Given an array arr[] of N integers and matrix Queries[][] consisting of Q queries of the form {m, a, b}. For each query the task is to find the sum of array elements according to the following conditions:
- If m = 1: Find the sum of the array elements in the range [a, b].
- If m = 2: Rearrange the array elements in increasing order and find the sum of the elements in the range [a, b] in the new array.
Examples:
Input: arr[] = {6, 4, 2, 7, 2, 7}, Q = 3, Queries[][3] = {{2, 3, 6}, {1, 3, 4}, {1, 1, 6}}
Output: 24 9 28
Explanation:
For Query 1:
m is 2, then array after sorting is arr[] = {2, 2, 4, 6, 7, 7} and sum of element in the range [3, 6] is 4 + 6 + 7 + 7 = 24.
For Query 2:
m is 1, then original array is arr[] = {6, 4, 2, 7, 2, 7} and sum of element in the range [3, 4] is 2 + 7 = 9.
For Query 3:
m is 1, then original array is arr[] = {6, 4, 2, 7, 2, 7} and sum of element in the range [1, 6] is 6 + 4 + 2 + 7 + 2 + 7 = 28.
Input: arr[] = {5, 5, 2, 3}, Q = 3, Queries[][10] = {{1, 2, 4}, {2, 1, 4}, {1, 1, 1}, {2, 1, 4}, {2, 1, 2}, {1, 1, 1}, {1, 3, 3}, {1, 1, 3}, {1, 4, 4}, {1, 2, 2}}
Output: 10 15 5 15 5 5 2 12 3 5
Naive Approach: The idea is to traverse the given queries and find the sum of all the elements according to the given conditions:
- Choose the array according to the given m, if m is equal to 1 then choose the original array otherwise choose the other array where all the elements of the array arr[] are sorted.
- Now calculate the sum of the array between the range [a, b].
- Iterate a loop over the range to find the sum.
- Print the sum for each query.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int range_sum(vector<int> v, int a,
int b)
{
int sum = 0;
for (int i = a - 1; i < b; i++) {
sum += v[i];
}
return sum;
}
void findSum(vector<int>& v1, int q,
int Queries[][3])
{
vector<int> v2;
int n = sizeof(v1) / sizeof(int);
for (int i = 0; i < n; i++) {
v2.push_back(v1[i]);
}
sort(v2.begin(), v2.end());
for (int i = 0; i < q; i++) {
int m = Queries[i][0];
int a = Queries[i][1];
int b = Queries[i][2];
if (m == 1) {
cout << range_sum(v1, a, b)
<< ' ';
}
else if (m == 2) {
cout << range_sum(v2, a, b)
<< ' ';
}
}
}
int main()
{
vector<int> arr = { 6, 4, 2, 7, 2, 7 };
int Q = 1;
int Queries[][3] = { { 2, 3, 6 } };
findSum(arr, Q, Queries);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int range_sum(Vector<Integer> v, int a,
int b)
{
int sum = 0;
for(int i = a - 1; i < b; i++)
{
sum += v.get(i);
}
return sum;
}
static int range_sum(int []v, int a,
int b)
{
int sum = 0;
for(int i = a - 1; i < b; i++)
{
sum += v[i];
}
return sum;
}
static void findSum(int []v1, int q,
int Queries[][])
{
Vector<Integer> v2 = new Vector<Integer>();
int n = v1.length;
for(int i = 0; i < n; i++)
{
v2.add(v1[i]);
}
Collections.sort(v2);
for(int i = 0; i < q; i++)
{
int m = Queries[i][0];
int a = Queries[i][1];
int b = Queries[i][2];
if (m == 1)
{
System.out.print(range_sum(
v1, a, b) + " ");
}
else if (m == 2)
{
System.out.print(range_sum(
v2, a, b) + " ");
}
}
}
public static void main(String[] args)
{
int []arr = { 6, 4, 2, 7, 2, 7 };
int Q = 1;
int Queries[][] = { { 2, 3, 6 } };
findSum(arr, Q, Queries);
}
}
|
Python3
def range_sum(v, a, b):
Sum = 0
for i in range(a - 1, b):
Sum += v[i]
return Sum
def findSum(v1, q, Queries):
