Given an array arr[] of n integers, construct a Sum Array sum[] (of same size) such that sum[i] is equal to the sum of all the elements of arr[] except arr[i]. Solve it without subtraction operator and in O(n).
Examples:
Input : arr[] = {3, 6, 4, 8, 9}
Output : sum[] = {27, 24, 26, 22, 21}
Input : arr[] = {4, 5, 7, 3, 10, 1}
Output : sum[] = {26, 25, 23, 27, 20, 29}
Brute Force Approach:
The brute force approach to solve this problem is to iterate through the array for each element, calculate the sum of all other elements, and store it in the sum array. This can be done using two nested loops where the outer loop iterates through all the elements, and the inner loop calculates the sum of all other elements.
- Create an empty sum array of size n to store the sum of all elements except the current element.
- Iterate through the array for each element using a for loop.
- Initialize the sum of the current element as 0.
- For each element, iterate through the array using another for loop.
- If the current index of the inner loop is not equal to the index of the outer loop, add the value of the element at the current index to the sum of the current element.
- Store the sum of all other elements in the sum array at the index of the current element.
- Iterate through the sum array and print its elements.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
void sumArray(int arr[], int n)
{
int sum[n];
for(int i=0; i<n; i++){
sum[i] = 0;
for(int j=0; j<n; j++){
if(j!=i)
sum[i] += arr[j];
}
}
for(int i=0; i<n; i++)
cout<<sum[i]<<" ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 3, 6, 4, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
sumArray(arr, n);
return 0;
}
import java.util.Arrays;
public class Main {
// Function to calculate the sum of elements in the array except the current element
static void sumArray(int[] arr, int n) {
int[] sum = new int[n];
// Calculate the sum of elements excluding the current element
for (int i = 0; i < n; i++) {
sum[i] = 0;
for (int j = 0; j < n; j++) {
if (j != i) {
sum[i] += arr[j];
}
}
}
// Print the calculated sums
for (int i = 0; i < n; i++) {
System.out.print(sum[i] + " ");
}
}
public static void main(String[] args) {
int[] arr = {3, 6, 4, 8, 9};
int n = arr.length;
sumArray(arr, n);
}
}
# Function to calculate the sum of elements in the list except the current element
def sumArray(arr):
n = len(arr)
sum_result = [0] * n
# Calculate the sum of elements excluding the current element
for i in range(n):
sum_result[i] = 0
for j in range(n):
if j != i:
sum_result[i] += arr[j]
# Print the sum of elements for each position
for i in range(n):
print(sum_result[i])
# Driver program to test the sumArray function
arr = [3, 6, 4, 8, 9]
sumArray(arr)
using System;
class Program
{
// Function to calculate the sum of elements except the current element
static void SumArray(int[] arr, int n)
{
int[] sum = new int[n];
for (int i = 0; i < n; i++)
{
sum[i] = 0;
for (int j = 0; j < n; j++)
{
if (j != i)
sum[i] += arr[j];
}
}
for (int i = 0; i < n; i++)
Console.Write(sum[i] + " ");
}
// Main driver program
static void Main()
{
int[] arr = { 3, 6, 4, 8, 9 };
int n = arr.Length;
SumArray(arr, n);
// Output: 27 24 26 22 21
}
}
// Function to calculate the sum of elements in the array except the current element
function sumArray(arr) {
const n = arr.length;
const sum = new Array(n);
// Calculate the sum of elements excluding the current element
for (let i = 0; i < n; i++) {
sum[i] = 0;
for (let j = 0; j < n; j++) {
if (j !== i) {
sum[i] += arr[j];
}
}
}
// Print the sum of elements for each position
for (let i = 0; i < n; i++) {
console.log(sum[i]);
}
}
// Driver program to test the sumArray function
const arr = [3, 6, 4, 8, 9];
sumArray(arr);
Time Complexity: O(n^2), as we are using nested loops to iterate through the array.
Space Complexity: O(n) because we are creating an additional array of the same size as the input array to store the sum of all elements except the current element.
Algorithm:
- Construct a temporary array leftSum[] such that leftSum[i] contains sum of all elements on left of arr[i] excluding arr[i].
