Range queries for alternatively addition and subtraction on given Array
Last Updated :
07 Apr, 2022
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 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 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# 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
<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>
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 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
# 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# 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
<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)
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