Given three array A[], B[] and C[] of N integers each. The task is to find the count of triplets (A[i], B[j], C[k]) such that A[i] < B[j] < C[k].
Input: A[] = {1, 5}, B[] = {2, 4}, C[] = {3, 6}
Output: 3
Triplets are (1, 2, 3), (1, 4, 6) and (1, 2, 6)
Input: A[] = {1, 1, 1}, B[] = {2, 2, 2}, C[] = {3, 3, 3}
Output: 27
Approach: Sort all the given arrays. Now fix an element say X in array B[] and for each X, the answer will be the product of the count of elements in array A[] which are less than X and the count of elements in array C[] which are greater than X. We can compute both of these counts using modified binary search.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count// of elements in arr[] which are// less than the given keyint countLessThan(int arr[], int n, int key){ int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] < key) { l = m + 1; index = m; } else { r = m - 1; } } return (index + 1);}// Function to return the count// of elements in arr[] which are// greater than the given keyint countGreaterThan(int arr[], int n, int key){ int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] <= key) { l = m + 1; } else { r = m - 1; index = m; } } if (index == -1) return 0; return (n - index);}// Function to return the count// of the required tripletsint countTriplets(int n, int* a, int* b, int* c){ // Sort all three arrays sort(a, a + n); sort(b, b + n); sort(c, c + n); int count = 0; // Iterate for all the elements of array B for (int i = 0; i < n; ++i) { int current = b[i]; int a_index = -1, c_index = -1; // Count of elements in A[] // which are less than the // chosen element from B[] int low = countLessThan(a, n, current); // Count of elements in C[] // which are greater than the // chosen element from B[] int high = countGreaterThan(c, n, current); // Update the count count += (low * high); } return count;}// Driver codeint main(){ int a[] = { 1, 5 }; int b[] = { 2, 4 }; int c[] = { 3, 6 }; int size = sizeof(a) / sizeof(a[0]); cout << countTriplets(size, a, b, c); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG { // Function to return the count // of elements in arr[] which are // less than the given key static int countLessThan(int arr[], int n, int key) { int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] < key) { l = m + 1; index = m; } else { r = m - 1; } } return (index + 1); } // Function to return the count // of elements in arr[] which are // greater than the given key static int countGreaterThan(int arr[], int n, int key) { int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] <= key) { l = m + 1; } else { r = m - 1; index = m; } } if (index == -1) return 0; return (n - index); } // Function to return the count // of the required triplets static int countTriplets(int n, int a[], int b[], int c[]) { // Sort all three arrays Arrays.sort(a) ; Arrays.sort(b); Arrays.sort(c); int count = 0; // Iterate for all the elements of array B for (int i = 0; i < n; ++i) { int current = b[i]; // Count of elements in A[] // which are less than the // chosen element from B[] int low = countLessThan(a, n, current); // Count of elements in C[] // which are greater than the // chosen element from B[] int high = countGreaterThan(c, n, current); // Update the count count += (low * high); } return count; } // Driver code public static void main(String args[]) { int a[] = { 1, 5 }; int b[] = { 2, 4 }; int c[] = { 3, 6 }; int size = a.length; System.out.println(countTriplets(size, a, b, c)); } } // This code is contributed by Arnab Kundu |
Python 3
# Python 3 implementation of the approach# Function to return the count# of elements in arr[] which are# less than the given keydef countLessThan(arr, n, key): l = 0 r = n - 1 index = -1 # Modified binary search while (l <= r): m = (l + r) // 2 if (arr[m] < key) : l = m + 1 index = m else : r = m - 1 return (index + 1)# Function to return the count# of elements in arr[] which are# greater than the given keydef countGreaterThan(arr, n, key): l = 0 r = n - 1 index = -1 # Modified binary search while (l <= r) : m = (l + r) // 2 if (arr[m] <= key) : l = m + 1 else : r = m - 1 index = m if (index == -1): return 0 return (n - index)# Function to return the count# of the required tripletsdef countTriplets(n, a, b, c): # Sort all three arrays a.sort b.sort() c.sort() count = 0 # Iterate for all the elements of array B for i in range(n): current = b[i] a_index = -1 c_index = -1 # Count of elements in A[] # which are less than the # chosen element from B[] low = countLessThan(a, n, current) # Count of elements in C[] # which are greater than the # chosen element from B[] high = countGreaterThan(c, n, current) # Update the count count += (low * high) return count# Driver codeif __name__ == "__main__": a = [ 1, 5 ] b = [ 2, 4 ] c = [ 3, 6 ] size = len(a) print( countTriplets(size, a, b, c)) # This code is contributed by ChitraNayal |
C#