- Construct another temporary array rightSum[] such that rightSum[i] contains sum of all elements on right of arr[i] excluding arr[i].
- To get sum[], sum left[] and right[].
Implementation:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
void sumArray(int arr[], int n)
{
/* Allocate memory for temporary arrays leftSum[],
rightSum[] and Sum[]*/
int leftSum[n], rightSum[n], Sum[n], i, j;
/* Left most element of left array is always 0 */
leftSum[0] = 0;
/* Rightmost most element of right
array is always 0 */
rightSum[n - 1] = 0;
/* Construct the left array*/
for (i = 1; i < n; i++)
leftSum[i] = arr[i - 1] + leftSum[i - 1];
/* Construct the right array*/
for (j = n - 2; j >= 0; j--)
rightSum[j] = arr[j + 1] + rightSum[j + 1];
/* Construct the sum array using
left[] and right[] */
for (i = 0; i < n; i++)
Sum[i] = leftSum[i] + rightSum[i];
/* print the constructed prod array */
for (i = 0; i < n; i++)
cout << Sum[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 3, 6, 4, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
sumArray(arr, n);
return 0;
}
// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class Geeks {
public static void sumArray(int arr[], int n)
{
/* Allocate memory for temporary arrays
leftSum[], rightSum[] and Sum[]*/
int leftSum[] = new int[n];
int rightSum[] = new int[n];
int Sum[] = new int[n];
int i = 0, j = 0;
/* Left most element of left array is
always 0 */
leftSum[0] = 0;
/* Right most element of right array
is always 0 */
rightSum[n - 1] = 0;
/* Construct the left array*/
for (i = 1; i < n; i++)
leftSum[i] = arr[i - 1] + leftSum[i - 1];
/* Construct the right array*/
for (j = n - 2; j >= 0; j--)
rightSum[j] = arr[j + 1] + rightSum[j + 1];
/* Construct the sum array using
left[] and right[] */
for (i = 0; i < n; i++)
Sum[i] = leftSum[i] + rightSum[i];
/*print the sum array*/
for (i = 0; i < n; i++)
System.out.print(Sum[i] + " ");
}
/* Driver function to test above function*/
public static void main(String[] args)
{
int arr[] = { 3, 6, 4, 8, 9 };
int n = arr.length;
sumArray(arr, n);
}
}
# Python3 implementation of above approach
def sumArray(arr, n):
# Allocate memory for temporary arrays
# leftSum[], rightSum[] and Sum[]
leftSum = [0 for i in range(n)]
rightSum = [0 for i in range(n)]
Sum = [0 for i in range(n)]
i, j = 0, 0
# Left most element of left
# array is always 0
leftSum[0] = 0
# Rightmost most element of right
# array is always 0
rightSum[n - 1] = 0
# Construct the left array
for i in range(1, n):
leftSum[i] = arr[i - 1] + leftSum[i - 1]
# Construct the right array
for j in range(n - 2, -1, -1):
rightSum[j] = arr[j + 1] + rightSum[j + 1]
# Construct the sum array using
# left[] and right[]
for i in range(0, n):
Sum[i] = leftSum[i] + rightSum[i]
# print the constructed prod array
for i in range(n):
print(Sum[i], end = " ")
# Driver Code
arr = [3, 6, 4, 8, 9]
n = len(arr)
sumArray(arr, n)
# This code is contributed by
# mohit kumar 29
// C# implementation of above approach
