if (remainder != 0)             return 0;         int average = sum / count;            // If we find an average return 1         if (average == K)             return 1;     }        // If we have already calculated this function     // simply return it instead of calculating it again     if (dp[index][sum][count] != -1)         return dp[index][sum][count];        // If we don't pick the current element     // simple recur for index -1     int dontpick = waysutil(index - 1,                             sum, count, arr, K);        // If we pick the current element add it to     // our current sum and increment count by 1     int pick = waysutil(index - 1,                         sum + arr[index],                         count + 1, arr, K);     int total = pick + dontpick;        // Store the value for the current function     dp[index][sum][count] = total;     return total; }    // Function to return the number of ways int ways(int N, int K, int* arr) {     vector Arr;        // Push -1 at the beginning to     // make it 1-based indexing     Arr.push_back(-1);     for (int i = 0; i < N; ++i) {         Arr.push_back(arr[i]);     }        // Initialize dp array by -1     memset(dp, -1, sizeof dp);        // Call recursive function     // waysutil to calculate total ways     int answer = waysutil(N, 0, 0, Arr, K);     return answer; }    // Driver code int main() {     int arr[] = { 3, 6, 2, 8, 7, 6, 5, 9 };     int N = sizeof(arr) / sizeof(arr[0]);     int K = 5;     cout << ways(N, K, arr);        return 0; }

## Java

 // Java implementation of the above approach import java.util.*;    class GFG  {        static int MAX_INDEX = 51;     static int MAX_SUM = 2505;        // This dp array is used to store our values     // so that we don't have to calculate same     // values again and again     static int[][][] dp = new int[MAX_INDEX][MAX_SUM][MAX_INDEX];        static int waysutil(int index, int sum, int count,                         Vector arr, int K)     {         // Base cases         // Index can't be less than 0         if (index < 0)          {             return 0;         }            if (index == 0)          {                // No element is picked hence             // average cannot be calculated             if (count == 0)             {                 return 0;             }             int remainder = sum % count;                // If remainder is non zero, we cannot             // divide the sum by count i.e. the average             // will not be an integer             if (remainder != 0)              {                 return 0;             }             int average = sum / count;                // If we find an average return 1             if (average == K)              {                 return 1;             }         }            // If we have already calculated this function         // simply return it instead of calculating it again         if (dp[index][sum][count] != -1)          {             return dp[index][sum][count];         }            // If we don't pick the current element         // simple recur for index -1         int dontpick = waysutil(index - 1,                 sum, count, arr, K);            // If we pick the current element add it to         // our current sum and increment count by 1         int pick = waysutil(index - 1,                 sum + arr.get(index),                 count + 1, arr, K);         int total = pick + dontpick;            // Store the value for the current function         dp[index][sum][count] = total;         return total;     }        // Function to return the number of ways     static int ways(int N, int K, int[] arr)     {         Vector Arr = new Vector<>();            // Push -1 at the beginning to         // make it 1-based indexing         Arr.add(-1);         for (int i = 0; i < N; ++i)         {             Arr.add(arr[i]);         }            // Initialize dp array by -1         for (int i = 0; i < MAX_INDEX; i++)         {             for (int j = 0; j < MAX_SUM; j++)             {                 for (int l = 0; l < MAX_INDEX; l++)                  {                     dp[i][j][l] = -1;                 }             }         }            // Call recursive function         // waysutil to calculate total ways         int answer = waysutil(N, 0, 0, Arr, K);         return answer;     }        // Driver code     public static void main(String args[])     {         int arr[] = {3, 6, 2, 8, 7, 6, 5, 9};         int N = arr.length;         int K = 5;         System.out.println(ways(N, K, arr));     } }    /* This code contributed by PrinciRaj1992 */

