Given an array of n elements. The task is to find number of sextuplets that satisfy the below equation such that a, b, c, d, e and f belong to the given array:
a * b + c - e = f
d
Examples:
Input : arr[] = { 1 }.
Output : 1
a = b = c = d = e = f = 1 satisfy
the equation.
Input : arr[] = { 2, 3 }
Output : 4
Input : arr[] = { 1, -1 }
Output : 24
First, reorder the equation, a * b + c = (f + e) * d.
Now, make two arrays, one for LHS (Left Hand Side) of the equation and one for the RHS (Right Hand Side) of the equation. Search each element of RHS’s array in the LHS’s array. Whenever you find a value of RHS in LHS, check how many times it is repeated in LHS and add that count to the total. Searching can be done using binary search, by sorting the LHS array.
Below is the implementation of this approach:
C++
// C++ program to count of 6 values from an array// that satisfy an equation with 6 variables#include<bits/stdc++.h>using namespace std;// Returns count of 6 values from arr[]// that satisfy an equation with 6 variablesint findSextuplets(int arr[], int n){ // Generating possible values of RHS of the equation int index = 0; int RHS[n*n*n + 1]; for (int i = 0; i < n; i++) if (arr[i]) // Checking d should be non-zero. for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) RHS[index++] = arr[i] * (arr[j] + arr[k]); // Sort RHS[] so that we can do binary search in it. sort(RHS, RHS + n); // Generating all possible values of LHS of the equation // and finding the number of occurrences of the value in RHS. int result = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { for(int k = 0; k < n; k++) { int val = arr[i] * arr[j] + arr[k]; result += (upper_bound(RHS, RHS + index, val) - lower_bound(RHS, RHS + index, val)); } } } return result;}// Driven Programint main(){ int arr[] = {2, 3}; int n = sizeof(arr)/sizeof(arr[0]); cout << findSextuplets(arr, n) << endl; return 0;} |
Java
// Java program to count of 6 values from an array // that satisfy an equation with 6 variables import java.util.Arrays; class GFG{static int upper_bound(int[] array, int length, int value) { int low = 0; int high = length; while (low < high) { final int mid = (low + high) / 2; if (value >= array[mid]) { low = mid + 1; } else { high = mid; } } return low; } static int lower_bound(int[] array, int length, int value) { int low = 0; int high = length; while (low < high) { final int mid = (low + high) / 2; if (value <= array[mid]) { high = mid; } else { low = mid + 1; } } return low; } static int findSextuplets(int[] arr, int n){ // Generating possible values of RHS of the equation int index = 0; int[] RHS = new int[n * n * n + 1]; for (int i = 0; i < n; i++) { if (arr[i] != 0) // Checking d should be non-zero. { for (int j = 0; j < n; j++) { for (int k = 0; k < n; k++) { RHS[index++] = arr[i] * (arr[j] + arr[k]); } } } } // Sort RHS[] so that we can do binary search in it. Arrays.sort(RHS); // Generating all possible values of LHS of the equation // and finding the number of occurrences of the value in RHS. int result = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < n; k++) { int val = arr[i] * arr[j] + arr[k]; result += (upper_bound(RHS, index, val)-lower_bound(RHS, index, val)); } } } return result;}// Driven Program public static void main(String[] args){ int[] arr = {2, 3}; int n = arr.length; System.out.println(findSextuplets(arr, n));}}// This code is contributed by mits |
Python3
# Python3 program to count of 6 values # from an array that satisfy an equation # with 6 variables def upper_bound(array, length, value): low = 0; high = length; while (low < high): mid = int((low + high) / 2); if (value >= array[mid]): low = mid + 1; else: high = mid; return low; def lower_bound(array, length, value): low = 0; high = length; while (low < high): mid = int((low + high) / 2); if (value <= array[mid]): high = mid; else: low = mid + 1; return low; def findSextuplets(arr, n): # Generating possible values of # RHS of the equation index = 0; RHS = [0] * (n * n * n + 1); for i in range(n): if (arr[i] != 0): # Checking d should be non-zero. for j in range(n): for k in range(n): RHS[index] = arr[i] * (arr[j] + arr[k]); index += 1; # Sort RHS[] so that we can do # binary search in it. RHS.sort(); # Generating all possible values of # LHS of the equation and finding the # number of occurrences of the value in RHS. result = 0; for i in range(n): for j in range(n): for k in range(n): val = arr[i] * arr[j] + arr[k]; result += (upper_bound(RHS, index, val) - lower_bound(RHS, index, val)); return result;# Driver Codearr = [2, 3];n = len(arr);print(findSextuplets(arr, n));# This code is contributed by mits |
C#