using System ;
class Geeks {
public static void sumArray(int []arr, int n)
{
/* Allocate memory for temporary arrays
leftSum[], rightSum[] and Sum[]*/
int []leftSum = new int[n];
int []rightSum = new int[n];
int []Sum = new int[n];
int i = 0, j = 0;
/* Left most element of left array is
always 0 */
leftSum[0] = 0;
/* Right most element of right array
is always 0 */
rightSum[n - 1] = 0;
/* Construct the left array*/
for (i = 1; i < n; i++)
leftSum[i] = arr[i - 1] + leftSum[i - 1];
/* Construct the right array*/
for (j = n - 2; j >= 0; j--)
rightSum[j] = arr[j + 1] + rightSum[j + 1];
/* Construct the sum array using
left[] and right[] */
for (i = 0; i < n; i++)
Sum[i] = leftSum[i] + rightSum[i];
/*print the sum array*/
for (i = 0; i < n; i++)
Console.Write(Sum[i] + " ");
}
/* Driver function to test above function*/
public static void Main()
{
int []arr = { 3, 6, 4, 8, 9 };
int n = arr.Length;
sumArray(arr, n);
}
// This code is contributed by Ryuga
}
<script>
// JavaScript implementation of above approach
function sumArray(arr, n)
{
/* Allocate memory for temporary arrays
leftSum[], rightSum[] and Sum[]*/
let leftSum = new Array(n);
let rightSum = new Array(n);
let Sum = new Array(n);
let i = 0, j = 0;
/* Left most element of left array is
always 0 */
leftSum[0] = 0;
/* Right most element of right array
is always 0 */
rightSum[n - 1] = 0;
/* Construct the left array*/
for (i = 1; i < n; i++)
leftSum[i] = arr[i - 1] + leftSum[i - 1];
/* Construct the right array*/
for (j = n - 2; j >= 0; j--)
rightSum[j] = arr[j + 1] + rightSum[j + 1];
/* Construct the sum array using
left[] and right[] */
for (i = 0; i < n; i++)
Sum[i] = leftSum[i] + rightSum[i];
/*print the sum array*/
for (i = 0; i < n; i++)
document.write(Sum[i] + " ");
}
let arr = [ 3, 6, 4, 8, 9 ];
let n = arr.length;
sumArray(arr, n);
</script>
<?php
// PHP implementation of above approach
function sumArray($arr, $n)
{
/* Allocate memory for temporary
arrays leftSum[], rightSum[] and Sum[]*/
$leftSum = array_fill(0, $n, 0);
$rightSum = array_fill(0, $n, 0);
$Sum = array_fill(0, $n, 0);
/* Left most element of left
array is always 0 */
$leftSum[0] = 0;
/* Rightmost most element of right
array is always 0 */
$rightSum[$n - 1] = 0;
/* Construct the left array*/
for ($i = 1; $i < $n; $i++)
$leftSum[$i] = $arr[$i - 1] +
$leftSum[$i - 1];
/* Construct the right array*/
for ($j = $n - 2; $j >= 0; $j--)
$rightSum[$j] = $arr[$j + 1] +
$rightSum[$j + 1];
/* Construct the sum array using
left[] and right[] */
for ($i = 0; $i < $n; $i++)
$Sum[$i] = $leftSum[$i] + $rightSum[$i];
/* print the constructed prod array */
for ($i = 0; $i < $n; $i++)
echo $Sum[$i]." ";
}
// Driver Code
$arr = array( 3, 6, 4, 8, 9 );
$n = count($arr);
sumArray($arr, $n);
// This code is contributed
// by chandan_jnu
?>
Complexity Analysis:
- Time Complexity: O(n)
- Space Complexity: O(n)
The above method can be optimized to work in O(1) auxiliary space.
