Dada una array arr[] y un entero K , la tarea es encontrar la diferencia absoluta entre el techo de la suma total de la array dividida por K y la suma del techo de cada elemento de la array dividida por K.
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4, 5, 6}, K = 4
Salida: 2
Explicación: Suma de la array = 21. Ceil of ( Suma de la array ) / K = 6.
Suma de techo de elementos de array dividido por K = (1/4) + (2/4) + (3/4) + (4/4) + (5/4) + (6/4) = 1 + 1 + 1 + 1 + 2 + 2 = 8.
Por lo tanto, diferencia absoluta = 8 – 6 = 2.Entrada: arr[] = {1, 2, 3}, K = 2
Salida: 1
Enfoque: siga los pasos a continuación para resolver el problema dado:
- Inicialice dos variables, digamos totalSum y perElementSum , para almacenar la suma total de la array y la suma del techo de cada elemento de la array dividido por K .
- Atraviese la array y realice lo siguiente:
- Agregue el elemento actual arr[i] a totalSum .
- Agregue el techo del elemento actual dividido por K , es decir, arr[i]/K .
- Después de los pasos anteriores, imprima el valor absoluto de totalSum y perElementSum como resultado.
A continuación se muestra la implementación del enfoque anterior:
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
// 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
# 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#
// 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
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>
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(1)