You are given n integers and you want to select exactly k integers such the total sum of all the selected integers is greater than s.
Answers
Step-by-step explanation:
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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:
#include <bits/stdc++.h>
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<int>& 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<int> 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;
}
Output:
19
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