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 xint 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 Codeint 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
// Java approach for the above approachpublic 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
# Python3 program for the above approachfrom math import ceil# Function to find absolute difference# between array sum divided by x and# sum of ceil of array elements divided by xdef 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 Codeif __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#
// C# approach for the above approachusing System;class GFG{// Function to find absolute difference// between array sum divided by x and// sum of ceil of array elements divided by xstatic 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 Codepublic 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 |
Javascript
<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> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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