Dada una array A[] y un entero positivo X . La tarea es encontrar la diferencia absoluta entre el piso de la suma total dividida por X y la suma del piso de cada elemento de A[] dividido por X.
Ejemplos:
Entrada: A[] = {1, 2, 3, 4, 5, 6}, X = 4
Salida: 2
Explicación :
- Suma de A[] = 1 + 2 + 3 + 4 + 5 + 6 = 21
- Suma de A[] dividida por X = 21 / 4 = 5
- Suma de piso de cada elemento dividido por X = 1/4 + 2/4 + 3/4 + 4/4 + 5/4 + 6/4 = 0 + 0 + 0 + 1 + 1 + 1 = 3
- Diferencia absoluta = 5 – 3 = 2
Entrada: A[] = {1, 2}, X = 2
Salida: 0
Enfoque : siga los pasos dados para resolver el problema
- Inicialice dos variables, totalFloorSum = 0 y FloorSumPerElement = 0
- Atraviesa la array , para i en el rango [0, N – 1]
- Actualice totalFloorSum = totalFloorSum + A[i] y FloorSumPerElement = FloorSumPerElement + floor(A[i] / X)
- Actualizar totalFloorSum = totalFloorSum / N
- Después de completar los pasos anteriores, imprima la diferencia absoluta de totalFloorSum y FloorSumPerElement
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 the two sum values
int floorDifference(int A[], int N, int X)
{
// Variable to store total sum
int totalSum = 0;
// Variable to store sum of A[i] / X
int perElementSum = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update totalSum
totalSum += A[i];
// Update perElementSum
perElementSum += A[i] / X;
}
// Floor of total sum divided by X
int totalFloorSum = totalSum / X;
// Return the absolute difference
return abs(totalFloorSum - perElementSum);
}
// Driver Code
int main()
{
// Input
int A[] = { 1, 2, 3, 4, 5, 6 };
int X = 4;
// Size of Array
int N = sizeof(A) / sizeof(A[0]);
// Function call to find absolute difference
// between the two sum values
cout << floorDifference(A, N, X);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find absolute difference
// between the two sum values
static int floorDifference(int A[], int N, int X)
{
// Variable to store total sum
int totalSum = 0;
// Variable to store sum of A[i] / X
int perElementSum = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update totalSum
totalSum += A[i];
// Update perElementSum
perElementSum += A[i] / X;
}
// Floor of total sum divided by X
int totalFloorSum = totalSum / X;
// Return the absolute difference
return Math.abs(totalFloorSum - perElementSum);
}
// Driver Code
public static void main(String[] args)
{
// Input
int A[] = { 1, 2, 3, 4, 5, 6 };
int X = 4;
// Size of Array
int N = A.length;
// Function call to find absolute difference
// between the two sum values
System.out.print( floorDifference(A, N, X));
}
}
// This code is contributed by code_hunt.
Python3
# Python3 program for the above approach # Function to find absolute difference # between the two sum values def floorDifference(A, N, X): # Variable to store total sum totalSum = 0 # Variable to store sum of A[i] / X perElementSum = 0 # Traverse the array for i in range(N): # Update totalSum totalSum += A[i] # Update perElementSum perElementSum += A[i] // X # Floor of total sum divided by X totalFloorSum = totalSum // X # Return the absolute difference return abs(totalFloorSum - perElementSum) # Driver Code if __name__ == '__main__': # Input A = [ 1, 2, 3, 4, 5, 6 ] X = 4 # Size of Array N = len(A) # Function call to find absolute difference # between the two sum values print (floorDifference(A, N, X)) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find absolute difference
// between the two sum values
static int floorDifference(int[] A, int N, int X)
{
// Variable to store total sum
int totalSum = 0;
// Variable to store sum of A[i] / X
int perElementSum = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update totalSum
totalSum += A[i];
// Update perElementSum
perElementSum += A[i] / X;
}
// Floor of total sum divided by X
int totalFloorSum = totalSum / X;
// Return the absolute difference
return Math.Abs(totalFloorSum - perElementSum);
}
// Driver code
static void Main()
{
// Input
int[] A = { 1, 2, 3, 4, 5, 6 };
int X = 4;
// Size of Array
int N = A.Length;
// Function call to find absolute difference
// between the two sum values
Console.Write( floorDifference(A, N, X));
}
}
// This code is contributed by sanjoy_62.
Javascript
<script>
// Javascript program for the above approach
// Function to find absolute difference
// between the two sum values
function floorDifference(A,N,X)
{
// Variable to store total sum
let totalSum = 0;
// Variable to store sum of A[i] / X
let perElementSum = 0;
// Traverse the array
for (let i = 0; i < N; i++) {
// Update totalSum
totalSum += A[i];
// Update perElementSum
perElementSum += Math.floor(A[i] / X);
}
// Floor of total sum divided by X
let totalFloorSum = Math.floor(totalSum / X);
// Return the absolute difference
return Math.abs(totalFloorSum - perElementSum);
}
// Driver Code
// Input
let A=[1, 2, 3, 4, 5, 6];
let X = 4;
// Size of Array
let N = A.length;
// Function call to find absolute difference
// between the two sum values
document.write( floorDifference(A, N, X));
// This code is contributed by unknown2108
</script>
Producción
2
Complejidad de Tiempo : O(N)
Espacio Auxiliar : O(1)