Difference between ceil of array sum divided by K and sum of ceil of array elements divided by K
Last Updated :
23 Jul, 2025
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++
// 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>
Time Complexity: O(N)
Auxiliary Space: O(1)
Explore
DSA Fundamentals
Data Structures
Algorithms
Advanced
Interview Preparation
Practice Problem