v2 = []
n = len(v1)
for i in range(n):
v2.append(v1[i])
v2.sort()
for i in range(q):
m = Queries[i][0]
a = Queries[i][1]
b = Queries[i][2]
if (m == 1):
print(range_sum(v1, a, b), end = " ")
elif (m == 2):
print(range_sum(v2, a, b), end = " ")
arr = [ 6, 4, 2, 7, 2, 7 ]
Q = 1
Queries = [ [ 2, 3, 6 ] ]
findSum(arr, Q, Queries)
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
class GFG{
static int range_sum(ArrayList v, int a,
int b)
{
int sum = 0;
for(int i = a - 1; i < b; i++)
{
sum += (int)v[i];
}
return sum;
}
static void findSum(ArrayList v1, int q,
int [,]Queries)
{
ArrayList v2 = new ArrayList();
int n = v1.Count;
for(int i = 0; i < n; i++)
{
v2.Add(v1[i]);
}
v2.Sort();
for(int i = 0; i < q; i++)
{
int m = Queries[i, 0];
int a = Queries[i, 1];
int b = Queries[i, 2];
if (m == 1)
{
Console.Write(range_sum(v1, a, b));
Console.Write(' ');
}
else if (m == 2)
{
Console.Write(range_sum(v2, a, b));
Console.Write(' ');
}
}
}
public static void Main(string[] args)
{
ArrayList arr=new ArrayList(){ 6, 4, 2,
7, 2, 7 };
int Q = 1;
int [,]Queries = { { 2, 3, 6 } };
findSum(arr, Q, Queries);
}
}
|
Javascript
<script>
function range_sum(v, a, b)
{
let sum = 0;
for(let i = a - 1; i < b; i++)
{
sum += v[i];
}
return sum;
}
function findSum(v1, q, Queries)
{
let v2 = [];
let n = v1.length;
for(let i = 0; i < n; i++)
{
v2.push(v1[i]);
}
v2.sort(function(a, b){return a - b;});
for(let i = 0; i < q; i++)
{
let m = Queries[i][0];
let a = Queries[i][1];
let b = Queries[i][2];
if (m == 1)
{
document.write(range_sum(
v1, a, b) + " ");
}
else if (m == 2)
{
document.write(range_sum(
v2, a, b) + " ");
}
}
}
let arr = [ 6, 4, 2, 7, 2, 7 ];
let Q = 1;
let Queries = [ [ 2, 3, 6 ] ];
findSum(arr, Q, Queries);
</script>
|
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized by reducing one loop using the prefix sum array. Below are the steps:
- Create the other array brr[] to store the elements of the given array arr[] in sorted order.
- Find the prefix sum of both the arrays arr[] and brr[].
- Traverse the given queries and if the query is of type 1 then the sum of the element in the range [a, b] is given by:
arr[b – 1] – arr[a – 2]
- If the query is of type 2 then the sum of the element in the range [a, b] is given by:
brr[b – 1] – arr[a – 2]
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int range_sum(vector<int>& arr, int a,
int b)
{
int sum = 0;
if (a - 2 < 0)
sum = arr[b - 1];
else
sum = arr[b - 1] - arr[a - 2];
return sum;
}
void precompute_sum(vector<int>& arr,
vector<int>& brr)
{
int N = (int)arr.size();
for (int i = 1; i <= N; i++) {
arr[i] = arr[i] + arr[i - 1];
brr[i] = brr[i] + brr[i - 1];
}
}
void find_sum(vector<int>& arr, int q,
int Queries[][3])
{
vector<int> brr(arr);
int N = (int)arr.size();
sort(brr.begin(), brr.end());
precompute_sum(arr, brr);
for (int i = 0; i < q; i++) {
int m = Queries[i][0];
int a = Queries[i][1];
int b = Queries[i][2];
if (m == 1) {
cout << range_sum(arr, a, b)
<< ' ';
}
else if (m == 2) {
cout << range_sum(brr, a, b)
<< ' ';
}
}
}
int main()
{
vector<int> arr = { 0, 6, 4, 2, 7, 2, 7 };
int Q = 1;
int Queries[][3] = { { 2, 3, 6 } };
find_sum(arr, Q, Queries);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int range_sum(int []arr, int a,
int b)
{
int sum = 0;
if (a - 2 < 0)
