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# Sum of odd Array elements after each update query

• Difficulty Level : Medium
• Last Updated : 02 Jan, 2023

Given an integer array Arr[] of length N and an array Queries[] of length Q where Queries[i] = [valuei, indexi]. For each query i, update Arr[indexi] = Arr[indexi] + valuei, then print the sum of the all the odd values of Arr[].

Examples:

Input: N = 4, Arr = [1,2,3,5], Q = 4, Queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: 8 7 7 9
Explanation: At the beginning, the array is [1,2,3,5].
After adding 1 to Arr, the array is [2,2,3,5], and the sum of odd values is 3 + 5 = 8.
After adding -3 to Arr, the array is [2,-1,3,5], and the sum of odd values is -1 + 3 + 5 = 7.
After adding -4 to Arr, the array is [-2,-1,3,5], and the sum of odd values is -1 + 3 + 5 = 7.
After adding 2 to Arr, the array is [-2,-1,3,7], and the sum of odd values is -1 + 3 + 7 = 9.

Input: N = 1 , Arr = , Q = 1,Queries = [[4,0]]
Output: 
Explanation: At the beginning, the array is .
After adding 4 to Arr, the array is , and the sum of odd values is 0.

Naive Approach: The problem can be solved by implementing the operations as mentioned.

For each query first update the value and then iterate through the array to find the sum of odd integers.

Follow the steps mentioned below to implement the idea:

• Run a loop form j = 0 to Q-1:
• Increment arr[idx] (where idx is the index mentioned in the query) by the given value in the query.
• Initialize the sum to 0.
• Run a nested loop from i = 0 to N-1:
• If the element is odd, then add that value to the sum.

Below is the implementation of the above approach.

## C++

```// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function performing calculations
vector<int> Sumodd(int N, int Arr[], int Q,
int Queries[])
{
vector<int> ans;
for (int j = 0; j < Q; j++) {

// Incrementing the array element
// for every query
Arr[Queries[j]] += Queries[j];

// Initializing sum variable to
int sum = 0;
for (int i = 0; i < N; i++) {

// If the element is odd then
// increment the sum
if (Arr[i] % 2 == 1 || Arr[i] % 2 == -1) {
sum += Arr[i];
}
}

// Pushing result of every query
ans.push_back(sum);
}

return ans;
}

//  Driver code
int main()
{
int Arr[] = { 1, 2, 3, 5 };
int N = sizeof(Arr) / sizeof(Arr);
int Q = 4;
int Queries[Q]
= { { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
vector<int> ans = Sumodd(N, Arr, Q, Queries);
for (int x : ans)
cout << x << " ";
return 0;
}```

## Java

```// Java code to implement the approach
import java.io.*;
import java.util.*;

class GFG
{

// Function performing calculations
public static ArrayList<Integer>
Sumodd(int N, int Arr[], int Q, int Queries[][])
{

ArrayList<Integer> ans = new ArrayList<Integer>();
for (int j = 0; j < Q; j++) {

// Incrementing the array element
// for every query
Arr[Queries[j]] += Queries[j];

// Initializing sum variable to
int sum = 0;
for (int i = 0; i < N; i++) {

// If the element is odd then
// increment the sum
if (Arr[i] % 2 == 1 || Arr[i] % 2 == -1) {
sum += Arr[i];
}
}

// Pushing result of every query
}

return ans;
}

// Driver Code
public static void main(String[] args)
{
int Arr[] = { 1, 2, 3, 5 };
int N = Arr.length;
int Q = 4;
int Queries[][]
= { { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
ArrayList<Integer> ans = Sumodd(N, Arr, Q, Queries);
for (int x : ans)
System.out.print(x + " ");
}
}

// This code is contributed by Rohit Pradhan```

## Python3

```# Python program for the above approach

# Function performing calculations
def Sumodd(N, Arr, Q, Queries):

ans = []
for j in range(Q):

# Incrementing the array element for every query
Arr[Queries[j]] += Queries[j]

# Initializing sum variable to 0
Sum = 0
for i in range(N):

# If the element is odd then increment the sum
if(Arr[i] % 2 is 1 or Arr[i] % 2 is -1):
Sum += Arr[i]

# Pushing result of every query
ans.append(Sum)

return ans

Arr = [1, 2, 3, 5]
N = len(Arr)
Q = 4
Queries = [[1, 0],
[-3, 1],
[-4, 0],
[2, 3]]

# Function call
ans = Sumodd(N, Arr, Q, Queries)
for x in range(len(ans)):
print(ans[x], end=" ")

# This code is contributed by lokeshmvs21.
```

