Count ways to choose Triplets of Pairs such that either first or second values are distinct
Last Updated :
23 Jan, 2023
Given an array of pairs arr[] of size N (N ? 3) where each element of pair is at most N and each pair is unique, the task is to determine the number of ways to select triplets from the given N pairs that satisfy at least one of the following conditions:
- The first value (a) of each pair should be distinct.
- The second value (b) of each pair should be distinct.
Examples:
Input: N = 4, arr[] = {{2, 4}, {3, 4}, {2, 1}, {1, 4}}
Output: 2
Explanation: The valid triplets are {{2, 4}. {3, 4}, {1, 4}}
and {{3, 4}, {2, 1}, {1, 4}}
Input: N = 5, arr[] = {{1, 4}, {2, 4}, {3, 3}, {4, 2}, {1, 2}}
Output: 8
Approach: Use the below idea to solve the problem
The total number of triplets possible is n*(n - 1)*(n - 2)/6. So we can count the number of invalid pairs for the triplets and subtract that count from the total triplets possible
Follow the below steps to solve the problem:
- Create two 2D vectors to keep a record of the pairs in the given array.
- Store pairs with the same first value in the first 2D vector and pairs with the same second value in the second 2D vector.
- Initialize ans = N*(N - 1)*(N - 2)/6, the count of total pairs possible.
- Traverse over first 2D vector and calculate the number of triplets that are not possible with each value in the given array.
- Subtract the count of invalid pairs from the ans.
- Return ans as the final answer.
Below is the implementation of the ab
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
void solve(vector<vector<int> >& details, int N)
{
// Vector to store the pairs with
// common values
vector<int> adj_a[N + 1], adj_b[N + 1];
for (int i = 0; i < N; i++) {
// Storing pairs with same first values
adj_a[details[i][0]].push_back(details[i][1]);
// Storing pairs with same second values
adj_b[details[i][1]].push_back(details[i][0]);
}
// Total ways to choose a triplet from
// n pairs is nc3
int ans = (N * (N - 1) * (N - 2)) / 6;
for (int i = 1; i < N + 1; i++) {
// Size to keep record of the elements
// with same first values
int size = adj_a[i].size();
for (auto& u : adj_a[i]) {
// Count of pairs that forms
// invalid triplets
int same_cnt = adj_b[u].size() - 1;
// Subtracting the invalid values
ans -= (size - 1) * same_cnt;
}
}
// Printing the answer
cout << ans << endl;
}
// Driver's code
int main()
{
// Given input
int n = 5;
vector<vector<int> > details{
{ 1, 4 }, { 2, 4 }, { 3, 3 }, { 4, 2 }, { 1, 2 }
};
// Function Call
solve(details, n);
return 0;
}
// Java code for the above approach
import java.io.*;
import java.util.*;
class GFG {
@SuppressWarnings("unchecked")
public static void solve(int details[][], int N)
{
// ArrayList to store the pairs with
// common values
ArrayList<Integer>[] adj_a = new ArrayList[N + 1];
ArrayList<Integer>[] adj_b = new ArrayList[N + 1];
for (int i = 0; i < N + 1; i++) {
adj_a[i] = new ArrayList<Integer>();
adj_b[i] = new ArrayList<Integer>();
}
// ArrayList<ArrayList<Integer>> adj_a=new
// ArrayList<ArrayList<Integer>>(N+1);
// ArrayList<ArrayList<Integer>> adj_b=new
// ArrayList<ArrayList<Integer>>(N+1);
for (int i = 0; i < N; i++) {
// Storing pairs with same first values
adj_a[details[i][0]].add(details[i][1]);
// Storing pairs with same second values
adj_b[details[i][1]].add(details[i][0]);
}
// Total ways to choose a triplet from
// n pairs is nc3
int ans = (N * (N - 1) * (N - 2)) / 6;
for (int i = 1; i < N + 1; i++) {
// Size to keep record of the elements
// with same first values
int size = adj_a[i].size();
for (Integer u : adj_a[i]) {
// Count of pairs that forms
// invalid triplets
int same_cnt = adj_b[u].size() - 1;
// Subtracting the invalid values
ans -= (size - 1) * same_cnt;
}
}
// Printing the answer
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
// Given input
int n = 5;
int details[][] = {
{ 1, 4 }, { 2, 4 }, { 3, 3 }, { 4, 2 }, { 1, 2 }
};
// Function Call
solve(details, n);
}
}
// This code is contributed by Rohit Pradhan
# Python code for the above approach
def solve(details, N):
adj_a = [0]*(N+1)
adj_b = [0]*(N+1)
for i in range(N+1):
adj_a[i] = []
adj_b[i] = []
for i in range(N):
# Storing pairs with same first values
adj_a[details[i][0]].append(details[i][1])
# Storing pairs with same second values
adj_b[details[i][1]].append(details[i][0])
# Total ways to choose a triplet from n pairs is nC3
ans = (N * (N-1)*(N-2))/6
for i in range(1, N+1):
# Size to keep record of the elements with same first values
size = len(adj_a[i])
for u in adj_a[i]:
# Count of pairs that forms invalid triplets
same_cnt = len(adj_b[u]) - 1
# Subtracting the invalid values
ans -= (size-1) * same_cnt
# Printing the answer
print(int(ans))
n = 5
details = [[1, 4],
[2, 4],
[3, 3],
[4, 2],
[1, 2]]
# Function call
solve(details, n)
