Print the sum of array after doing k queries on the array

Given an array, arr[] of size n. Calculate the sum of arr[] by performing k queries on arr[], such that we have to add an element given in array y[], on the basis of the boolean value given in array check[]. 

For every index i (0<= i <= n - 1):

• If check[i] = true, Add y[i] to all the odd numbers present in the array.
• If check[i] = false, Add y[i] to all the even numbers present in the array.

Examples:

Input: n = 3, k = 3, arr[] = {1, 2, 4}, check[] = {false, true, false}, y[] = {2, 3, 5}
Output: 29
Explanation:

• for, k = 1
  check[0] = false, y[0] = 2
  So, as per operation add to the even number
  arr = [1, 2+2, 4+2] = {1, 4, 6}
• for, k = 2
  check[1] = true, y[1] = 3
  so, add to the odd number
  arr = [1+3, 4, 6] = {4, 4, 6}
• for, k = 3
  check[2] = false, y[2] = 5
  so, add to the even number
  arr = [4+5, 4+5, 6+5] = {9, 9, 11}

So, finally, print the sum of the array = 9 + 9 + 11 = 29

Input: n = 2, k = 1, arr[] = {1, 2}, check[] = {false}, y[] = {0}
Output: 3
Explanation: As check[0] = false, we have to y[0] to all the even elements present in the array. Therefore, the final sum is 1 + 2 =3.

Approach: The basic idea to solve the above problem is:

Iterate through whole the array arr[] for k times, check whether the element present is odd or even, and add an element accordingly. 

Time complexity: O(k*n), where k = no of queries , n = length of array
Auxiliary Space: O(1)

Efficient Approach: The above idea can be optimized as:

Iterate over the arr[], count the number of even and odd elements present. Now, Run a loop for k queries along by keep a check on array check[]. If check[i] == true, Add the total sum by count odd * y[i], otherwise if it's false, add  count even * y[i]. Also, change the count of  even and odd according to y[i], if it's odd. 

Steps involved in the implementation of the code:

Step 1: First calculate the total sum of the array.

Step 2: count of number of odd and even elements in the array.

Step 3: Run the loop for the k queries.

Step 4: If (check[first loop index] = true) then, 

• Update the total sum => total sum += (count odd * y[first loop index])
• If the number which is added y[first loop index] is odd then,
• Update even count => count even += odd count, and  odd count = 0

Step 5: If(check[first loop index] = false) then,

• Update the total sum => total sum += (count even* y[first loop index])
• If the number which is added y[first loop index] is odd then, 
• Update odd count => odd count += even count, and even count = 0

Step 6: Finally print the total sum.

Below is the implementation of the above approach:

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the final sum
long long sum(int n, int k, vector<int> arr,
              vector<bool> check, vector<int> y)
{
    // It will store the final answer
    long long totalSum = 0;

    int countOdd = 0, countEven = 0;

    // Loop for finding the totalSum
    // and count odd and even
    for (int i = 0; i < n; i++) {
        totalSum = totalSum + arr[i];
        if (arr[i] & 1)
            countOdd++;
        else
            countEven++;
    }

    // Loop for k queries
    for (int q = 0; q < k; q++) {

        if (check[q]) {
            // Adding to odd number
            totalSum = totalSum + (countOdd * 1LL * y[q]);

            if (y[q] & 1) {
                countEven = countEven + countOdd;
                countOdd = 0;
            }
        }
        else {
            // Adding to even number
            totalSum = totalSum + (countEven * 1LL * y[q]);

            if (y[q] & 1) {
                countOdd = countOdd + countEven;
                countEven = 0;
            }
        }
    }
    // Returning the final answer
    return totalSum;
}

// Driver code
int main()
{

    // Input 1
    int n1 = 3, k1 = 3;
    vector<int> arr1 = { 1, 2, 4 };
    vector<bool> check1 = { false, true, false };
    vector<int> y1 = { 2, 3, 5 };

    // Function call
    cout << sum(n1, k1, arr1, check1, y1) << endl;

    // Input 2
    int n2 = 6, k2 = 7;
    vector<int> arr2 = { 1, 3, 2, 4, 10, 48 };
    vector<bool> check2
        = { true, false, false, false, true, false, false };
    vector<int> y2 = { 6, 5, 4, 5, 3, 12, 1 };

    // Function call
    cout << sum(n2, k2, arr2, check2, y2) << endl;
    return 0;
}

import java.util.*;

class Main {
    // Function to calculate the final sum
    static long sum(int n, int k, int[] arr,
                    boolean[] check, int[] y)
    {
        // It will store the final answer
        long totalSum = 0;

        int countOdd = 0, countEven = 0;

        // Loop for finding the totalSum
        // and count odd and even
        for (int i = 0; i < n; i++) {
            totalSum = totalSum + arr[i];
            if ((arr[i] & 1) == 1)
                countOdd++;
            else
                countEven++;
        }

        // Loop for k queries
        for (int q = 0; q < k; q++) {

            if (check[q]) {
                // Adding to odd number
                totalSum
                    = totalSum + (countOdd * 1L * y[q]);

                if ((y[q] & 1) == 1) {
                    countEven = countEven + countOdd;
                    countOdd = 0;
                }
            }
            else {
                // Adding to even number
                totalSum
                    = totalSum + (countEven * 1L * y[q]);

                if ((y[q] & 1) == 1) {
                    countOdd = countOdd + countEven;
                    countEven = 0;
                }
            }
        }
        // Returning the final answer
        return totalSum;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Input 1
        int n1 = 3, k1 = 3;
        int[] arr1 = { 1, 2, 4 };
        boolean[] check1 = { false, true, false };
        int[] y1 = { 2, 3, 5 };

