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Range queries for alternatively addition and subtraction on given Array
  • Last Updated : 27 Aug, 2020

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: 
 

C++




// 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;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Structure to represent a range query
static class Query
{
    int L, R;
    public Query(int l, int r)
    {
        super();
        L = l;
        R = r;
    }
};
 
// Function to find the result of
// alternatively adding and subtracting
// elements in the range [L, R]
static int findResultUtil(int arr[],
                          int L, int R)
{
    int result = 0;
 
    // A boolean variable flag to
    // alternatively add and subtract
    boolean 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
static void findResult(int arr[], int n,
                       Query q[], int m)
{
 
    // Iterate for each query
    for(int i = 0; i < m; i++)
    {
        System.out.print(findResultUtil(arr,
              q[i].L, q[i].R) + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given array
    int arr[] = { 10, 13, 15, 2, 45,
                  31, 22, 3, 27 };
 
    int n = arr.length;
 
    // Given Queries
    Query q[] = { new Query(2, 5), new Query(6, 8),
                  new Query(1, 7), new Query(4, 8),
                  new Query(0, 5) };
 
    int m = q.length;
 
    // Function call
    findResult(arr, n, q, m);
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program for the above approach
 
# Function to find the result of
# alternatively adding and subtracting
# elements in the range [L, R]
def findResultUtil(arr, L, R):
     
    result = 0
 
    # A boolean variable flag to
    # alternatively add and subtract
    flag = False
 
    # Iterate from [L, R]
    for i in range(L, R + 1):
         
        # 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
def findResult(arr, n, q, m):
 
    # Iterate for each query
    for i in range(m):
        print(findResultUtil(arr,
                             q[i][0],
                             q[i][1]),
                             end = " ")
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    arr = [ 10, 13, 15, 2, 45,
            31, 22, 3, 27 ]
 
    n = len(arr)
 
    # Given Queries
    q = [ [ 2, 5 ], [ 6, 8 ],
          [ 1, 7 ], [ 4, 8 ],
          [ 0, 5 ] ]
 
    m = len(q)
 
    # Function Call
    findResult(arr, n, q, m)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
class GFG{
 
// Structure to represent a range query
class Query
{
    public int L, R;
    public Query(int l, int r)
    {
        L = l;
        R = r;
    }
};
 
// Function to find the result of
// alternatively adding and subtracting
// elements in the range [L, R]
static int findResultUtil(int []arr,
                          int L, int R)
{
    int result = 0;
 
    // A bool 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 readonly result
    return result;
}
 
// Function to find the value
// for each query
static void findResult(int []arr, int n,
                       Query []q, int m)
{
 
    // Iterate for each query
    for(int i = 0; i < m; i++)
    {
        Console.Write(findResultUtil(arr,
              q[i].L, q[i].R) + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Given array
    int []arr = { 10, 13, 15, 2, 45,
                  31, 22, 3, 27 };
 
    int n = arr.Length;
 
    // Given Queries
    Query []q = { new Query(2, 5), new Query(6, 8),
                  new Query(1, 7), new Query(4, 8),
                  new Query(0, 5) };
 
    int m = q.Length;
 
    // Function call
    findResult(arr, n, q, m);
}
}
 
// This code is contributed by gauravrajput1
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.

Below is the implementation of the above approach: 
 

CPP




// 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;
}

Java




// Java program for the above approach
import java.util.*;
class GFG{
 
// Structure to represent a query
static class Query
{
    int L, R;
 
    public Query(int l, int r)
    {
        super();
        L = l;
        R = r;
    }
};
 
// This function fills the Prefix Array
static 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]
static 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 % 2 == 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]
static void findResult(int arr[], int n,
                         Query q[], int m)
{
 
    // Declare prefix array
    int []prefixArray = new int[n];
 
    // Function Call to fill prefix arr[]
    fillPrefixArray(arr, n, prefixArray);
 
    // Iterate for each query
    for (int i = 0; i < m; i++)
    {
        System.out.print(findResultUtil(
                           prefixArray, q[i].L,
                                        q[i].R) + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given array
    int arr[] = { 10, 13, 15, 2, 45,
                31, 22, 3, 27 };
 
    int n = arr.length;
 
    // Given Queries
    Query q[] = {new Query( 2, 5 ), new Query( 6, 8 ),
                 new Query( 1, 7 ), new Query( 4, 8 ),
                 new Query( 0, 5 )};
 
    int m = q.length;
 
    // Function Call
    findResult(arr, n, q, m);
}
}
 
// This code is contributed by PrinciRaj1992

C#




// C# program for the above approach
using System;
class GFG{
 
// Structure to represent a query
class Query
{
   public  int L, R;
 
    public Query(int l, int r)
    {
        L = l;
        R = r;
    }
};
 
// This function fills the Prefix Array
static 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]
static 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 % 2 == 1)
    {
        result = result * (-1);
    }
 
    // Return the readonly result
    return result;
}
 
// Function to find the sum of all
// alternative add and subtract
// between ranges [L, R]
static void findResult(int []arr, int n,
                       Query []q, int m)
{
 
    // Declare prefix array
    int []prefixArray = new int[n];
 
    // Function Call to fill prefix []arr
    fillPrefixArray(arr, n, prefixArray);
 
    // Iterate for each query
    for (int i = 0; i < m; i++)
    {
        Console.Write(findResultUtil(
                           prefixArray, q[i].L,
                                        q[i].R) + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Given array
    int []arr = { 10, 13, 15, 2, 45,
                31, 22, 3, 27 };
 
    int n = arr.Length;
 
    // Given Queries
    Query []q = {new Query( 2, 5 ), new Query( 6, 8 ),
                 new Query( 1, 7 ), new Query( 4, 8 ),
                 new Query( 0, 5 )};
 
    int m = q.Length;
 
    // Function Call
    findResult(arr, n, q, m);
}
}
 
// This code is contributed by Rohit_ranjan

Output: 
 

27 46 -33 60 24






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

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