# This code is contributed by lokeshmvs21.
using System;
using System.Collections.Generic;
public class GFG {
public static void solve(int[, ] details, int N)
{
// ArrayList to store the pairs with
// common values
/*List<int>[] adj_a = new List[N + 1];
List<int>[] adj_b = new List[N + 1];*/
List<int>[] adj_a = new List<int>[ N + 1 ];
List<int>[] adj_b = new List<int>[ N + 1 ];
for (int i = 0; i < N + 1; i++) {
adj_a[i] = new List<int>();
adj_b[i] = new List<int>();
}
for (int i = 0; i < N; i++) {
// Storing pairs with same first values
adj_a[details[i, 0]].Add(details[i, 1]);
// Storing pairs with same second values
adj_b[details[i, 1]].Add(details[i, 0]);
}
// Total ways to choose a triplet from
// n pairs is nc3
int ans = (N * (N - 1) * (N - 2)) / 6;
for (int i = 1; i < N + 1; i++) {
// Size to keep record of the elements
// with same first values
int size = adj_a[i].Count;
for (int u = 0; u < adj_a[i].Count; u++) {
// Count of pairs that forms
// invalid triplets
int same_cnt = adj_b[adj_a[i][u]].Count - 1;
// Subtracting the invalid values
ans -= (size - 1) * same_cnt;
}
}
// Printing the answer
Console.WriteLine(ans);
}
static public void Main()
{
// Given input
int n = 5;
int[, ] details = {
{ 1, 4 }, { 2, 4 }, { 3, 3 }, { 4, 2 }, { 1, 2 }
};
// Function Call
solve(details, n);
}
}
// This code is contributed by akashish__
// JavaScript code for the above approach
function solve(details, N)
{
// Vector to store the pairs with
// common values
let adj_a = new Array(N + 1);
let adj_b = new Array(N + 1);
for (let i = 0; i <= N; i++) {
adj_a[i] = [];
adj_b[i] = [];
}
for (let i = 0; i < N; i++) {
// Storing pairs with same first values
adj_a[details[i][0]].push(details[i][1]);
// Storing pairs with same second values
adj_b[details[i][1]].push(details[i][0]);
}
// Total ways to choose a triplet from
// n pairs is nc3
let ans = (N * (N - 1) * (N - 2)) / 6;
for (let i = 1; i < N + 1; i++) {
// Size to keep record of the elements
// with same first values
let size = adj_a[i].length;
for (let u of adj_a[i]) {
// Count of pairs that forms
// invalid triplets
let same_cnt = adj_b[u].length - 1;
// Subtracting the invalid values
ans -= (size - 1) * same_cnt;
}
}
// Printing the answer
console.log(ans);
}
// Driver's code
// Given input
let n = 5;
let details = [
[1, 4],
[2, 4],
[3, 3],
[4, 2],
[1, 2],
];
// Function Call
solve(details, n);
// This code is contributed by Ishankhandelwals.
above approach:
Time Complexity: O(N), the loop run once for every first value.
Auxiliary Space: O(N), where N is the size of the adjacency list
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