        // Function call
        System.out.println(sum(n1, k1, arr1, check1, y1));

        // Input 2
        int n2 = 6, k2 = 7;
        int[] arr2 = { 1, 3, 2, 4, 10, 48 };
        boolean[] check2 = { true, false, false, false,
                             true, false, false };
        int[] y2 = { 6, 5, 4, 5, 3, 12, 1 };

        // Function call
        System.out.println(sum(n2, k2, arr2, check2, y2));
    }
}

# Python implementation for the above approach

def sum(n, k, arr, check, y):
    # It will store the final answer
    total_sum = 0

    count_odd, count_even = 0, 0

    # Loop for finding the totalSum
    # and count odd and even
    for i in range(n):
        total_sum += arr[i]
        if arr[i]%2 != 0:
            count_odd += 1
        else:
            count_even += 1

    # Loop for k queries
    for q in range(k):
        if check[q]:
            # Adding to odd number
            total_sum += count_odd*y[q]
            if y[q]%2 != 0:
                count_even += count_odd
                count_odd = 0
        else:
            # Adding to even number
            total_sum += count_even*y[q]
            if y[q]%2 != 0:
                count_odd += count_even
                count_even = 0

    # Returning the final answer
    return total_sum


# Input 1
n1, k1 = 3, 3
arr1 = [1, 2, 4]
check1 = [False, True, False]
y1 = [2, 3, 5]

# function call
print(sum(n1, k1, arr1, check1, y1))

# Input 2
n2, k2 = 6, 7
arr2 = [1, 3, 2, 4, 10, 48]
check2 = [True, False, False, False, True, False, False]
y2 = [6, 5, 4, 5, 3, 12, 1]

# function call
print(sum(n2, k2, arr2, check2, y2))

using System;
using System.Collections.Generic;

class Program 
{
  
  // Function to calculate the final sum
  static long sum(int n, int k, List<int> arr,
                  List<bool> check, List<int> y)
  {
    
    // It will store the final answer
    long totalSum = 0;
    int countOdd = 0, countEven = 0;

    // Loop for finding the totalSum
    // and count odd and even
    for (int i = 0; i < n; i++) {
      totalSum = totalSum + arr[i];
      if ((arr[i] & 1) != 0)
        countOdd++;
      else
        countEven++;
    }

    // Loop for k queries
    for (int q = 0; q < k; q++) {

      if (check[q]) {
        // Adding to odd number
        totalSum
          = totalSum + (countOdd * (long)y[q]);

        if ((y[q] & 1) != 0) {
          countEven = countEven + countOdd;
          countOdd = 0;
        }
      }
      else {
        // Adding to even number
        totalSum
          = totalSum + (countEven * (long)y[q]);

        if ((y[q] & 1) != 0) {
          countOdd = countOdd + countEven;
          countEven = 0;
        }
      }
    }
    // Returning the final answer
    return totalSum;
  }

  // Driver code
  static void Main(string[] args)
  {

    // Input 1
    int n1 = 3, k1 = 3;
    List<int> arr1 = new List<int>{ 1, 2, 4 };
    List<bool> check1
      = new List<bool>{ false, true, false };
    List<int> y1 = new List<int>{ 2, 3, 5 };

    // Function call
    Console.WriteLine(sum(n1, k1, arr1, check1, y1));

    // Input 2
    int n2 = 6, k2 = 7;
    List<int> arr2
      = new List<int>{ 1, 3, 2, 4, 10, 48 };
    List<bool> check2
      = new List<bool>{ true, false, false, false,
                       true, false, false };
    List<int> y2
      = new List<int>{ 6, 5, 4, 5, 3, 12, 1 };

    // Function call
    Console.WriteLine(sum(n2, k2, arr2, check2, y2));
  }
}

// This code is contributed by divya_p123.

// JavaScript implementation of the above approach

// Function to calculate the final sum
function sum(n, k, arr, check, y) {
    // It will store the final answer
    var totalSum = 0;

    var countOdd = 0, countEven = 0;

    // Loop for finding the totalSum 
    // and count odd and even
    for (var i = 0; i < n; i++) {
        totalSum = totalSum + arr[i];
        if (arr[i] & 1) {
            countOdd++;
        }
        else {
            countEven++;
        }
    }

    // Loop for k queries
    for (var q = 0; q < k; q++) {
        if (check[q]) {
            // Adding to odd number
            totalSum = totalSum + countOdd * y[q];

            if (y[q] & 1) {
                countEven = countEven + countOdd;
                countOdd = 0;
            }
        }
        else {
            // Adding to even number
            totalSum = totalSum + countEven * y[q];

            if (y[q] & 1) {
                countOdd = countOdd + countEven;
                countEven = 0;
            }
        }
    }
    // Returning the final answer
    return totalSum;
}

// Driver code

// Input 1
var n1 = 3, k1 = 3;
var arr1 = [1, 2, 4];
var check1 = [false, true, false];
var y1 = [2, 3, 5];

// Function call
console.log(sum(n1, k1, arr1, check1, y1));

// Input 2
var n2 = 6, k2 = 7;
var arr2 = [1, 3, 2, 4, 10, 48];
var check2 = [true, false, false, false, true, false, false];
var y2 = [6, 5, 4, 5, 3, 12, 1];

// Function call
console.log(sum(n2, k2, arr2, check2, y2));

// This code is contributed by Prasad Kandekar(prasad264)

29
196

Time Complexity: O(max(n,  k)), where n = size of array, k = no of queries
Auxiliary Space: O(1)