Given an array arr[], the task is to calculate the count of possible triplets such that they can be removed from the array without changing the arithmetic mean of the array.
Example:
Input: arr[] = {8, 7, 4, 6, 3, 0, 7}
Output: 3
Explanation: The given array has 3 possible triplets such that removing them will not affect the arithmetic mean of the array. There are {7, 3, 0}, {4, 6, 0} and {3, 0, 7}.Input: arr[] = {5, 5, 5, 5}
Output: 4
Approach: The given problem can be solved using the observation that for the mean of the remaining array to be constant, the mean of the removed triplet must be equal to the mean of the initial array. Hence the given problem is reduced to finding the count of triplets with the given sum which can be solved using hashing by following the below steps:
- Iterate the given array arr[] for all possible values of pairs (a, b) and insert their sum into a map.
- While iterating the array, check if (TargetSum – (a + b)) already exists in the map. If yes, then increment the value of the required count by its frequency.
Below is the implementation of the above approach:
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count the number of // triplets with the given sum int countTriplets( int arr[], int n, int sum)
{ // Stores the final count
int cnt = 0;
// Map to store occurred elements
unordered_map< int , int > m;
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// Check if Sum - (a + b)
// is present in map
int k = sum - (arr[i] + arr[j]);
if (m.find(k) != m.end())
// Increment count
cnt += m[k];
}
// Store the occurrences
m[arr[i]]++;
}
// Return Answer
return cnt;
} // Function to C=find count of triplets // that can be removed without changing // arithmetic mean of the given array int count_triplets( int arr[], int n)
{ // Stores sum of all elements
// of the given array
int sum = 0;
// Calculate the sum of the array
for ( int i = 0; i < n; i++) {
sum = sum + arr[i];
}
// Store the arithmetic mean
int mean = sum / n;
int reqSum = 3 * mean;
if ((3 * sum) % n != 0)
return 0;
// Return count
return countTriplets(arr, n, reqSum);
} // Driver Code int main()
{ int arr[] = { 5, 5, 5, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << count_triplets(arr, N);
return 0;
} |
// Java code for the above approach import java.util.HashMap;
class GFG {
// Function to count the number of
// triplets with the given sum
static int countTriplets( int [] arr, int n, int sum)
{
// Stores the final count
int cnt = 0 ;
// Map to store occurred elements
HashMap<Integer, Integer> m = new HashMap<>();
for ( int i = 0 ; i < n - 1 ; i++) {
for ( int j = i + 1 ; j < n; j++) {
// Check if Sum - (a + b)
// is present in map
int k = sum - (arr[i] + arr[j]);
if (m.containsKey(k))
// Increment count
cnt += m.get(k);
}
// Store the occurrences
if (m.containsKey(arr[i]))
m.put(arr[i],m.get(arr[i])+ 1 );
else
m.put(arr[i], 1 );
}
// Return Answer
return cnt;
}
// Function to C=find count of triplets
// that can be removed without changing
// arithmetic mean of the given array
static int count_triplets( int [] arr, int n)
{
// Stores sum of all elements
// of the given array
int sum = 0 ;
// Calculate the sum of the array
for ( int i = 0 ; i < n; i++) {
sum = sum + arr[i];
}
// Store the arithmetic mean
int mean = sum / n;
int reqSum = 3 * mean;
if (( 3 * sum) % n != 0 )
return 0 ;
// Return count
return countTriplets(arr, n, reqSum);
}
// Driver Code
public static void main (String[] args)
{
int [] arr = { 5 , 5 , 5 , 5 };
int N = arr.length;
System.out.println(count_triplets(arr, N));
}
} // This code is contributed by Shubham Singh. |
# python code for the above approach # Function to count the number of # triplets with the given sum def countTriplets(arr, n, sum ):
# Stores the final count
cnt = 0
# Map to store occurred elements
m = {}
for i in range ( 0 , n - 1 ):
for j in range (i + 1 , n):
# Check if Sum - (a + b)
# is present in map
k = sum - (arr[i] + arr[j])
if (k in m):
# Increment count
cnt + = m[k]
# Store the occurrences
if arr[i] in m:
m[arr[i]] + = 1
else :
m[arr[i]] = 1
# Return Answer
return cnt
# Function to C=find count of triplets # that can be removed without changing # arithmetic mean of the given array def count_triplets(arr, n):
# Stores sum of all elements
# of the given array
sum = 0
# Calculate the sum of the array
for i in range ( 0 , n):
sum = sum + arr[i]
# Store the arithmetic mean
mean = sum / / n
reqSum = 3 * mean
if (( 3 * sum ) % n ! = 0 ):
return 0
# Return count
return countTriplets(arr, n, reqSum)
# Driver Code if __name__ = = "__main__" :
arr = [ 5 , 5 , 5 , 5 ]
N = len (arr)
print (count_triplets(arr, N))
# This code is contributed by rakeshsahni
|
// C# code for the above approach using System;
using System.Collections.Generic;
class GFG {
// Function to count the number of
// triplets with the given sum
static int countTriplets( int [] arr, int n, int sum)
{
// Stores the final count
int cnt = 0;
// Map to store occurred elements
Dictionary< int , int > m = new Dictionary< int , int >();
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// Check if Sum - (a + b)
// is present in map
int k = sum - (arr[i] + arr[j]);
if (m.ContainsKey(k))
// Increment count
cnt += m[k];
}
// Store the occurrences
if (m.ContainsKey(arr[i]))
m[arr[i]]++;
else
m[arr[i]] = 1;
}
// Return Answer
return cnt;
}
// Function to C=find count of triplets
// that can be removed without changing
// arithmetic mean of the given array
static int count_triplets( int [] arr, int n)
{
// Stores sum of all elements
// of the given array
int sum = 0;
// Calculate the sum of the array
for ( int i = 0; i < n; i++) {
sum = sum + arr[i];
}
// Store the arithmetic mean
int mean = sum / n;
int reqSum = 3 * mean;
if ((3 * sum) % n != 0)
return 0;
// Return count
return countTriplets(arr, n, reqSum);
}
// Driver Code
public static void Main()
{
int [] arr = { 5, 5, 5, 5 };
int N = arr.Length;
Console.WriteLine(count_triplets(arr, N));
}
} // This code is contributed by ukasp. |
<script> // JavaScript code for the above approach
// Function to count the number of
// triplets with the given sum
const countTriplets = (arr, n, sum) => {
// Stores the final count
let cnt = 0;
// Map to store occurred elements
let m = {};
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
// Check if Sum - (a + b)
// is present in map
let k = sum - (arr[i] + arr[j]);
if (k in m)
// Increment count
cnt += m[k];
}
// Store the occurrences
if (arr[i] in m) m[arr[i]]++;
else m[arr[i]] = 1;
}
// Return Answer
return cnt;
}
// Function to C=find count of triplets
// that can be removed without changing
// arithmetic mean of the given array
const count_triplets = (arr, n) => {
// Stores sum of all elements
// of the given array
let sum = 0;
// Calculate the sum of the array
for (let i = 0; i < n; i++) {
sum = sum + arr[i];
}
// Store the arithmetic mean
let mean = parseInt(sum / n);
let reqSum = 3 * mean;
if ((3 * sum) % n != 0)
return 0;
// Return count
return countTriplets(arr, n, reqSum);
}
// Driver Code
let arr = [5, 5, 5, 5];
let N = arr.length;
document.write(count_triplets(arr, N));
// This code is contributed by rakeshsahni </script> |
Output:
4
Time Complexity: O(N2), since there runs a nested loop both for N times.
Auxiliary Space: O(N), since N extra space is taken for hashing values.