Given two arrays H and S. The array H[] contains the length of the hypotenuse and the array S[] contains Area of a right-angled triangle. The task is to find all possible pairs of (H, S) such that we can construct a right-angled triangle with hypotenuse H and area S.
Examples:
Input : H[] = {1, 6, 4} ; S[] = {23, 3, 42, 14}
Output : 2
Possible pairs are {6, 3} {4, 3}
Input : H[] = {1, 6, 4, 3} ; S[] = {23, 3, 42, 5}
Output : 3
Possible pairs are {6, 3} {6, 5} {4, 3}
Say, 
= Base of Right Angled Triangle 
= Height of the Right Angled Triangle
Therefore,
area S = (a*b)/2 or, 4*S*S=a*a*b*b
Also,
a2 + b2 = H2
Therefore,
4*S2 = a2(H2-a2)
Solving this quadratic equation in a2 and putting discriminant>=0 (condition for a to exist). We will get,
H2 >= 4*S For a right-angled triangle to exist with hypotenuse H and area S.
Naive Approach: The naive approach is to find all possible pairs of (H,S) and check if they satisfy the condition, H2 >= 4*S. Count the number of pairs that satisfies this condition and print the count.
Below is the implementation of the naive approach:
C++
#include <iostream>using namespace std;// Function to check the conditionbool check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsint findPairs(int H[], int n, int S[], int m){ int count = 0; // Checking all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (check(H[i], S[j])) count++; } } return count;}// Driver codeint main(){ int H[] = { 1, 6, 4 }; int n = sizeof(H)/sizeof(H[0]); int S[] = { 23, 3, 42, 14 }; int m = sizeof(S)/sizeof(S[0]); cout<<findPairs(H, n, S, m); return 0;} |
Java
class GFG {// Function to check the conditionstatic boolean check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsstatic int findPairs(int H[], int n, int S[], int m){ int count = 0; // Checkinag all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (check(H[i], S[j])) count++; } } return count;}// Driver codepublic static void main(String args[]){ int H[] = { 1, 6, 4 }; int n = H.length; int S[] = { 23, 3, 42, 14 }; int m = S.length; System.out.println(findPairs(H, n, S, m));}}// This code is contributed // by ankita_saini |
Python3
# Python 3 implementation # of above approach# Function to check the condition def check(H, S) : # Condition for triangle to exist return H * H >= 4 * S# Function to find all pairsdef findPairs(H, n, S, m): count = 0 # Checking all possible pairs for i in range(n) : for j in range(m) : if check(H[i], S[j]) : count += 1 return count# Driver Codeif __name__ == "__main__" : H = [ 1, 6, 4] n = len(H) S = [ 23, 3, 42, 14] m = len(S) # function calling print(findPairs(H, n, S,m)) # This code is contributed by ANKITRAI1 |
C#
// C# implementation of above approachusing System;class GFG {// Function to check the conditionstatic bool check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsstatic int findPairs(int[] H, int n, int[] S, int m){ int count = 0; // Checkinag all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (check(H[i], S[j])) count++; } } return count;}// Driver codepublic static void Main(){ int[] H = { 1, 6, 4 }; int n = H.Length; int[] S = { 23, 3, 42, 14 }; int m = S.Length; Console.Write(findPairs(H, n, S, m));}}// This code is contributed// by ChitraNayal |
PHP
<?php// PHP implementation of above approach// Function to check the conditionfunction check($H, $S){ // Condition for triangle to exist return $H * $H >= 4 * $S;}// Function to find all pairsfunction findPairs($H, $n, $S, $m){ $count = 0; // Checking all possible pairs for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $m; $j++) { if (check($H[$i], $S[$j])) $count++; } } return $count;}// Driver code$H = array( 1, 6, 4 );$n = count($H);$S = array( 23, 3, 42, 14 );$m = count($S);echo findPairs($H, $n, $S, $m); // This code is contributed by mits?> |
Javascript
<script>// Function to check the condition function check(H , S) { // Condition for triangle to exist return H * H >= 4 * S; } // Function to find all pairs function findPairs(H , n , S , m) { var count = 0; // Checkinag all possible pairs for (let i = 0; i < n; i++) { for (let j = 0; j < m; j++) { if (check(H[i], S[j])) count++; } } return count; } // Driver code var H = [ 1, 6, 4 ]; var n = H.length; var S = [ 23, 3, 42, 14 ]; var m = S.length; document.write(findPairs(H, n, S, m));// This code is contributed by Rajput-Ji </script> |
Output:
2
Time Complexity: O(n*m) where n and m are the sizes of the array H and S respectively.
Auxiliary Space: O(1)
Efficient Approach: An efficient approach is to sort both the arrays available in increasing order. Then, for every possible length of the hypotenuse, apply Binary search to find the maximum area which satisfies the necessary condition.
Say, after the Binary search maximum possible area is available at index 4 in the array S[]. Then we can form 4 such possible pairs since all area less than that at index 4 will also be satisfying the condition.