## C#

```// C# implementation
using System;
public class HelloWorld
{

// Function performing calculations
static int[] Sumodd(int N, int[] Arr, int Q,
int[,] Queries)
{

int[] ans=new int[Q];
for (int j = 0; j < Q; j++) {

// Incrementing the array element
// for every query
Arr[Queries[j,1]] += Queries[j,0];

// Initializing sum variable to
int sum = 0;
for (int i = 0; i < N; i++) {

// If the element is odd then
// increment the sum
if (Arr[i] % 2 == 1 || Arr[i] % 2 == -1) {
sum += Arr[i];
}
}
// Pushing result of every query
ans[j]=(sum);
}
return ans;
}

public static void Main(string[] args)
{
int[] Arr = { 1, 2, 3, 5 };
int N = Arr.Length;
int Q = 4;
int[,] Queries = new int[4,2]{ { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
int[] ans = Sumodd(N, Arr, Q, Queries);
for (int x = 0; x < ans.Length; x++){
Console.Write(ans[x] + " ");
}
}
}

// This code is contributed by ksam24000```

## Javascript

```<script>
// JavaScript program for the above approach

// Function performing calculations
function Sumodd(N, Arr, Q, Queries)
{

let ans = [];
for (let j = 0; j < Q; j++) {

// Incrementing the array element
// for every query
Arr[Queries[j]] += Queries[j];

// Initializing sum variable to
let sum = 0;
for (let i = 0; i < N; i++) {

// If the element is odd then
// increment the sum
if (Arr[i] % 2 == 1 || Arr[i] % 2 == -1) {
sum += Arr[i];
}
}

// Pushing result of every query
ans.push(sum);
}

return ans;
}

// Driver Code
let Arr = [ 1, 2, 3, 5 ];
let N = Arr.length;
let Q = 4;
let Queries
= [[ 1, 0 ], [ -3, 1 ], [ -4, 0 ], [ 2, 3 ]];

// Function call
let ans = Sumodd(N, Arr, Q, Queries);
for (let x in ans)
document.write(ans[x] + " ");

// This code is contributed by sanjoy_62.
</script>
```
Output

`8 7 7 9 `

Time Complexity: O(Q*N).
Auxiliary Space: O(Q) i.e answer array of Q size.

## Efficient Approach using Precomputation:

The problem can be solved using precomputation based on the following observation:

Say the sum of odd elements was S before performing an update query. So, for update query, there can be four cases:

• Element changes from odd to odd: In this case the sum increases by value (where value is the amount needed to be updated as per the query). This can be interpreted as subtracting the previous value from S and then adding the new updated value.
• Element changes from even to odd: In this case the increase in sum is the same as the new odd value.
• Element changes from odd to even: In this case the sum decreases by an amount same as the previous value of the element.
• Element changes from even to even: In this case there is no change in sum.

So we can see that whenever the element was odd before update, initially there is a decrement in the sum and if the new updated value is odd, there is an increment of that amount in S.

Follow the steps mentioned below to implement the idea:

• Iterate throughout the array and calculate the sum of the odd values.
• Run a loop from j = 0 to Q-1:
• Store the previous value of Arr[indexj] in temp variable.
• If the temp value is odd then decrement the sum by temp.
• Incrementing the value Arr[indexj] by valuej and update temp.
• If the temp value is odd then increment the sum by temp.
• Push the sum into the answer array.

Below is the implementation of the above approach.

## C++

```// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function performing calculations.
vector<int> Sumodd(int N, int Arr[], int Q,
int Queries[])
{
vector<int> ans;

// Initializing the sum variable to 0.
int sum = 0;
for (int i = 0; i < N; i++) {

// Loop to precomputing
// the sum of odd elements
if (Arr[i] % 2 != 0) {
sum += Arr[i];
}
}

for (int j = 0; j < Q; j++) {
int temp = Arr[Queries[j]];

// If the element was odd then
// decrement it from the sum
if (temp % 2 != 0) {
sum -= temp;
}

// Update the array element for every query
Arr[Queries[j]] += Queries[j];

temp = Arr[Queries[j]];

// If the updated value of temp is odd
// then increment the sum
if (temp % 2 != 0) {
sum += temp;
}

// Push result of every query
ans.push_back(sum);
}

return ans;
}

//  Driver code
int main()
{
int Arr[] = { 1, 2, 3, 5 };
int N = sizeof(Arr) / sizeof(Arr);
int Q = 4;
int Queries[Q]
= { { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
vector<int> ans = Sumodd(N, Arr, Q, Queries);
for (int x : ans)
cout << x << " ";
return 0;
}```

