Given an array arr[] of N integers and Q queries of the form {X, Y} of the following two types:
- If X = 1, rotate the given array to the left by Y positions.
- If X = 2, print the maximum sum subarray of length Y in the current state of the array.
Examples:
Input: N = 5, arr[] = {1, 2, 3, 4, 5}, Q = 2, Query[][] = {{1, 2}, {2, 3}}
Output:
Query 1: 3 4 5 1 2
Query 2: 12
Explanation:
Query 1: Shift array to the left 2 times: {1, 2, 3, 4, 5} -> {2, 3, 4, 5, 1} -> {3, 4, 5, 1, 2}
Query 2: Maximum sum subarray of length 3 is {3, 4, 5} and the sum is 12Input: N = 5, arr[] = {3, 4, 5, 1, 2}, Q = 3, Query[][] = {{1, 3}, {1, 1}, {2, 4}}
Output:
Query 1: 1 2 3 4 5
Query 2: 2 3 4 5 1
Query 3: 14
Explanation:
Query 1: Shift array to the left 3 times: {3, 4, 5, 1, 2} -> {4, 5, 1, 2, 3} -> {5, 1, 2, 3, 4} -> {1, 2, 3, 4, 5}
Query 2: Shift array to the left 1 time: {1, 2, 3, 4, 5} -> {2, 3, 4, 5, 1}
Query 3: Maximum sum subarray of length 4 is {2, 3, 4, 5} and sum is 14
Naive Approach: The simplest approach is to rotate the array by shifting elements one by one up to distance Y for queries is of type 1 and generating the sum of all the subarrays of length Y and print the maximum sum if the query is of type 2.
Time Complexity: O(Q*N*Y)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use the Juggling Algorithm for array rotation and for finding the maximum sum subarray of length Y, use the Sliding Window Technique. Follow the steps below to solve the problem:
- If X = 1, rotate the array by Y, using the Juggling Algorithm.
- Otherwise, if X = 2, find the maximum sum subarray of length Y using the Sliding Window Technique.
- Print the array if query X is 1.
- Otherwise, print the maximum sum subarray of size Y.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the maximum // sum of length k int MaxSum(vector< int > arr, int n,
int k)
{ int i, max_sum = 0, sum = 0;
// Calculating the max sum for
// the first k elements
for (i = 0; i < k; i++) {
sum += arr[i];
}
max_sum = sum;
// Find subarray with maximum sum
while (i < n) {
// Update the sum
sum = sum - arr[i - k] + arr[i];
if (max_sum < sum) {
max_sum = sum;
}
i++;
}
// Return maximum sum
return max_sum;
} // Function to calculate gcd of the // two numbers n1 and n2 int gcd( int n1, int n2)
{ // Base Case
if (n2 == 0) {
return n1;
}
// Recursively find the GCD
else {
return gcd(n2, n1 % n2);
}
} // Function to rotate the array by Y vector< int > RotateArr(vector< int > arr,
int n, int d)
{ // For handling k >= N
int i = 0, j = 0;
d = d % n;
// Dividing the array into
// number of sets
int no_of_sets = gcd(d, n);
for (i = 0; i < no_of_sets; i++) {
int temp = arr[i];
j = i;
// Rotate the array by Y
while ( true ) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break ;
arr[j] = arr[k];
j = k;
}
// Update arr[j]
arr[j] = temp;
}
// Return the rotated array
return arr;
} // Function that performs the queries // on the given array void performQuery(vector< int >& arr,
int Q[][2], int q)
{ int N = ( int )arr.size();
// Traverse each query
for ( int i = 0; i < q; i++) {
// If query of type X = 1
if (Q[i][0] == 1) {
arr = RotateArr(arr, N,
Q[i][1]);
// Print the array
for ( auto t : arr) {
cout << t << " " ;
}
cout << "\n" ;
}
// If query of type X = 2
else {
cout << MaxSum(arr, N, Q[i][1])
<< "\n" ;
}
}
} // Driver Code int main()
{ // Given array arr[]
vector< int > arr = { 1, 2, 3, 4, 5 };
int q = 5;
// Given Queries
int Q[][2] = { { 1, 2 }, { 2, 3 },
{ 1, 3 }, { 1, 1 },
{ 2, 4 }
};
// Function Call
performQuery(arr, Q, q);
return 0;
} |
// Java program for // the above approach class GFG{
// Function to calculate the maximum // sum of length k static int MaxSum( int []arr,
