# Count unordered pairs (i,j) such that product of a[i] and a[j] is power of two

Given an array of N elements. The task is to count **unordered pairs** (i, j) in the array such that the product of a[i] and a[j] can be expressed as a power of two.**Examples**:

Input: arr[] = {2, 3, 4, 8, 10}Output: 3Explanation:The pair of array element will be (2, 4), (2, 8), (4, 8) whose product are 8, 16, 32 respectively which can be expressed as power of 2, like 2^3, 2^4, 2^5.Input: arr[] = { 2, 5, 8, 16, 128 }Output: 6

If you multiply and and their product become , then z=x*y, now if it’s possible to express as power of two then it can be proved that both and can be expressed as power of two. Basically **z= 2 ^{a} = 2^{(b+c)} = 2^{b} * 2^{c} = x * y**, where and both

can hold a minimum value 0.

So now we have to count the number of elements in the array which can be expressed as a power of two. If the count is k, then answer will be kC2 = k*(k-1)/2, as we need the count of unordered pairs.

Below is the implementation of above approach:

## C++

`// C++ program to Count unordered pairs (i, j)` `// in array such that product of a[i] and a[j]` `// can be expressed as power of two` `#include <bits/stdc++.h>` `using` `namespace` `std;` `/* Function to check if x is power of 2*/` `bool` `isPowerOfTwo(` `int` `x)` `{` ` ` `/* First x in the below expression is` ` ` `for the case when x is 0 */` ` ` `return` `x && (!(x&(x-1)));` `}` `// Function to Count unordered pairs` `void` `Count_pairs(` `int` `a[], ` `int` `n)` `{` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// is a number can be expressed` ` ` `// as power of two` ` ` `if` `(isPowerOfTwo(a[i]))` ` ` `count++;` ` ` `}` ` ` `// count total number` ` ` `// of unordered pairs` ` ` `int` `ans = (count * (count - 1)) / 2;` ` ` `cout << ans << ` `"\n"` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `a[] = { 2, 5, 8, 16, 128 };` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]);` ` ` `Count_pairs(a, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to Count unordered pairs (i, j)` `// in array such that product of a[i] and a[j]` `// can be expressed as power of two` `import` `java.io.*;` `class` `GFG {` `/* Function to check if x is power of 2*/` `static` `boolean` `isPowerOfTwo(` `int` `x)` `{` `/* First x in the below expression is` ` ` `for the case when x is 0 */` `return` `(x >` `0` `&& (!((x&(x-` `1` `))>` `0` `)));` `}` `// Function to Count unordered pairs` `static` `void` `Count_pairs(` `int` `a[], ` `int` `n)` `{` ` ` `int` `count = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `// is a number can be expressed` ` ` `// as power of two` ` ` `if` `(isPowerOfTwo(a[i]))` ` ` `count++;` ` ` `}` ` ` `// count total number` ` ` `// of unordered pairs` ` ` `int` `ans = (count * (count - ` `1` `)) / ` `2` `;` ` ` `System.out.println( ans);` `}` `// Driver code` ` ` `public` `static` `void` `main (String[] args) {` ` ` `int` `a[] = { ` `2` `, ` `5` `, ` `8` `, ` `16` `, ` `128` `};` ` ` `int` `n = a.length;` ` ` `Count_pairs(a, n);` ` ` `}` `}` `// This code is contributed` `// by shs` |

## Python 3

`# Python3 program to Count unordered pairs` `# (i, j) in array such that product of a[i]` `# and a[j] can be expressed as power of two` `# Function to check if x is power of 2` `def` `isPowerOfTwo(x) :` ` ` `# First x in the below expression` ` ` `# is for the case when x is 0` ` ` `return` `(x ` `and` `(` `not` `(x & (x ` `-` `1` `))))` `# Function to Count unordered pairs` `def` `Count_pairs(a, n) :` ` ` `count ` `=` `0` ` ` `for` `i ` `in` `range` `(n) :` ` ` `# is a number can be expressed` ` ` `# as power of two` ` ` `if` `isPowerOfTwo(a[i]) :` ` ` `count ` `+` `=` `1` ` ` `# count total number` ` ` `# of unordered pairs` ` ` `ans ` `=` `(count ` `*` `(count ` `-` `1` `)) ` `/` `2` ` ` `print` `(ans)` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `a ` `=` `[ ` `2` `, ` `5` `, ` `8` `, ` `16` `, ` `128` `]` ` ` `n ` `=` `len` `(a)` ` ` `Count_pairs(a, n)` ` ` `# This code is contributed by ANKITRAI1` |