## Python3

 # Python implementation of above approach import numpy as np    MAX_INDEX = 51 MAX_SUM = 2505    # This dp array is used to store our values  # so that we don't have to calculate same  # values again and again     # Initialize dp array by -1  dp = np.ones((MAX_INDEX,MAX_SUM,MAX_INDEX)) * -1;     def waysutil(index, sum, count, arr, K) :         # Base cases      # Index can't be less than 0      if (index < 0) :         return 0;         if (index == 0) :            # No element is picked hence          # average cannot be calculated          if (count == 0) :             return 0;                        remainder = sum % count;             # If remainder is non zero, we cannot          # divide the sum by count i.e. the average          # will not be an integer          if (remainder != 0) :             return 0;                         average = sum // count;             # If we find an average return 1          if (average == K) :             return 1;         # If we have already calculated this function      # simply return it instead of calculating it again      if (dp[index][sum][count] != -1) :         return dp[index][sum][count];         # If we don't pick the current element      # simple recur for index -1      dontpick = waysutil(index - 1,                              sum, count, arr, K);         # If we pick the current element add it to      # our current sum and increment count by 1      pick = waysutil(index - 1,                          sum + arr[index],                          count + 1, arr, K);                                 total = pick + dontpick;         # Store the value for the current function      dp[index][sum][count] = total;             return total;        # Function to return the number of ways  def ways(N, K, arr) :        Arr = [];         # Push -1 at the beginning to      # make it 1-based indexing      Arr.append(-1);      for i in range(N) :         Arr.append(arr[i]);         # Call recursive function      # waysutil to calculate total ways      answer = waysutil(N, 0, 0, Arr, K);      return answer;        # Driver code  if __name__ == "__main__" :         arr = [ 3, 6, 2, 8, 7, 6, 5, 9 ];      N =len(arr);      K = 5;      print(ways(N, K, arr));     # This code is contributed by AnkitRai01

## C#

 // C# implementation of the above approach using System; using System.Collections.Generic;        class GFG  {        static int MAX_INDEX = 51;     static int MAX_SUM = 2505;        // This dp array is used to store our values     // so that we don't have to calculate same     // values again and again     static int[,,] dp = new int[MAX_INDEX, MAX_SUM, MAX_INDEX];        static int waysutil(int index, int sum, int count,                         List arr, int K)     {         // Base cases         // Index can't be less than 0         if (index < 0)          {             return 0;         }            if (index == 0)          {                // No element is picked hence             // average cannot be calculated             if (count == 0)             {                 return 0;             }             int remainder = sum % count;                // If remainder is non zero, we cannot             // divide the sum by count i.e. the average             // will not be an integer             if (remainder != 0)              {                 return 0;             }             int average = sum / count;                // If we find an average return 1             if (average == K)              {                 return 1;             }         }            // If we have already calculated this function         // simply return it instead of calculating it again         if (dp[index,sum,count] != -1)          {             return dp[index, sum, count];         }            // If we don't pick the current element         // simple recur for index -1         int dontpick = waysutil(index - 1,                 sum, count, arr, K);            // If we pick the current element add it to         // our current sum and increment count by 1         int pick = waysutil(index - 1,                 sum + arr[index],                 count + 1, arr, K);         int total = pick + dontpick;            // Store the value for the current function         dp[index,sum,count] = total;         return total;     }        // Function to return the number of ways     static int ways(int N, int K, int[] arr)     {         List Arr = new List();            // Push -1 at the beginning to         // make it 1-based indexing         Arr.Add(-1);         for (int i = 0; i < N; ++i)         {             Arr.Add(arr[i]);         }            // Initialize dp array by -1         for (int i = 0; i < MAX_INDEX; i++)         {             for (int j = 0; j < MAX_SUM; j++)             {                 for (int l = 0; l < MAX_INDEX; l++)                  {                     dp[i, j, l] = -1;                 }             }         }            // Call recursive function         // waysutil to calculate total ways         int answer = waysutil(N, 0, 0, Arr, K);         return answer;     }        // Driver code     public static void Main(String []args)     {         int []arr = {3, 6, 2, 8, 7, 6, 5, 9};         int N = arr.Length;         int K = 5;         Console.WriteLine(ways(N, K, arr));     } }    // This code is contributed by Princi Singh

Output:

19

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