Implementation:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
void sumArray(int arr[], int n)
{
int i, temp = 0;
/* Allocate memory for the sum array */
int Sum[n];
/* Initialize the sum array as 0 */
memset(Sum, 0, n);
/* In this loop, temp variable contains
sum of elements on left side excluding
arr[i] */
for (i = 0; i < n; i++) {
Sum[i] = temp;
temp += arr[i];
}
/* Initialize temp to 0 for sum on right
side */
temp = 0;
/* In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */
for (i = n - 1; i >= 0; i--) {
Sum[i] += temp;
temp += arr[i];
}
for (i = 0; i < n; i++)
cout << Sum[i] << " ";
}
/* Driver program to test above function */
int main()
{
int arr[] = { 3, 6, 4, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
sumArray(arr, n);
return 0;
}
// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class Geeks {
public static void sumArray(int arr[], int n)
{
int i = 0, temp = 0;
int Sum[] = new int[n];
Arrays.fill(Sum, 0);
/* In this loop, temp variable contains
sum of elements on left side excluding
arr[i] */
for (i = 0; i < n; i++) {
Sum[i] = temp;
temp += arr[i];
}
/* Initialize temp to 0 for sum on right side */
temp = 0;
/* In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */
for (i = n - 1; i >= 0; i--) {
Sum[i] += temp;
temp += arr[i];
}
for (i = 0; i < n; i++)
System.out.print(Sum[i] + " ");
}
/* Driver function to test above function*/
public static void main(String[] args)
{
int arr[] = { 3, 6, 4, 8, 9 };
int n = arr.length;
sumArray(arr, n);
}
}
# Python3 implementation of above approach
def sumArray(arr, n):
i, temp = 0, 0
# Allocate memory for the sum array */
Sum = [0 for i in range(n)]
'''In this loop, temp variable contains
sum of elements on left side excluding
arr[i] '''
for i in range(n):
Sum[i] = temp
temp += arr[i]
# Initialize temp to 0 for sum
# on right side */
temp = 0
''' In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */'''
for i in range(n - 1, -1, -1):
Sum[i] += temp
temp += arr[i]
for i in range(n):
print(Sum[i], end = " ")
# Driver Code
arr = [ 3, 6, 4, 8, 9 ]
n = len(arr)
sumArray(arr, n)
# This code is contributed by
# Mohit Kumar 29
// C# implementation of above approach
using System;
class Geeks {
public static void sumArray(int []arr, int n)
{
int i = 0, temp = 0;
int []Sum = new int[n];
for( i=0;i<n;i++)
Sum[i] = 0;
/* In this loop, temp variable contains
sum of elements on left side excluding
arr[i] */
for (i = 0; i < n; i++) {
Sum[i] = temp;
temp += arr[i];
}
/* Initialize temp to 0 for sum on right side */
temp = 0;
/* In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */
for (i = n - 1; i >= 0; i--) {
Sum[i] += temp;
temp += arr[i];
}
for (i = 0; i < n; i++)
Console.Write(Sum[i] + " ");
}
/* Driver function to test above function*/
public static void Main()
{
int []arr = { 3, 6, 4, 8, 9 };
int n = arr.Length;
sumArray(arr, n);
}
}
// This code is contributed by inder_verma..
<script>
// Javascript implementation of above approach
function sumArray(arr, n)
{
let i = 0, temp = 0;
let Sum = new Array(n);
for( i=0;i<n;i++)
Sum[i] = 0;
/* In this loop, temp variable contains
sum of elements on left side excluding
arr[i] */
for (i = 0; i < n; i++) {
Sum[i] = temp;
temp += arr[i];
}
/* Initialize temp to 0 for sum on right side */
temp = 0;
/* In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */
for (i = n - 1; i >= 0; i--) {
Sum[i] += temp;
temp += arr[i];
}
for (i = 0; i < n; i++)
document.write(Sum[i] + " ");
}
let arr = [ 3, 6, 4, 8, 9 ];
let n = arr.length;
sumArray(arr, n);
// This code is contributed by decode2207.
</script>
<?php
// PHP implementation of above approach
function sumArray($arr, $n)
{
$temp = 0;
/* Allocate memory for the sum array */
/* Initialize the sum array as 0 */
$Sum = array_fill(0, $n, 0);
/* In this loop, temp variable contains
sum of elements on left side excluding
arr[i] */
for ($i = 0; $i < $n; $i++)
{
$Sum[$i] = $temp;
$temp += $arr[$i];
}
/* Initialize temp to 0 for sum
on right side */
$temp = 0;
/* In this loop, temp variable contains
sum of elements on right side excluding
arr[i] */
for ($i = $n - 1; $i >= 0; $i--)
{
$Sum[$i] += $temp;
$temp += $arr[$i];
}
for ($i = 0; $i < $n; $i++)
echo $Sum[$i] . " ";
}
// Driver Code
$arr = array( 3, 6, 4, 8, 9 );
$n = count($arr);
sumArray($arr, $n);
// This code is contributed by
// chandan_jnu
?>
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)
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