## C#

`// C# program to Count unordered pairs (i, j)` `// in array such that product of a[i] and a[j]` `// can be expressed as power of two` `using` `System;` `public` `class` `GFG{` ` ` ` ` `/* Function to check if x is power of 2*/` `static` `bool` `isPowerOfTwo(` `int` `x)` `{` `/* First x in the below expression is` ` ` `for the case when x is 0 */` `return` `(x >0&& (!((x&(x-1))>0)));` `}` `// Function to Count unordered pairs` `static` `void` `Count_pairs(` `int` `[]a, ` `int` `n)` `{` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// is a number can be expressed` ` ` `// as power of two` ` ` `if` `(isPowerOfTwo(a[i]))` ` ` `count++;` ` ` `}` ` ` `// count total number` ` ` `// of unordered pairs` ` ` `int` `ans = (count * (count - 1)) / 2;` ` ` `Console.WriteLine( ans);` `}` `// Driver code` ` ` `static` `public` `void` `Main (){` ` ` `int` `[]a = { 2, 5, 8, 16, 128 };` ` ` `int` `n = a.Length;` ` ` `Count_pairs(a, n);` ` ` `}` `}` `// This code is contributed` `// by Sach_Code` |

## PHP

`<?php` `// PHP program to Count unordered` `// pairs (i, j) in array such that` `// product of a[i] and a[j] can be` `// expressed as power of two` `/* Function to check if x is power of 2*/` `function` `isPowerOfTwo(` `$x` `)` `{` ` ` `/* First x in the below expression is` ` ` `for the case when x is 0 */` ` ` `return` `(` `$x` `&& (!(` `$x` `& (` `$x` `- 1))));` `}` `// Function to Count unordered pairs` `function` `Count_pairs(` `$a` `, ` `$n` `)` `{` ` ` `$count` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `// is a number can be expressed` ` ` `// as power of two` ` ` `if` `(isPowerOfTwo(` `$a` `[` `$i` `]))` ` ` `$count` `++;` ` ` `}` ` ` `// count total number` ` ` `// of unordered pairs` ` ` `$ans` `= (` `$count` `* (` `$count` `- 1)) / 2;` ` ` `echo` `$ans` `, ` `"\n"` `;` `}` `// Driver code` `$a` `= ` `array` `( 2, 5, 8, 16, 128 );` `$n` `= sizeof(` `$a` `);` `Count_pairs(` `$a` `, ` `$n` `);` `// This code is contributed` `// by Sach_code` `?>` |

## Javascript

`<script>` `// JavaScript program to` `// Count unordered pairs (i, j)` `// in array such that product of a[i] and a[j]` `// can be expressed as power of two` `/* Function to check if x is power of 2*/` `function` `isPowerOfTwo( x)` `{` `/* First x in the below expression is` ` ` `for the case when x is 0 */` `return` `(x >0&& (!((x&(x-1))>0)));` `}` `// Function to Count unordered pairs` `function` `Count_pairs(a,n)` `{` ` ` `let count = 0;` ` ` `for` `(let i = 0; i < n; i++) {` ` ` `// is a number can be expressed` ` ` `// as power of two` ` ` `if` `(isPowerOfTwo(a[i]))` ` ` `count++;` ` ` `}` ` ` `// count total number` ` ` `// of unordered pairs` ` ` `let ans = (count * (count - 1)) / 2;` ` ` `document.write( ans);` `}` `// Driver code` ` ` `let a = [ 2, 5, 8, 16, 128 ];` ` ` `let n = a.length;` ` ` `Count_pairs(a, n);` `// This code is contributed by sravan kumar` `</script>` |

**Output:**

6

**Time Complexity: **O(N), where N is the number of elements in the array.

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