Sum of the updated array after performing the given operation
Given an array arr[] of N elements, the task is to update all the array elements such that an element arr[i] is updated as arr[i] = arr[i] – X where X = arr[i + 1] + arr[i + 2] + … + arr[N – 1] and finally print the sum of the updated array.
Examples:
Input: arr[] = {40, 25, 12, 10}
Output: 8
The updated array will be {-7, 3, 2, 10}.
-7 + 3 + 2 + 10 = 8
Input: arr[] = {50, 30, 10, 2, 0}
Output: 36
Approach: A simple solution is for every possible value of i, update arr[i] = arr[i] – sum(arr[i+1…N-1]).
C++
#include <iostream>
using namespace std;
int sumArr( int arr[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
return sum;
}
int sumModArr( int arr[], int n)
{
for ( int i = 0; i < n - 1; i++) {
int subSum = 0;
for ( int j = i + 1; j < n; j++) {
subSum += arr[j];
}
arr[i] -= subSum;
}
return sumArr(arr, n);
}
int main()
{
int arr[] = { 40, 25, 12, 10 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumModArr(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int sumArr( int arr[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += arr[i];
return sum;
}
static int sumModArr( int arr[], int n)
{
for ( int i = 0 ; i < n - 1 ; i++)
{
int subSum = 0 ;
for ( int j = i + 1 ; j < n; j++)
{
subSum += arr[j];
}
arr[i] -= subSum;
}
return sumArr(arr, n);
}
public static void main(String []args)
{
int arr[] = { 40 , 25 , 12 , 10 };
int n = arr.length;
System.out.println(sumModArr(arr, n));
}
}
|
Python3
def sumArr(arr, n):
sum = 0
for i in range (n):
sum + = arr[i]
return sum
def sumModArr(arr, n):
for i in range (n - 1 ):
subSum = 0
for j in range (i + 1 , n):
subSum + = arr[j]
arr[i] - = subSum
return sumArr(arr, n)
arr = [ 40 , 25 , 12 , 10 ]
n = len (arr)
print (sumModArr(arr, n))
|
C#
using System;
class GFG
{
static int sumArr( int []arr, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
return sum;
}
static int sumModArr( int []arr, int n)
{
for ( int i = 0; i < n - 1; i++)
{
int subSum = 0;
for ( int j = i + 1; j < n; j++)
{
subSum += arr[j];
}
arr[i] -= subSum;
}
return sumArr(arr, n);
}
public static void Main()
{
int []arr = { 40, 25, 12, 10 };
int n = arr.Length;
Console.WriteLine(sumModArr(arr, n));
}
}
|
Javascript
<script>
function sumArr(arr , n) {
var sum = 0;
for (i = 0; i < n; i++)
sum += arr[i];
return sum;
}
function sumModArr(arr , n) {
for (i = 0; i < n - 1; i++) {
var subSum = 0;
for (j = i + 1; j < n; j++) {
subSum += arr[j];
}
arr[i] -= subSum;
}
return sumArr(arr, n);
}
var arr = [ 40, 25, 12, 10 ];
var n = arr.length;
document.write(sumModArr(arr, n));
</script>
|
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: An efficient solution is to traverse the array from the end so that the sum of the subarray till now i.e. sum(arr[i+1…n-1]) can be used to calculate the sum of the current subarray arr[i…n-1] i.e. sum(arr[i…n-1]) = arr[i] + sum(arr[i+1…n-1]).
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int sumArr( int arr[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
return sum;
}
int sumModArr( int arr[], int n)
{
int subSum = arr[n - 1];
for ( int i = n - 2; i >= 0; i--) {
int curr = arr[i];
arr[i] -= subSum;
subSum += curr;
}
return sumArr(arr, n);
}
int main()
{
int arr[] = { 40, 25, 12, 10 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumModArr(arr, n);
return 0;
}
|
Java
class GFG
{
static int sumArr( int arr[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += arr[i];
return sum;
}
static int sumModArr( int arr[], int n)
{
int subSum = arr[n - 1 ];
for ( int i = n - 2 ; i >= 0 ; i--)
{
int curr = arr[i];
arr[i] -= subSum;
subSum += curr;
}
return sumArr(arr, n);
}
public static void main (String[] args)
{
int []arr = { 40 , 25 , 12 , 10 };
int n = arr.length;
System.out.println(sumModArr(arr, n));
}
}
|
Python3
def sumModArr(arr, n):
subSum = arr[n - 1 ];
for i in range (n - 2 , - 1 , - 1 ):
curr = arr[i];
arr[i] - = subSum;
subSum + = curr;
return sum (arr)
if __name__ = = "__main__" :
arr = [ 40 , 25 , 12 , 10 ];
n = len (arr);
print (sumModArr(arr, n));
|
C#
using System;
class GFG
{
static int sumArr( int []arr, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
return sum;
}
static int sumModArr( int []arr, int n)
{
int subSum = arr[n - 1];
for ( int i = n - 2; i >= 0; i--)
{
int curr = arr[i];
arr[i] -= subSum;
subSum += curr;
}
return sumArr(arr, n);
}
public static void Main (String[] args)
{
int []arr = { 40, 25, 12, 10 };
int n = arr.Length;
Console.WriteLine(sumModArr(arr, n));
}
}
|
Javascript
<script>
function sumArr(arr, n) {
var sum = 0;
for ( var i = 0; i < n; i++) sum += arr[i];
return sum;
}
function sumModArr(arr, n) {
var subSum = arr[n - 1];
for ( var i = n - 2; i >= 0; i--) {
var curr = arr[i];
arr[i] -= subSum;
subSum += curr;
}
return sumArr(arr, n);
}
var arr = [40, 25, 12, 10];
var n = arr.length;
document.write(sumModArr(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
29 Mar, 2023
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