// C# program to count of 6 values from an array // that satisfy an equation with 6 variables using System;using System.Collections;class GFG{static int upper_bound(int[] array, int length, int value) { int low = 0; int high = length; while (low < high) { int mid = (low + high) / 2; if (value >= array[mid]) { low = mid + 1; } else { high = mid; } } return low; }static int lower_bound(int[] array, int length, int value) { int low = 0; int high = length; while (low < high) { int mid = (low + high) / 2; if (value <= array[mid]) { high = mid; } else { low = mid + 1; } } return low; } static int findSextuplets(int[] arr, int n){ // Generating possible values of RHS of the equation int index = 0; int[] RHS = new int[n * n * n + 1]; for (int i = 0; i < n; i++) { if (arr[i] != 0) // Checking d should be non-zero. { for (int j = 0; j < n; j++) { for (int k = 0; k < n; k++) { RHS[index++] = arr[i] * (arr[j] + arr[k]); } } } } // Sort RHS[] so that we can do binary search in it. Array.Sort(RHS); // Generating all possible values of LHS of the equation // and finding the number of occurrences of the value in RHS. int result = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < n; k++) { int val = arr[i] * arr[j] + arr[k]; result += (upper_bound(RHS, index, val)-lower_bound(RHS, index, val)); } } } return result;}// Driven Program static void Main(){ int[] arr = {2, 3}; int n = arr.Length; Console.WriteLine(findSextuplets(arr, n));}}// This code is contributed by mits |
PHP
<?php// PHP program to count of 6 values from // an array that satisfy an equation // with 6 variables function upper_bound($array, $length, $value){ $low = 0; $high = $length; while ($low < $high) { $mid = (int)(($low + $high) / 2); if ($value >= $array[$mid]) $low = $mid + 1; else $high = $mid; } return $low;}function lower_bound($array, $length, $value){ $low = 0; $high = $length; while ($low < $high) { $mid = (int)(($low + $high) / 2); if ($value <= $array[$mid]) $high = $mid; else $low = $mid + 1; } return $low;}// Returns count of 6 values from arr[]// that satisfy an equation with 6 variablesfunction findSextuplets($arr, $n){ // Generating possible values of // RHS of the equation $index = 0; $RHS = array_fill(0, $n * $n * $n + 1, 0); for ($i = 0; $i < $n; $i++) if ($arr[$i] != 0) // Checking d should be non-zero. for ($j = 0; $j < $n; $j++) for ($k = 0; $k < $n; $k++) $RHS[$index++] = $arr[$i] * ($arr[$j] + $arr[$k]); // Sort RHS[] so that we can do // binary search in it. sort($RHS); // Generating all possible values of LHS // of the equation and finding the number // of occurrences of the value in RHS. $result = 0; for ($i = 0; $i < $n; $i++) for ($j = 0; $j < $n; $j++) for ($k = 0; $k < $n; $k++) { $val = $arr[$i] * $arr[$j] + $arr[$k]; $result += (upper_bound($RHS, $index, $val) - lower_bound($RHS, $index, $val)); } return $result;}// Driver Code$arr = array(2, 3);$n = count($arr);print(findSextuplets($arr, $n));// This code is contributed by mits?> |
Javascript
<script>// Javascript program to count of// 6 values from an array // that satisfy an equation with // 6 variables function upper_bound(array , length , value){ var low = 0; var high = length; while (low < high) { var mid = parseInt((low + high) / 2); if (value >= array[mid]) { low = mid + 1; } else { high = mid; } } return low; } function lower_bound(array , length , value) { var low = 0; var high = length; while (low < high) { var mid = parseInt((low + high) / 2); if (value <= array[mid]) { high = mid; } else { low = mid + 1; } } return low; } function findSextuplets(arr , n){ // Generating possible values of // RHS of the equation var index = 0; var RHS = Array.from({length: n * n * n + 1}, (_, i) => 0); for (i = 0; i < n; i++) { // Checking d should be non-zero. if (arr[i] != 0) { for (j = 0; j < n; j++) { for (k = 0; k < n; k++) { RHS[index++] = arr[i] * (arr[j] + arr[k]); } } } } // Sort RHS so that we can do // binary search in it. RHS.sort((a,b)=>a-b); // Generating all possible values // of LHS of the equation // and finding the number of occurrences // of the value in RHS. var result = 0; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { for (k = 0; k < n; k++) { var val = arr[i] * arr[j] + arr[k]; result += (upper_bound(RHS, index, val)- lower_bound(RHS, index, val)); } } } return result;}// Driven Program var arr = [2, 3];var n = arr.length;document.write(findSextuplets(arr, n));// This code is contributed by 29AjayKumar </script> |
Output
4
Time Complexity : O(N3 log N)
Auxiliary Space: O(N3) as it is using extra space for array RHS
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