int n, int k)
{ int i, max_sum = 0 , sum = 0 ;
// Calculating the max sum for
// the first k elements
for (i = 0 ; i < k; i++)
{
sum += arr[i];
}
max_sum = sum;
// Find subarray with maximum sum
while (i < n)
{
// Update the sum
sum = sum - arr[i - k] +
arr[i];
if (max_sum < sum)
{
max_sum = sum;
}
i++;
}
// Return maximum sum
return max_sum;
} // Function to calculate gcd // of the two numbers n1 and n2 static int gcd( int n1, int n2)
{ // Base Case
if (n2 == 0 )
{
return n1;
}
// Recursively find the GCD
else
{
return gcd(n2, n1 % n2);
}
} // Function to rotate the array by Y static int []RotateArr( int []arr,
int n, int d)
{ // For handling k >= N
int i = 0 , j = 0 ;
d = d % n;
// Dividing the array into
// number of sets
int no_of_sets = gcd(d, n);
for (i = 0 ; i < no_of_sets; i++)
{
int temp = arr[i];
j = i;
// Rotate the array by Y
while ( true )
{
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break ;
arr[j] = arr[k];
j = k;
}
// Update arr[j]
arr[j] = temp;
}
// Return the rotated array
return arr;
} // Function that performs the queries // on the given array static void performQuery( int []arr,
int Q[][], int q)
{ int N = arr.length;
// Traverse each query
for ( int i = 0 ; i < q; i++)
{
// If query of type X = 1
if (Q[i][ 0 ] == 1 )
{
arr = RotateArr(arr, N,
Q[i][ 1 ]);
// Print the array
for ( int t : arr)
{
System.out.print(t + " " );
}
System.out.print( "\n" );
}
// If query of type X = 2
else
{
System.out.print(MaxSum(arr, N,
Q[i][ 1 ]) + "\n" );
}
}
} // Driver Code public static void main(String[] args)
{ // Given array arr[]
int []arr = { 1 , 2 , 3 , 4 , 5 };
int q = 5 ;
// Given Queries
int Q[][] = {{ 1 , 2 }, { 2 , 3 },
{ 1 , 3 }, { 1 , 1 },
{ 2 , 4 }};
// Function Call
performQuery(arr, Q, q);
} } // This code is contributed by Rajput-Ji |
# Python3 program for the above approach # Function to calculate the maximum # sum of length k def MaxSum(arr, n, k):
i, max_sum = 0 , 0
sum = 0
# Calculating the max sum for
# the first k elements
while i < k:
sum + = arr[i]
i + = 1
max_sum = sum
# Find subarray with maximum sum
while (i < n):
# Update the sum
sum = sum - arr[i - k] + arr[i]
if (max_sum < sum ):
max_sum = sum
i + = 1
# Return maximum sum
return max_sum
# Function to calculate gcd of the # two numbers n1 and n2 def gcd(n1, n2):
# Base Case
if (n2 = = 0 ):
return n1
# Recursively find the GCD
else :
return gcd(n2, n1 % n2)
# Function to rotate the array by Y def RotateArr(arr, n, d):
# For handling k >= N
i = 0
j = 0
d = d % n
# Dividing the array into
# number of sets
no_of_sets = gcd(d, n)
for i in range (no_of_sets):
temp = arr[i]
j = i
# Rotate the array by Y
while ( True ):
k = j + d
if (k > = n):
k = k - n
if (k = = i):
break
arr[j] = arr[k]
j = k
# Update arr[j]
arr[j] = temp
# Return the rotated array
return arr
# Function that performs the queries # on the given array def performQuery(arr, Q, q):
N = len (arr)
# Traverse each query
for i in range (q):
# If query of type X = 1
if (Q[i][ 0 ] = = 1 ):
arr = RotateArr(arr, N, Q[i][ 1 ])
# Print the array
for t in arr:
print (t, end = " " )
print ()
# If query of type X = 2
else :
print (MaxSum(arr, N, Q[i][ 1 ]))
# Driver Code if __name__ = = '__main__' :
# Given array arr[]
arr = [ 1 , 2 , 3 , 4 , 5 ]
q = 5
# Given Queries
Q = [ [ 1 , 2 ], [ 2 , 3 ],
[ 1 , 3 ], [ 1 , 1 ],
[ 2 , 4 ] ]
# Function call
performQuery(arr, Q, q)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
class GFG{
// Function to calculate the maximum // sum of length k static int MaxSum( int [] arr, int n,
int k)
{ int i, max_sum = 0, sum = 0;
// Calculating the max sum for
// the first k elements
