Given an array arr[] and an integer K, the task is to find the absolute difference between the ceil of the total sum of the array divided by K and the sum of the ceil of every array element divided by K.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 4
Output: 2
Explanation: Sum of the array = 21. Ceil of ( Sum of the array ) / K = 6.
Sum of ceil of array elements divided by K = (1/4) + (2/4) + (3/4) + (4/4) + (5/4) + (6/4) = 1 + 1 + 1 + 1 + 2 + 2 = 8.
Therefore, absolute difference = 8 – 6 = 2.Input: arr[] = {1, 2, 3}, K = 2
Output: 1
Approach: Follow the steps below to solve the given problem:
- Initialize two variables, say totalSum and perElementSum, to store the total sum of the array and the sum of the ceil of every array element divided by K.
-
Traverse the array and perform the following:
- Add the current element arr[i] to the totalSum.
- Add the ceil of the current element divided by K i.e., arr[i]/K.
- After the above steps print the absolute value of totalSum and perElementSum as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find absolute difference // between array sum divided by x and // sum of ceil of array elements divided by x int ceilDifference( int arr[], int n,
int x)
{ // Stores the total sum
int totalSum = 0;
// Stores the sum of ceil of
// array elements divided by x
int perElementSum = 0;
// Traverse the array
for ( int i = 0; i < n; i++) {
// Adding each array element
totalSum += arr[i];
// Add the value ceil of arr[i]/x
perElementSum
+= ceil (( double )(arr[i])
/ ( double )(x));
}
// Find the ceil of the
// total sum divided by x
int totalCeilSum
= ceil (( double )(totalSum)
/ ( double )(x));
// Return absolute difference
return abs (perElementSum
- totalCeilSum);
} // Driver Code int main()
{ int arr[] = { 1, 2, 3, 4, 5, 6 };
int K = 4;
int N = sizeof (arr) / sizeof (arr[0]);
cout << ceilDifference(arr, N, K);
return 0;
} |
// Java approach for the above approach public class GFG{
// Function to find absolute difference
// between array sum divided by x and
// sum of ceil of array elements divided by x
static int ceilDifference( int arr[], int n,
int x)
{
// Stores the total sum
int totalSum = 0 ;
// Stores the sum of ceil of
// array elements divided by x
int perElementSum = 0 ;
// Traverse the array
for ( int i = 0 ; i < n; i++) {
// Adding each array element
totalSum += arr[i];
// Add the value ceil of arr[i]/x
perElementSum
+= Math.ceil(( double )(arr[i])
/ ( double )(x));
}
// Find the ceil of the
// total sum divided by x
int totalCeilSum
= ( int ) Math.ceil(( double )(totalSum)
/ ( double )(x));
// Return absolute difference
return Math.abs(perElementSum
- totalCeilSum);
}
// Driver Code
public static void main(String[] args) {
int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 };
int K = 4 ;
int N = arr.length;
System.out.println(ceilDifference(arr, N, K));
}
} // This code is contributed by abhinavjain194 |
# Python3 program for the above approach from math import ceil
# Function to find absolute difference # between array sum divided by x and # sum of ceil of array elements divided by x def ceilDifference(arr, n, x):
# Stores the total sum
totalSum = 0
# Stores the sum of ceil of
# array elements divided by x
perElementSum = 0
# Traverse the array
for i in range (n):
# Adding each array element
totalSum + = arr[i]
# Add the value ceil of arr[i]/x
perElementSum + = ceil(arr[i] / x)
# Find the ceil of the
# total sum divided by x
totalCeilSum = ceil(totalSum / x)
# Return absolute difference
return abs (perElementSum - totalCeilSum)
# Driver Code if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 4 , 5 , 6 ]
K = 4
N = len (arr)
print (ceilDifference(arr, N, K))
# This code is contributed by mohit kumar 29. |
// C# approach for the above approach using System;
class GFG{
// Function to find absolute difference // between array sum divided by x and // sum of ceil of array elements divided by x static int ceilDifference( int [] arr, int n, int x)
{ // Stores the total sum
int totalSum = 0;
// Stores the sum of ceil of
// array elements divided by x
int perElementSum = 0;
// Traverse the array
for ( int i = 0; i < n; i++)
{
// Adding each array element
totalSum += arr[i];
// Add the value ceil of arr[i]/x
perElementSum += ( int )Math.Ceiling(
( double )(arr[i]) / ( double )(x));
}
// Find the ceil of the
// total sum divided by x
int totalCeilSum = ( int )Math.Ceiling(
( double )(totalSum) / ( double )(x));
// Return absolute difference
return Math.Abs(perElementSum - totalCeilSum);
} // Driver Code public static void Main( string [] args)
{ int [] arr = { 1, 2, 3, 4, 5, 6 };
int K = 4;
int N = arr.Length;
Console.Write(ceilDifference(arr, N, K));
} } // This code is contributed by ukasp |
<script> // Javascript approach for the above approach // Function to find absolute difference
// between array sum divided by x and
// sum of ceil of array elements divided by x
function ceilDifference(arr , n, x)
{
// Stores the total sum
var totalSum = 0;
// Stores the sum of ceil of
// array elements divided by x
var perElementSum = 0;
// Traverse the array
for ( var i = 0; i < n; i++) {
// Adding each array element
totalSum += arr[i];
// Add the value ceil of arr[i]/x
perElementSum
+= parseInt(Math.ceil((arr[i])
/ (x)));
}
// Find the ceil of the
// total sum divided by x
var totalCeilSum
= parseInt( Math.ceil((totalSum)
/ (x)));
// Return absolute difference
return Math.abs(perElementSum
- totalCeilSum);
}
// Driver Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var K = 4;
var N = arr.length;
document.write(ceilDifference(arr, N, K));
// This code contributed by shikhasingrajput </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)