Range queries for alternatively addition and subtraction on given Array

0 0 14 minutes read

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 the 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 are 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

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to find the result of
// alternatively adding and subtracting
// elements in the range [L, R]
function findResultUtil(arr, L, R)
{
    var result = 0;
 
    // A boolean variable flag to
    // alternatively add and subtract
    var flag = false;
 
    // Iterate from [L, R]
    for (var 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
function findResult(arr, n, q, m)
{
 
    // Iterate for each query
    for (var i = 0; i < m; i++) {
        document.write( findResultUtil(arr,
                               q[i][0],
                               q[i][1])
             + " ");
    }
}
 
// Driver Code
// Given array
var arr = [10, 13, 15, 2, 45,
              31, 22, 3, 27 ];
var n = arr.length;
// Given Queries
var q = [ [ 2, 5 ], [ 6, 8 ], 
              [ 1, 7 ], [ 4, 8 ], 
              [ 0, 5 ] ];
var m = q.length;
// Function Call
findResult(arr, n, q, m);
 
 
</script>

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:

Initialize the first element of prefix array(say pre[]) to first element of the arr[].
Traverse through an 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.
Now, Iterate through each query from 1 to Q, and find the result on the basis of the 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:

C++

// 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

Python3

# Python program for the above approach
 
# Structure to represent a query
class Query:
    def __init__(self,l,r):
        self.L = l
        self.R = r
 
# This function fills the Prefix Array
def fillPrefixArray(arr, n, prefixArray):
     
    # Initialise the prefix array
    prefixArray[0] = arr[0]
 
    # Iterate all the element of arr[]
    # and update the prefix array
    for i in range(1,n):
         
        # 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]
def findResultUtil(prefixArray, L, R):
 
    # 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]
def findResult(arr, n, q, m):
     
    # Declare prefix array
    prefixArray = [0 for i in range(n)]
 
    # Function Call to fill prefix arr[]
    fillPrefixArray(arr, n, prefixArray)
     
    # Iterate for each query
    for i in range(m):
        print(findResultUtil(prefixArray, q[i].L,q[i].R),end = " ")
 
# Driver Code
 
# Given array
arr = [ 10, 13, 15, 2, 45, 31, 22, 3, 27 ]
n = len(arr)
 
# Given Queries
q = [ Query(2, 5), Query(6, 8), Query(1, 7), Query(4, 8), Query(0, 5)]
m = len(q)
 
# Function Call
findResult(arr, n, q, m)
 
# This code is contributed by shinjanpatra

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

Javascript

<script>
 
// Javascript program for the above approach
 
// Structure to represent a query
class Query
{
    constructor(l, r)
    {
        this.L = l;
        this.R = r;
    }
}
 
// This function fills the Prefix Array
function fillPrefixArray(arr, n, prefixArray)
{
     
    // Initialise the prefix array
    prefixArray[0] = arr[0];
  
    // Iterate all the element of arr[]
    // and update the prefix array
    for(let 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]
function findResultUtil(prefixArray, L, R)
{
     let 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]
function findResult(arr, n, q, m)
{
     
    // Declare prefix array
    let prefixArray = new Array(n);
  
    // Function Call to fill prefix arr[]
    fillPrefixArray(arr, n, prefixArray);
  
    // Iterate for each query
    for(let i = 0; i < m; i++)
    {
        document.write(findResultUtil(
                        prefixArray, q[i].L,
                                     q[i].R) + " ");
    }
}
 
// Driver Code
 
// Given array
let arr = [ 10, 13, 15, 2, 45,
            31, 22, 3, 27 ];
let n = arr.length;
 
// Given Queries
let q = [ new Query(2, 5), new Query(6, 8),
          new Query(1, 7), new Query(4, 8),
          new Query(0, 5)];
let m = q.length;
 
// Function Call
findResult(arr, n, q, m);
 
// This code is contributed by unknown2108
 
</script>

Output:

27 46 -33 60 24

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!