Count the triplets such that A[i] < B[j] < C[k]
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)
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