Given an array A[] and a positive integer X. The task is to find the absolute difference between the floor of the total sum divided by X and the sum of the floor of each element of A[] divided by X.
Examples:
Input: A[] = {1, 2, 3, 4, 5, 6}, X = 4
Output: 2
Explanation :
- Sum of A[] = 1 + 2 + 3 + 4 + 5 + 6 = 21
- Sum of A[] divided by X = 21 / 4 = 5
- Sum of floor of every element divided by X = 1 / 4 + 2 / 4 + 3 / 4 + 4 / 4 + 5 / 4 + 6 / 4 = 0 + 0 + 0 + 1 + 1 + 1 = 3
- Absolute Difference = 5 – 3 = 2
Input: A[] = {1, 2}, X = 2
Output: 0
Approach : Follow the given steps to solve the problem
- Initialize two variables, totalFloorSum = 0 and FloorSumPerElement = 0
- Traverse the array, for i in range [0, N – 1]
- Update totalFloorSum = totalFloorSum + A[i] and FloorSumPerElement = FloorSumPerElement + floor(A[i] / X)
- Update totalFloorSum = totalFloorSum / N
- After completing the above steps, print the absolute difference of totalFloorSum and FloorSumPerElement
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 the two sum valuesint 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 Codeint 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 approachimport java.util.*;class GFG{ // Function to find absolute difference// between the two sum valuesstatic 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 Codepublic 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 valuesdef 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 Codeif __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 approachusing System;class GFG{// Function to find absolute difference// between the two sum valuesstatic 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 codestatic 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 valuesfunction 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// Inputlet A=[1, 2, 3, 4, 5, 6];let X = 4;// Size of Arraylet N = A.length;// Function call to find absolute difference// between the two sum valuesdocument.write( floorDifference(A, N, X));// This code is contributed by unknown2108</script> |
Output
2
Time Complexity : O(N)
Auxiliary Space : O(1)
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