Given a array a[] of non-negative integers. Count the number of pairs (i, j) in the array such that a[i] + a[j] < a[i]*a[j]. (the pair (i, j) and (j, i) are considered same and i should not be equal to j)
Examples:
Input : a[] = {3, 4, 5}
Output : 3
Pairs are (3, 4) , (4, 5) and (3,5)
Input : a[] = {1, 1, 1}
Output : 0
Naive approach: For each value a[i] count the number of a[j] (i > j) such that a[i]*a[j] > a[i] + a[j]
Implementation:
C++
// Naive C++ program to count number of pairs// such that their sum is more than product.#include<bits/stdc++.h>using namespace std;// Returns the number of valid pairsint countPairs (int arr[], int n){ int ans = 0; // initializing answer // Traversing the array. For each array // element, checking its predecessors that // follow the condition for (int i = 0; i<n; i++) for (int j = i-1; j>= 0; j--) if (arr[i]*arr[j] > arr[i] + arr[j]) ans++; return ans;}// Driver functionint main(){ int arr[] = {3, 4, 5}; int n = sizeof(arr)/sizeof(arr[0]); cout << countPairs(arr, n); return 0;} |
Java
// Naive java program to count number of pairs// such that their sum is more than product.import java.*;public class GFG { // Returns the number of valid pairs static int countPairs (int arr[], int n) { int ans = 0; // initializing answer // Traversing the array. For each array // element, checking its predecessors that // follow the condition for (int i = 0; i<n; i++) for (int j = i-1; j>= 0; j--) if (arr[i]*arr[j] > arr[i] + arr[j]) ans++; return ans; } // Driver code public static void main(String args[]) { int arr[] = {3, 4, 5}; int n = arr.length; System.out.println(countPairs(arr, n)); }}// This code is contributed by Sam007 |
Python3
# Naive Python program to count number# of pairs such that their sum is more# than product.# Returns the number of valid pairsdef countPairs(arr, n): # initializing answer ans = 0 # Traversing the array. For each # array element, checking its # predecessors that follow the # condition for i in range(0, n): j = i-1 while(j >= 0): if (arr[i] * arr[j] > arr[i] + arr[j]): ans = ans + 1 j = j - 1 return ans # Driver program to test above function.arr = [3, 4, 5]n = len(arr) k = countPairs(arr, n)print(k) # This code is contributed by Sam007. |
C#
// Naive C# program to count number of pairs// such that their sum is more than product.using System; public class GFG { // Returns the number of valid pairs static int countPairs (int []arr, int n) { int ans = 0; // initializing answer // Traversing the array. For each array // element, checking its predecessors that // follow the condition for (int i = 0; i<n; i++) for (int j = i-1; j>= 0; j--) if (arr[i]*arr[j] > arr[i] + arr[j]) ans++; return ans; } // driver program public static void Main() { int []arr = {3, 4, 5}; int n = arr.Length; Console.Write( countPairs(arr, n)); }}// This code is contributed by Sam007 |
PHP
<?php// Naive PHP program to // count number of pairs// such that their sum // is more than product.// Returns the number// of valid pairsfunction countPairs ($arr, $n){ // initializing answer $ans = 0; // Traversing the array. // For each array // element, checking // its predecessors that // follow the condition for ($i = 0; $i < $n; $i++) for ($j = $i - 1; $j >= 0; $j--) if ($arr[$i] * $arr[$j] > $arr[$i] + $arr[$j]) $ans++; return $ans;}// Driver Code$arr = array(3, 4, 5);$n = sizeof($arr);echo(countPairs($arr, $n));// This code is contributed by Ajit.?> |
Javascript
<script> // Naive Javascript program to count number of pairs // such that their sum is more than product. // Returns the number of valid pairs function countPairs(arr, n) { let ans = 0; // initializing answer // Traversing the array. For each array // element, checking its predecessors that // follow the condition for (let i = 0; i<n; i++) for (let j = i-1; j>= 0; j--) if (arr[i]*arr[j] > (arr[i] + arr[j])) ans++; return ans; } let arr = [3, 4, 5]; let n = arr.length; document.write( countPairs(arr, n));</script> |
Output
3
Efficient approach:
When a[i] = 0 : a[i]*a[j] = 0 and a[i] + a[j] >= 0 so if a[i] = 0 no pairs can be found.