sum = arr[b - 1];
else
sum = arr[b - 1] - arr[a - 2];
return sum;
}
static void precompute_sum(int []arr,
int []brr)
{
int N = (int)arr.length;
for(int i = 1; i < N; i++)
{
arr[i] = arr[i] + arr[i - 1];
brr[i] = brr[i] + brr[i - 1];
}
}
static void find_sum(int []arr, int q,
int Queries[][])
{
int []brr = arr.clone();
int N = (int)arr.length;
Arrays.sort(brr);
precompute_sum(arr, brr);
for(int i = 0; i < q; i++)
{
int m = Queries[i][0];
int a = Queries[i][1];
int b = Queries[i][2];
if (m == 1)
{
System.out.print(range_sum(
arr, a, b) + " ");
}
else if (m == 2)
{
System.out.print(range_sum(
brr, a, b) + " ");
}
}
}
public static void main(String[] args)
{
int []arr = { 0, 6, 4, 2, 7, 2, 7 };
int Q = 1;
int Queries[][] = { { 2, 3, 6 } };
find_sum(arr, Q, Queries);
}
}
|
Python3
def range_sum(arr, a, b):
sum = 0
if (a - 2 < 0):
sum = arr[b - 1]
else:
sum = (arr[b - 1] -
arr[a - 2])
return sum
def precompute_sum(arr, brr):
N = len(arr)
for i in range(1, N):
arr[i] = arr[i] + arr[i - 1]
brr[i] = brr[i] + brr[i - 1]
def find_sum(arr, q, Queries):
brr = arr.copy()
N = len(arr)
brr.sort()
precompute_sum(arr, brr)
for i in range(q):
m = Queries[i][0]
a = Queries[i][1]
b = Queries[i][2]
if (m == 1):
print (range_sum(arr,
a, b),
end = ' ')
elif (m == 2):
print (range_sum(brr,
a, b),
end = ' ')
if __name__ == "__main__":
arr = [0, 6, 4,
2, 7, 2, 7]
Q = 1
Queries = [[2, 3, 6]]
find_sum(arr, Q, Queries)
|
C#
using System;
class GFG{
static int range_sum(int []arr,
int a,
int b)
{
int sum = 0;
if (a - 2 < 0)
sum = arr[b - 1];
else
sum = arr[b - 1] - arr[a - 2];
return sum;
}
static void precompute_sum(int []arr,
int []brr)
{
int N = (int)arr.Length;
for(int i = 1; i < N; i++)
{
arr[i] = arr[i] + arr[i - 1];
brr[i] = brr[i] + brr[i - 1];
}
}
static void find_sum(int []arr, int q,
int [,]Queries)
{
int []brr = new int[arr.Length];
arr.CopyTo(brr, 0);
int N = (int)arr.Length;
Array.Sort(brr);
precompute_sum(arr, brr);
for(int i = 0; i < q; i++)
{
int m = Queries[i, 0];
int a = Queries[i, 1];
int b = Queries[i, 2];
if (m == 1)
{
Console.Write(range_sum(
arr, a, b) + " ");
}
else if (m == 2)
{
Console.Write(range_sum(
brr, a, b) + " ");
}
}
}
public static void Main(String[] args)
{
int []arr = {0, 6, 4, 2, 7, 2, 7};
int Q = 1;
int [,]Queries = {{2, 3, 6}};
find_sum(arr, Q, Queries);
}
}
|
Javascript
<script>
function range_sum(arr,a,b)
{
let sum = 0;
if (a - 2 < 0)
sum = arr[b - 1];
else
sum = arr[b - 1] - arr[a - 2];
return sum;
}
function precompute_sum(arr,brr)
{
let N = arr.length;
for(let i = 1; i < N; i++)
{
arr[i] = arr[i] + arr[i - 1];
brr[i] = brr[i] + brr[i - 1];
}
}
function find_sum(arr,q,Queries)
{
let brr = [...arr];
let N = arr.length;
brr.sort(function(a,b){return a-b;});
precompute_sum(arr, brr);
for(let i = 0; i < q; i++)
{
let m = Queries[i][0];
let a = Queries[i][1];
let b = Queries[i][2];
if (m == 1)
{
document.write(range_sum(
arr, a, b) + " ");
}
else if (m == 2)
{
document.write(range_sum(
brr, a, b) + " ");
}
}
}
let arr=[0, 6, 4, 2, 7, 2, 7];
let Q = 1;
let Queries = [[ 2, 3, 6 ]];
find_sum(arr, Q, Queries);
</script>
|
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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Last Updated :
17 Jun, 2021
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