for (i = 0; i < k; i++)
{
sum += arr[i];
}
max_sum = sum;
// Find subarray with maximum sum
while (i < n)
{
// Update the sum
sum = sum - arr[i - k] +
arr[i];
if (max_sum < sum)
{
max_sum = sum;
}
i++;
}
// Return maximum sum
return max_sum;
} // Function to calculate gcd // of the two numbers n1 and n2 static int gcd( int n1, int n2)
{ // Base Case
if (n2 == 0)
{
return n1;
}
// Recursively find the GCD
else
{
return gcd(n2, n1 % n2);
}
} // Function to rotate the array by Y static int []RotateArr( int []arr, int n,
int d)
{ // For handling k >= N
int i = 0, j = 0;
d = d % n;
// Dividing the array into
// number of sets
int no_of_sets = gcd(d, n);
for (i = 0; i < no_of_sets; i++)
{
int temp = arr[i];
j = i;
// Rotate the array by Y
while ( true )
{
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break ;
arr[j] = arr[k];
j = k;
}
// Update arr[j]
arr[j] = temp;
}
// Return the rotated array
return arr;
} // Function that performs the queries // on the given array static void performQuery( int [] arr,
int [,] Q, int q)
{ int N = arr.Length;
// Traverse each query
for ( int i = 0; i < q; i++)
{
// If query of type X = 1
if (Q[i, 0] == 1)
{
arr = RotateArr(arr, N,
Q[i, 1]);
// Print the array
for ( int t = 0; t < arr.Length; t++)
{
Console.Write(arr[t] + " " );
}
Console.WriteLine();
}
// If query of type X = 2
else
{
Console.WriteLine(MaxSum(arr, N,
Q[i, 1]));
}
}
} // Driver code static void Main()
{ // Given array arr[]
int [] arr = { 1, 2, 3, 4, 5 };
int q = 5;
// Given Queries
int [,] Q = { { 1, 2 }, { 2, 3 },
{ 1, 3 }, { 1, 1 },
{ 2, 4 } };
// Function call
performQuery(arr, Q, q);
} } // This code is contributed by divyeshrabadiya07 |
<script> // javascript program for // the above approach // Function to calculate the maximum // sum of length k
function MaxSum(arr , n , k) {
var i, max_sum = 0, sum = 0;
// Calculating the max sum for
// the first k elements
for (i = 0; i < k; i++) {
sum += arr[i];
}
max_sum = sum;
// Find subarray with maximum sum
while (i < n) {
// Update the sum
sum = sum - arr[i - k] + arr[i];
if (max_sum < sum) {
max_sum = sum;
}
i++;
}
// Return maximum sum
return max_sum;
}
// Function to calculate gcd
// of the two numbers n1 and n2
function gcd(n1 , n2) {
// Base Case
if (n2 == 0) {
return n1;
}
// Recursively find the GCD
else {
return gcd(n2, n1 % n2);
}
}
// Function to rotate the array by Y
function RotateArr(arr , n , d) {
// For handling k >= N
var i = 0, j = 0;
d = d % n;
// Dividing the array into
// number of sets
var no_of_sets = gcd(d, n);
for (i = 0; i < no_of_sets; i++) {
var temp = arr[i];
j = i;
// Rotate the array by Y
while ( true ) {
var k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break ;
arr[j] = arr[k];
j = k;
}
// Update arr[j]
arr[j] = temp;
}
// Return the rotated array
return arr;
}
// Function that performs the queries
// on the given array
function performQuery(arr , Q , q) {
var N = arr.length;
// Traverse each query
for (i = 0; i < q; i++) {
// If query of type X = 1
if (Q[i][0] == 1) {
arr = RotateArr(arr, N, Q[i][1]);
// Print var the array
for ( var t =0 ;t< arr.length;t++) {
document.write(arr[t] + " " );
}
document.write( "<br/>" );
}
// If query of type X = 2
else {
document.write(MaxSum(arr, N, Q[i][1]) + "<br/>" );
}
}
}
// Driver Code
// Given array arr
var arr = [ 1, 2, 3, 4, 5 ];
var q = 5;
// Given Queries
var Q = [ [ 1, 2 ], [ 2, 3 ], [ 1, 3 ], [ 1, 1 ], [ 2, 4 ] ];
// Function Call
performQuery(arr, Q, q);
// This code contributed by aashish1995 </script> |
3 4 5 1 2 12 1 2 3 4 5 2 3 4 5 1 14
Time Complexity: O(Q*N), where Q is the number of queries, and N is the size of the given array.
Auxiliary Space: O(N)