Range queries for alternatively addition and subtraction on given Array

Given an array arr[] of N integers and Q queries where every row consists of two numbers L and R which denotes the range [L, R], the task is to find value of alternate addition and subtraction of the array element between range [L, R].

Examples:

Input: arr[] = {10, 13, 15, 2, 45, 31, 22, 3, 27}, Q[][] = {{2, 5}, {6, 8}, {1, 7}, {4, 8}, {0, 5}}
Output: 27 46 -33 60 24
Explanation:
Result of query {2, 5} is 27 ( 15 – 2 + 45 – 31)
Result of query {6, 8} is 46 ( 22 – 3 + 27)
Result of query {1, 7} is -33 ( 13 – 15 + 2 – 45 + 31 – 22 + 3)
Result of query {4, 8} is 60 ( 45 – 31 + 22 – 3 + 27)
Result of query {0, 5} is 24 ( 10 – 13 + 15 – 2 + 45 – 31)

Input: arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1}, Q[] = {{2, 5}, {6, 8}, {1, 7}, {4, 8}, {0, 5}}
Output: 0 1 1 1 0

Naive Approach: The naive idea is to iterate from index L to R for each query and find the value of alternatively adding and subtracting the elements of the array and print the value after the operations performed.



Below is the implementation of the above approach:

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Structure to represent a range query
struct Query {
    int L, R;
};
  
// Function to find the result of
// alternatively adding and subtracting
// elements in the range [L, R]
int findResultUtil(int arr[],
                   int L, int R)
{
    int result = 0;
  
    // A boolean variable flag to
    // alternatively add and subtract
    bool flag = false;
  
    // Iterate from [L, R]
    for (int i = L; i <= R; i++) {
  
        // if flag is false, then
        // add & toggle the flag
        if (flag == false) {
            result = result + arr[i];
            flag = true;
        }
  
        // if flag is true subtract
        // and toggle the flag
        else {
            result = result - arr[i];
            flag = false;
        }
    }
  
    // Return the final result
    return result;
}
  
// Function to find the value
// for each query
void findResult(int arr[], int n,
                Query q[], int m)
{
  
    // Iterate for each query
    for (int i = 0; i < m; i++) {
        cout << findResultUtil(arr,
                               q[i].L,
                               q[i].R)
             << " ";
    }
}
  
// Driver Code
int main()
{
  
    // Given array
    int arr[] = { 10, 13, 15, 2, 45,
                  31, 22, 3, 27 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Given Queries
    Query q[] = { { 2, 5 }, { 6, 8 }, 
                  { 1, 7 }, { 4, 8 }, 
                  { 0, 5 } };
  
    int m = sizeof(q) / sizeof(q[0]);
  
    // Function Call
    findResult(arr, n, q, m);
    return 0;
}

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Output:

27 46 -33 60 24

Time Complexity: O(N*Q)
Auxiliary Space: O(1)

Efficient Approach: The efficient approach is to use Prefix Sum Array to solve the above problem. Below are the steps:

  1. Initialize the first element of prefix array(say pre[]) to first element of the arr[].
  2. Traverse through index from 1 to N-1 and alternative add and subtract the elements of arr[i] from pre[i-1] and store it in pre[i] to make a prefix array.
  3. Now, Iterate through each query from 1 to Q, and find the result on the basis of below cases:
    • Case 1: If L is zero then result is pre[R].
    • Case 2: If L is non-zero, find the result using the equation:
      result = Query(L, R) = pre[R] – pre[L - 1]
      
    • If L is odd multiply the above result by -1.
  4. Below is the implementation of the above approach:

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    // C++ program for the above approach
    #include <bits/stdc++.h>
    using namespace std;
      
    // Structure to represent a query
    struct Query {
        int L, R;
    };
      
    // This function fills the Prefix Array
    void fillPrefixArray(int arr[], int n,
                         int prefixArray[])
    {
        // Initialise the prefix array
        prefixArray[0] = arr[0];
      
        // Iterate all the element of arr[]
        // and update the prefix array
        for (int i = 1; i < n; i++) {
      
            // If n is even then, add the
            // previous value of prefix array
            // with the current value of arr
            if (i % 2 == 0) {
      
                prefixArray[i]
                    = prefixArray[i - 1]
                      + arr[i];
            }
      
            // if n is odd, then subtract
            // the previous value of prefix
            // Array from current value
            else {
                prefixArray[i]
                    = prefixArray[i - 1]
                      - arr[i];
            }
        }
    }
      
    // Function to find the result of
    // alternatively adding and subtracting
    // elements in the range [L< R]
    int findResultUtil(int prefixArray[],
                       int L, int R)
    {
        int result;
      
        // Case 1 : when L is zero
        if (L == 0) {
            result = prefixArray[R];
        }
      
        // Case 2 : When L is non zero
        else {
            result = prefixArray[R]
                     - prefixArray[L - 1];
        }
      
        // If L is odd means range starts from
        // odd position multiply result by -1
        if (L & 1) {
            result = result * (-1);
        }
      
        // Return the final result
        return result;
    }
      
    // Function to find the sum of all
    // alternative add and subtract
    // between ranges [L, R]
    void findResult(int arr[], int n,
                    Query q[], int m)
    {
      
        // Declare prefix array
        int prefixArray[n];
      
        // Function Call to fill prefix arr[]
        fillPrefixArray(arr, n, prefixArray);
      
        // Iterate for each query
        for (int i = 0; i < m; i++) {
      
            cout << findResultUtil(prefixArray,
                                   q[i].L,
                                   q[i].R)
                 << " ";
        }
    }
      
    // Driver Code
    int main()
    {
      
        // Given array
        int arr[] = { 10, 13, 15, 2, 45,
                      31, 22, 3, 27 };
      
        int n = sizeof(arr) / sizeof(arr[0]);
      
        // Given Queries
        Query q[] = { { 2, 5 }, { 6, 8 },
                      { 1, 7 }, { 4, 8 }, 
                      { 0, 5 } };
      
        int m = sizeof(q) / sizeof(q[0]);
      
        // Function Call
        findResult(arr, n, q, m);
        return 0;
    }

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    Output:

    27 46 -33 60 24
    

    Time Complexity: O(N + Q)
    Auxiliary Space: O(N)

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