When a[i] = 1 : a[i]*a[j] = a[j] and a[i] + a[j] = 1 + a[j], so no pairs can be found when a[i] = 1
When a[i] = 2 and a[j] = 2 : a[i]*a[j] = a[i] + a[j] = 4
When a[i] = 2 and a[j] > 2 or a[i] > 2 and a[j] >= 2 : All such pairs are valid.
To solve this problem, count the number of 2s in the array say twoCount. Count the numbers greater than 2 in the array say twoGreaterCount. Answer will be twoCount * twoGreaterCount + twoGreaterCount * (twoGreaterCount-1)/2
Implementation:
C++
// C++ implementation of efficient approach// to count valid pairs.#include<bits/stdc++.h>using namespace std;// returns the number of valid pairsint CountPairs (int arr[], int n){ // traversing the array, counting the // number of 2s and greater than 2 // in array int twoCount = 0, twoGrCount = 0; for (int i = 0; i<n; i++) { if (arr[i] == 2) twoCount++; else if (arr[i] > 2) twoGrCount++; } return twoCount*twoGrCount + (twoGrCount*(twoGrCount-1))/2;}// Driver functionint main(){ int arr[] = {3, 4, 5}; int n = sizeof(arr)/sizeof(arr[0]); cout << CountPairs(arr, n); return 0;} |
Java
// Java implementation of efficient approach// to count valid pairs.import java.*;public class GFG { // Returns the number of valid pairs static int countPairs (int arr[], int n) { // traversing the array, counting the // number of 2s and greater than 2 // in array int twoCount = 0, twoGrCount = 0; for (int i = 0; i<n; i++) { if (arr[i] == 2) twoCount++; else if (arr[i] > 2) twoGrCount++; } return twoCount*twoGrCount + (twoGrCount*(twoGrCount-1))/2; } // Driver code public static void main(String args[]) { int arr[] = {3, 4, 5}; int n = arr.length; System.out.println(countPairs(arr, n)); }}// This code is contributed by Sam007 |
Python3
# python implementation of efficient approach# to count valid pairs.# returns the number of valid pairsdef CountPairs (arr,n): # traversing the array, counting the # number of 2s and greater than 2 # in array twoCount = 0 twoGrCount = 0 for i in range(0, n): if (arr[i] == 2): twoCount += 1 else if (arr[i] > 2): twoGrCount += 1 return ((twoCount * twoGrCount) + (twoGrCount * (twoGrCount - 1)) / 2)# Driver functionarr = [3, 4, 5]n = len(arr)print( CountPairs(arr, n))# This code is contributed by Sam007. |
C#
// C# implementation of efficient approach// to count valid pairs.using System; public class GFG { // Returns the number of valid pairs static int countPairs (int []arr, int n) { // traversing the array, counting the // number of 2s and greater than 2 // in array int twoCount = 0, twoGrCount = 0; for (int i = 0; i<n; i++) { if (arr[i] == 2) twoCount++; else if (arr[i] > 2) twoGrCount++; } return twoCount*twoGrCount + (twoGrCount*(twoGrCount-1))/2; } // driver program public static void Main() { int []arr = {3, 4, 5}; int n = arr.Length; Console.Write( countPairs(arr, n)); }}// This code is contributed by Sam007 |
PHP
<?php// PHP implementation of// efficient approach// to count valid pairs.// returns the number// of valid pairsfunction CountPairs ($arr, $n){ // traversing the array, counting // the number of 2s and greater // than 2 in array $twoCount = 0; $twoGrCount = 0; for ($i = 0; $i < $n; $i++) { if ($arr[$i] == 2) $twoCount++; else if ($arr[$i] > 2) $twoGrCount++; } return $twoCount * $twoGrCount + ($twoGrCount * ($twoGrCount - 1)) / 2;}// Driver Code$arr = array(3, 4, 5);$n = sizeof($arr);echo(CountPairs($arr, $n));// This code is contributed by Ajit.?> |
Javascript
<script> // Javascript implementation of efficient approach to count valid pairs. // returns the number of valid pairs function CountPairs(arr, n) { // traversing the array, counting the // number of 2s and greater than 2 // in array let twoCount = 0, twoGrCount = 0; for (let i = 0; i<n; i++) { if (arr[i] == 2) twoCount++; else if (arr[i] > 2) twoGrCount++; } return twoCount*twoGrCount + parseInt((twoGrCount*(twoGrCount-1))/2, 10); } let arr = [3, 4, 5]; let n = arr.length; document.write(CountPairs(arr, n));</script> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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