Given an array arr[] consisting of N integers, the task is to find the sum of array elements that are equidistant from the two consecutive powers of 2.
Examples:
Input: arr[] = {10, 24, 17, 3, 8}
Output: 27
Explanation:
Following array elements are equidistant from two consecutive powers of 2:
- arr[1] (= 24) is equally separated from 24 and 25.
- arr[3] (= 3) is equally separated from 21 and 22.
Therefore, the sum of 24 and 3 is 27.
Input: arr[] = {17, 5, 6, 35}
Output: 6
Approach: Follow the steps below to solve the problem:
- Initialize a variable, say res, that stores the sum of array elements.
-
Traverse the given array arr[] and perform the following steps:
- Find the value of log2(arr[i]) and store it in a variable, say power.
- Find the value of 2(power) and 2(power + 1)and store them in variables, say LesserValue and LargerValue, respectively.
- If the value of (arr[i] – LesserValue) equal to (LargerValue – arr[i]), then increment the value of res by arr[i].
- After completing the above steps, print the value of res as the resultant sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to print the sum of array // elements that are equidistant from // two consecutive powers of 2 int FindSum( int arr[], int N)
{ // Stores the resultant sum of the
// array elements
int res = 0;
// Traverse the array arr[]
for ( int i = 0; i < N; i++) {
// Stores the power of 2 of the
// number arr[i]
int power = log2(arr[i]);
// Stores the number which is
// power of 2 and lesser than
// or equal to arr[i]
int LesserValue = pow (2, power);
// Stores the number which is
// power of 2 and greater than
// or equal to arr[i]
int LargerValue = pow (2, power + 1);
// If arr[i] - LesserValue is the
// same as LargerValue-arr[i]
if ((arr[i] - LesserValue)
== (LargerValue - arr[i])) {
// Increment res by arr[i]
res += arr[i];
}
}
// Return the resultant sum res
return res;
} // Driver Code int main()
{ int arr[] = { 10, 24, 17, 3, 8 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << FindSum(arr, N);
return 0;
} |
Java
// Java program for the above approach class GFG {
// Function to print the sum of array
// elements that are equidistant from
// two consecutive powers of 2
static int FindSum( int [] arr, int N)
{
// Stores the resultant sum of the
// array elements
int res = 0 ;
// Traverse the array arr[]
for ( int i = 0 ; i < N; i++) {
// Stores the power of 2 of the
// number arr[i]
int power
= ( int )(Math.log(arr[i]) / Math.log( 2 ));
// Stores the number which is
// power of 2 and lesser than
// or equal to arr[i]
int LesserValue = ( int )Math.pow( 2 , power);
// Stores the number which is
// power of 2 and greater than
// or equal to arr[i]
int LargerValue = ( int )Math.pow( 2 , power + 1 );
// If arr[i] - LesserValue is the
// same as LargerValue-arr[i]
if ((arr[i] - LesserValue)
== (LargerValue - arr[i])) {
// Increment res by arr[i]
res += arr[i];
}
}
// Return the resultant sum res
return res;
}
// Driver Code
public static void main(String[] args)
{
int [] arr = { 10 , 24 , 17 , 3 , 8 };
int N = arr.length;
System.out.println(FindSum(arr, N));
}
} // This code is contributed by ukasp. |
Python3
# Python3 program for the above approach from math import log2
# Function to print sum of array # elements that are equidistant from # two consecutive powers of 2 def FindSum(arr, N):
# Stores the resultant sum of the
# array elements
res = 0
# Traverse the array arr[]
for i in range (N):
# Stores the power of 2 of the
# number arr[i]
power = int (log2(arr[i]))
# Stores the number which is
# power of 2 and lesser than
# or equal to arr[i]
LesserValue = pow ( 2 , power)
# Stores the number which is
# power of 2 and greater than
# or equal to arr[i]
LargerValue = pow ( 2 , power + 1 )
# If arr[i] - LesserValue is the
# same as LargerValue-arr[i]
if ((arr[i] - LesserValue) = =
(LargerValue - arr[i])):
# Increment res by arr[i]
res + = arr[i]
# Return the resultant sum res
return res
# Driver Code if __name__ = = '__main__' :
arr = [ 10 , 24 , 17 , 3 , 8 ]
N = len (arr)
print (FindSum(arr, N))
# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System;
class GFG{
// Function to print the sum of array // elements that are equidistant from // two consecutive powers of 2 static int FindSum( int [] arr, int N)
{ // Stores the resultant sum of the
// array elements
int res = 0;
// Traverse the array arr[]
for ( int i = 0; i < N; i++) {
// Stores the power of 2 of the
// number arr[i]
int power = ( int )(Math.Log(arr[i]) / Math.Log(2));
// Stores the number which is
// power of 2 and lesser than
// or equal to arr[i]
int LesserValue = ( int )Math.Pow(2, power);
// Stores the number which is
// power of 2 and greater than
// or equal to arr[i]
int LargerValue = ( int )Math.Pow(2, power + 1);
// If arr[i] - LesserValue is the
// same as LargerValue-arr[i]
if ((arr[i] - LesserValue)
== (LargerValue - arr[i])) {
// Increment res by arr[i]
res += arr[i];
}
}
// Return the resultant sum res
return res;
} // Driver Code public static void Main()
{ int [] arr= { 10, 24, 17, 3, 8 };
int N = arr.Length;
Console.WriteLine(FindSum(arr, N));
} } // This code is contributed by code_hunt. |
Javascript
<script> // Javascript program for the above approach // Function to print the sum of array // elements that are equidistant from // two consecutive powers of 2 function FindSum(arr, N)
{ // Stores the resultant sum of
// the array elements
let res = 0;
// Traverse the array arr[]
for (let i = 0; i < N; i++)
{
// Stores the power of 2 of the
// number arr[i]
let power = Math.floor(
Math.log2(arr[i]));
// Stores the number which is
// power of 2 and lesser than
// or equal to arr[i]
let LesserValue = Math.pow(2, power);
// Stores the number which is
// power of 2 and greater than
// or equal to arr[i]
let LargerValue = Math.pow(2, power + 1);
// If arr[i] - LesserValue is the
// same as LargerValue-arr[i]
if ((arr[i] - LesserValue) ==
(LargerValue - arr[i]))
{
// Increment res by arr[i]
res += arr[i];
}
}
// Return the resultant sum res
return res;
} // Driver Code let arr = [ 10, 24, 17, 3, 8 ]; let N = arr.length; document.write(FindSum(arr, N)); // This code is contributed by sanjoy_62 </script> |
Output:
27
Time Complexity: O(N)
Auxiliary Space: O(1)