// C# implementation of the approach using System;class GFG{ // Function to return the count // of elements in arr[] which are // less than the given key static int countLessThan(int []arr, int n, int key) { int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] < key) { l = m + 1; index = m; } else { r = m - 1; } } return (index + 1); } // Function to return the count // of elements in arr[] which are // greater than the given key static int countGreaterThan(int []arr, int n, int key) { int l = 0, r = n - 1; int index = -1; // Modified binary search while (l <= r) { int m = (l + r) / 2; if (arr[m] <= key) { l = m + 1; } else { r = m - 1; index = m; } } if (index == -1) return 0; return (n - index); } // Function to return the count // of the required triplets static int countTriplets(int n, int []a, int []b, int []c) { // Sort all three arrays Array.Sort(a) ; Array.Sort(b); Array.Sort(c); int count = 0; // Iterate for all the elements of array B for (int i = 0; i < n; ++i) { int current = b[i]; // Count of elements in A[] // which are less than the // chosen element from B[] int low = countLessThan(a, n, current); // Count of elements in C[] // which are greater than the // chosen element from B[] int high = countGreaterThan(c, n, current); // Update the count count += (low * high); } return count; } // Driver code public static void Main() { int []a = { 1, 5 }; int []b = { 2, 4 }; int []c = { 3, 6 }; int size = a.Length; Console.WriteLine(countTriplets(size, a, b, c)); } }// This code is contributed by AnkitRai01 |
PHP
<?php// PHP implementation of the approach// Function to return the count// of elements in arr[] which are// less than the given keyfunction countLessThan(&$arr, $n, $key){ $l = 0; $r = $n - 1; $index = -1; // Modified binary search while ($l <= $r) { $m = intval(($l + $r) / 2); if ($arr[$m] < $key) { $l = $m + 1; $index = $m; } else { $r = $m - 1; } } return ($index + 1);}// Function to return the count// of elements in arr[] which are// greater than the given keyfunction countGreaterThan(&$arr, $n, $key){ $l = 0; $r = $n - 1; $index = -1; // Modified binary search while ($l <= $r) { $m = intval(($l + $r) / 2); if ($arr[$m] <= $key) { $l = $m + 1; } else { $r = $m - 1; $index = $m; } } if ($index == -1) return 0; return ($n - $index);}// Function to return the count// of the required tripletsfunction countTriplets($n, &$a, &$b, &$c){ // Sort all three arrays sort($a); sort($b); sort($c); $count = 0; // Iterate for all the elements of array B for ($i = 0; $i < $n; ++$i) { $current = $b[$i]; $a_index = -1; $c_index = -1; // Count of elements in A[] // which are less than the // chosen element from B[] $low = countLessThan($a, $n, $current); // Count of elements in C[] // which are greater than the // chosen element from B[] $high = countGreaterThan($c, $n, $current); // Update the count $count += ($low * $high); } return $count;}// Driver code$a = array( 1, 5 );$b = array( 2, 4 );$c = array( 3, 6 );$size = sizeof($a);echo countTriplets($size, $a, $b, $c);// This code is contributed by ChitraNayal?> |
Javascript
<script>// JavaScript implementation of the approach // Function to return the count // of elements in arr[] which are // less than the given key function countLessThan(arr,n,key){ let l = 0, r = n - 1; let index = -1; // Modified binary search while (l <= r) { let m = Math.floor((l + r) / 2); if (arr[m] < key) { l = m + 1; index = m; } else { r = m - 1; } } return (index + 1); } // Function to return the count // of elements in arr[] which are // greater than the given key function countGreaterThan(arr,n,key){ let l = 0, r = n - 1; let index = -1; // Modified binary search while (l <= r) { let m = Math.floor((l + r) / 2); if (arr[m] <= key) { l = m + 1; } else { r = m - 1; index = m; } } if (index == -1) return 0; return (n - index); } // Function to return the count // of the required triplets function countTriplets(n,a,b,c){ // Sort all three arrays a.sort(function(e,f){return e-f;}) ; b.sort(function(e,f){return e-f;}) ; c.sort(function(e,f){return e-f;}) ; let count = 0; // Iterate for all the elements of array B for (let i = 0; i < n; ++i) { let current = b[i]; // Count of elements in A[] // which are less than the // chosen element from B[] let low = countLessThan(a, n, current); // Count of elements in C[] // which are greater than the // chosen element from B[] let high = countGreaterThan(c, n, current); // Update the count count += (low * high); } return count; }// Driver code let a=[1, 5 ];let b=[2, 4];let c=[3, 6 ];let size = a.length;document.write(countTriplets(size, a, b, c)); // This code is contributed by avanitrachhadiya2155</script> |
Output:
3
Time Complexity: O(nlog(n))
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