Below is the implementation of the efficient approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to check the conditionbool check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsint findPairs(int H[], int n, int S[], int m){ int count = 0; // Sort both the arrays sort(H, H + n); sort(S, S + m); // To keep track of last possible Area int index = -1; for (int i = 0; i < n; i++) { // Apply Binary Search for // each Hypotenuse Length int start = 0; int end = m - 1; while (start <= end) { int mid = start + (end - start) / 2; if (check(H[i], S[mid])) { index = mid; start = mid + 1; } else { end = mid - 1; } } // Check if we get any // possible Area or Not if (index != -1) { // All area less than area[index] // satisfy property count += index + 1; } } return count;}// Driver codeint main(){ int H[] = { 1, 6, 4 }; int n = sizeof(H)/sizeof(H[0]); int S[] = { 23, 3, 42, 14 }; int m = sizeof(S)/sizeof(S[0]); cout<<findPairs(H, n, S, m); return 0;} |
Java
/*package whatever //do not write package name here */import java.util.Arrays; import java.io.*;class GFG { // Function to check the conditionstatic boolean check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsstatic int findPairs(int H[], int n, int S[], int m){ int count = 0; // Sort both the arrays Arrays.sort(H); Arrays.sort(S); // To keep track of last possible Area int index = -1; for (int i = 0; i < n; i++) { // Apply Binary Search for // each Hypotenuse Length int start = 0; int end = m - 1; while (start <= end) { int mid = start + (end - start) / 2; if (check(H[i], S[mid])) { index = mid; start = mid + 1; } else { end = mid - 1; } } // Check if we get any // possible Area or Not if (index != -1) { // All area less than area[index] // satisfy property count += index + 1; } } return count;}// Driver code public static void main (String[] args) { int H[] = { 1, 6, 4 }; int n = H.length; int S[] = { 23, 3, 42, 14 }; int m = S.length; System.out.println(findPairs(H, n, S, m)); } // This code is contributed // by ajit...} |
Python3
# Function to check the condition def check(H, S): # Condition for triangle to exist return H * H >= 4 * S;# Function to find all pairs def findPairs(H, n, S, m): count = 0; # Sort both the arrays H.sort(); S.sort(); # To keep track of last possible Area index = -1; for i in range(n): # Apply Binary Search for # each Hypotenuse Length start = 0; end = m - 1; while (start <= end): mid = int(start + (end - start) / 2); if (check(H[i], S[mid])): index = mid; start = mid + 1; else: end = mid - 1; # Check if we get any possible # Area or Not if (index != -1): # All area less than area[index] # satisfy property count += index + 1; return count; # Driver code H = [ 1, 6, 4 ]; n = len(H); S= [ 23, 3, 42, 14 ]; m = len(S); print(findPairs(H, n, S, m));# This code is contributed by mits |
C#
/*package whatever //do not write package name here */using System;public class GFG{ // Function to check the conditionstatic bool check(int H, int S){ // Condition for triangle to exist return H * H >= 4 * S;}// Function to find all pairsstatic int findPairs(int []H, int n, int []S, int m){ int count = 0; // Sort both the arrays Array.Sort(H); Array.Sort(S); // To keep track of last possible Area int index = -1; for (int i = 0; i < n; i++) { // Apply Binary Search for // each Hypotenuse Length int start = 0; int end = m - 1; while (start <= end) { int mid = start + (end - start) / 2; if (check(H[i], S[mid])) { index = mid; start = mid + 1; } else { end = mid - 1; } } // Check if we get any // possible Area or Not if (index != -1) { // All area less than area[index] // satisfy property count += index + 1; } } return count;}// Driver code static public void Main (){ int []H = { 1, 6, 4 }; int n = H.Length; int []S = { 23, 3, 42, 14 }; int m = S.Length; Console.WriteLine(findPairs(H, n, S, m)); } // This code is contributed // by akt_mit...} |
PHP
<?php// Function to check the conditionfunction check($H, $S){ // Condition for triangle to exist return $H * $H >= 4 * $S;}// Function to find all pairsfunction findPairs($H, $n, $S, $m){ $count = 0; // Sort both the arrays sort($H); sort($S); // To keep track of last possible Area $index = -1; for ($i = 0; $i < $n; $i++) { // Apply Binary Search for // each Hypotenuse Length $start = 0; $end = $m - 1; while ($start <= $end) { $mid = $start + (int)($end - $start) / 2; if (check($H[$i], $S[$mid])) { $index = $mid; $start = $mid + 1; } else { $end = $mid - 1; } } // Check if we get any possible // Area or Not if ($index != -1) { // All area less than area[index] // satisfy property $count += $index + 1; } } return $count;}// Driver code$H = array( 1, 6, 4 );$n = sizeof($H);$S = array(23, 3, 42, 14 );$m = sizeof($S);echo findPairs($H, $n, $S, $m);// This code is contributed by Sach_Code?> |
Javascript
<script> // Function to check the condition function check(H, S) { // Condition for triangle to exist return H * H >= 4 * S; } // Function to find all pairs function findPairs(H, n, S, m) { let count = 0; // Sort both the arrays H.sort(function(a, b){return a - b}); S.sort(function(a, b){return a - b}); // To keep track of last possible Area let index = -1; for (let i = 0; i < n; i++) { // Apply Binary Search for // each Hypotenuse Length let start = 0; let end = m - 1; while (start <= end) { let mid = start + parseInt((end - start) / 2, 10); if (check(H[i], S[mid])) { index = mid; start = mid + 1; } else { end = mid - 1; } } // Check if we get any // possible Area or Not if (index != -1) { // All area less than area[index] // satisfy property count += index + 1; } } return count; } let H = [ 1, 6, 4 ]; let n = H.length; let S = [ 23, 3, 42, 14 ]; let m = S.length; document.write(findPairs(H, n, S, m)); </script> |
Output:
2
Time Complexity: O(n*log(n)+m*log(m)+n*log(m)) where n and m are the sizes of the array H and S respectively. Here n*log(n) and m*log(m) are for sorting the array and n*log(m) is for traversing and applying binary search each time.
Auxiliary Space: O(1)
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