## Java

```// Java code to implement the approach
import java.io.*;
import java.util.*;

class GFG {

// Function performing calculations
static List<Integer> Sumodd(int N, int[] Arr, int Q,
int[][] Queries)
{
List<Integer> ans = new ArrayList<>();

// Initializing the sum variable to 0.
int sum = 0;
for (int i = 0; i < N; i++) {

// Loop to precomputing
// the sum of odd elements
if (Arr[i] % 2 != 0) {
sum += Arr[i];
}
}

for (int j = 0; j < Q; j++) {
int temp = Arr[Queries[j]];

// If the element was odd then
// decrement it from the sum
if (temp % 2 != 0) {
sum -= temp;
}

// Update the array element for every query
Arr[Queries[j]] += Queries[j];

temp = Arr[Queries[j]];

// If the updated value of temp is odd
// then increment the sum
if (temp % 2 != 0) {
sum += temp;
}

// Push result of every query
}

return ans;
}

public static void main(String[] args)
{
int[] Arr = { 1, 2, 3, 5 };
int N = Arr.length;
int Q = 4;
int[][] Queries
= { { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
List<Integer> ans = Sumodd(N, Arr, Q, Queries);
for (int i = 0; i < ans.size(); i++) {
System.out.print(ans.get(i) + " ");
}
}
}

// This code is contributed by lokeshmvs21.```

## Python3

```# Python code to implement the approach

# Function performing calculations
def Sumodd(N, Arr, Q, Queries):
ans = []

# Initializing the sum variable to 0.
Sum = 0

for i in range(N):
# Loop to precomputing
# the sum of odd elements
if(Arr[i] % 2 != 0):
Sum += Arr[i]

for j in range(Q):
temp = Arr[Queries[j]]
# If the element was odd then
# decrement it from the sum
if(temp % 2 != 0):
Sum -= temp

# Update the array element for every query
Arr[Queries[j]] += Queries[j]
temp = Arr[Queries[j]]

# If the updated value of temp is odd
# then increment the sum
if(temp % 2 != 0):
Sum += temp

# Append result of every query
ans.append(Sum)

return ans

Arr = [1, 2, 3, 5]
N = len(Arr)
Q = 4
Queries = [[1, 0], [-3, 1], [-4, 0], [2, 3]]

# Function call
ans = Sumodd(N, Arr, Q, Queries)
for i in range(len(ans)):
print(ans[i], end=" ")

# This code is contributed by lokeshmvs21.
```

## C#

```// C# code to implement the approach
using System;
using System.Collections;
using System.Collections.Generic;

public class GFG {

// Function performing calculations
static ArrayList Sumodd(int N, int[] Arr, int Q,
int[, ] Queries)
{
ArrayList ans = new ArrayList();

// Initializing the sum variable to 0.
int sum = 0;
for (int i = 0; i < N; i++) {

// Loop to precomputing
// the sum of odd elements
if (Arr[i] % 2 != 0) {
sum += Arr[i];
}
}

for (int j = 0; j < Q; j++) {
int temp = Arr[Queries[j, 1]];

// If the element was odd then
// decrement it from the sum
if (temp % 2 != 0) {
sum -= temp;
}

// Update the array element for every query
Arr[Queries[j, 1]] += Queries[j, 0];

temp = Arr[Queries[j, 1]];

// If the updated value of temp is odd
// then increment the sum
if (temp % 2 != 0) {
sum += temp;
}

// Push result of every query
}

return ans;
}

static public void Main()
{

// Code
int[] Arr = { 1, 2, 3, 5 };
int N = Arr.Length;
int Q = 4;
int[, ] Queries
= { { 1, 0 }, { -3, 1 }, { -4, 0 }, { 2, 3 } };

// Function call
ArrayList ans = Sumodd(N, Arr, Q, Queries);
for (int i = 0; i < ans.Count; i++) {
Console.Write(ans[i] + " ");
}
}
}

// This code is contributed by lokesh.```

## Javascript

```// JS code to implement the approach

// Function performing calculations.
function Sumodd(N,  Arr, Q, Queries)
{
let ans = []

// Initializing the sum variable to 0.
let sum = 0;
for (let i = 0; i < N; i++) {

// Loop to precomputing
// the sum of odd elements
if (Arr[i] % 2 != 0) {
sum += Arr[i];
}
}

for (let j = 0; j < Q; j++) {
let temp = Arr[Queries[j]];

// If the element was odd then
// decrement it from the sum
if (temp % 2 != 0) {
sum -= temp;
}

// Update the array element for every query
Arr[Queries[j]] += Queries[j];

temp = Arr[Queries[j]];

// If the updated value of temp is odd
// then increment the sum
if (temp % 2 != 0) {
sum += temp;
}

// Push result of every query
ans.push(sum);
}

return ans;
}

//  Driver code
let Arr = [ 1, 2, 3, 5 ];
let N = Arr.length;
let Q = 4;
let Queries = [ [ 1, 0 ], [ -3, 1 ], [ -4, 0 ], [ 2, 3 ] ];

// Function call
let ans = Sumodd(N, Arr, Q, Queries);
console.log(ans);

// This code is contributed by akashish__```
Output

`8 7 7 9 `

Time Complexity: O(max(Q, N)).
Auxiliary Space: O(Q) i.e answer array of Q size.

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