You are given an array of n-elements and an odd-integer m. You have to construct a new sum_array from given array such that sum_array[i] = ?arr[j] for (i-(m/2)) < j (i+(m/2)).
note : for 0 > j or j >= n take arr[j] = 0.
Examples:
Input : arr[] = {1, 2, 3, 4, 5}, m = 3 Output : sum_array = {3, 6, 9, 12, 9} Explanation : sum_array[0] = arr[0] + arr[1] sum_array[1] = arr[0] + arr[1] + arr[2] sum_array[2] = arr[1] + arr[2] + arr[3] sum_array[3] = arr[2] + arr[3] + arr[4] sum_array[4] = arr[3] + arr[4] Input : arr[] = {2, 4, 3, 4, 2}, m = 1 Output : sum_array = {2, 4, 3, 4, 2} Explanation : sum_array[0] = arr[0] sum_array[1] = arr[1] sum_array[2] = arr[2] sum_array[3] = arr[3] sum_array[4] = arr[4]
Basic Approach : As per problem statement, we calculate sum_array[i] by iterating over i-(m/2) to i+(m/2). According to this approach, we have a nested loop which will result into time complexity of O(n*m).
Efficient Approach : For calculating sum_array is to use sliding window concept and thus can easily save our time. For Sliding window, the time complexity is O(n).
Algorithm
calculate sum of first (m/2)+1 elementssum_array[0] = sumfor i=1 to i<nif( (i-(m/2)-1) >= 0 ) sum -= arr[(i-(m/2)-1)]if( (i+m/2) < n) sum += arr[(i+m/2)]sum_array[i] = sumprint sum_array
Implementation:
// CPP Program to find sum array for a given // array. #include <bits/stdc++.h> using namespace std;
// function to calc sum_array and print void calcSum_array( int arr[], int n, int m)
{ int sum = 0;
int sum_array[n];
// calc 1st m/2 + 1 element for 1st window
for ( int i = 0; i < m / 2 + 1; i++)
sum += arr[i];
sum_array[0] = sum;
// use sliding window to
// calculate rest of sum_array
for ( int i = 1; i < n; i++) {
if (i - (m / 2) - 1 >= 0)
sum -= arr[i - (m / 2) - 1];
if (i + (m / 2) < n)
sum += arr[i + (m / 2)];
sum_array[i] = sum;
}
// print sum_array
for ( int i = 0; i < n; i++)
cout << sum_array[i] << " " ;
} // driver program int main()
{ int arr[] = { 3, 6, 2, 7, 3, 8, 4,
9, 1, 5, 0, 4 };
int m = 5;
int n = sizeof (arr) / sizeof ( int );
calcSum_array(arr, n, m);
return 0;
} |
// Java Program to find sum array // for a given array. class GFG
{ // function to calc sum_array and print
static void calcSum_array( int arr[], int n, int m)
{
int sum = 0 ;
int sum_array[] = new int [n];
// calc 1st m/2 + 1 element
// for 1st window
for ( int i = 0 ; i < m / 2 + 1 ; i++)
sum += arr[i];
sum_array[ 0 ] = sum;
// use sliding window to
// calculate rest of sum_array
for ( int i = 1 ; i < n; i++)
{
if (i - (m / 2 ) - 1 >= 0 )
sum -= arr[i - (m / 2 ) - 1 ];
if (i + (m / 2 ) < n)
sum += arr[i + (m / 2 )];
sum_array[i] = sum;
}
// print sum_array
for ( int i = 0 ; i < n; i++)
System.out.print(sum_array[i] + " " );
}
// Driver program
public static void main(String[] args)
{
int arr[] = { 3 , 6 , 2 , 7 , 3 , 8 , 4 , 9 , 1 , 5 , 0 , 4 };
int m = 5 ;
int n = arr.length;
calcSum_array(arr, n, m);
}
} // This code is contributed by prerna saini. |
# Python3 Program to find Sum array # for a given array. import math as mt
# function to calc Sum_array and print def calcSum_array(arr, n, m):
Sum = 0
Sum_array = [ 0 for i in range (n)]
# calc 1st m/2 + 1 element for 1st window
for i in range (m / / 2 + 1 ):
Sum + = arr[i]
Sum_array[ 0 ] = Sum
# use sliding window to
# calculate rest of Sum_array
for i in range ( 1 , n):
if (i - (m / / 2 ) - 1 > = 0 ):
Sum - = arr[i - (m / / 2 ) - 1 ]
if (i + (m / 2 ) < n):
Sum + = arr[i + (m / / 2 )]
Sum_array[i] = Sum
# print Sum_array
for i in range (n):
print (Sum_array[i], end = " " )
# Driver Code arr = [ 3 , 6 , 2 , 7 , 3 , 8 , 4 , 9 , 1 , 5 , 0 , 4 ]
m = 5
n = len (arr)
calcSum_array(arr, n, m) # This code is contributed by mohit kumar 29 |
// C# Program to find sum array // for a given array. using System;
class GFG
{ // function to calc sum_array and print
static void calcSum_array( int []arr, int n, int m)
{
int sum = 0;
int []sum_array = new int [n];
// calc 1st m/2 + 1 element
// for 1st window
for ( int i = 0; i < m / 2 + 1; i++)
sum += arr[i];
sum_array[0] = sum;
// use sliding window to
// calculate rest of sum_array
for ( int i = 1; i < n; i++)
{
if (i - (m / 2) - 1 >= 0)
sum -= arr[i - (m / 2) - 1];
if (i + (m / 2) < n)
sum += arr[i + (m / 2)];
sum_array[i] = sum;
}
// print sum_array
for ( int i = 0; i < n; i++)
Console.Write(sum_array[i] + " " );
}
// Driver program
public static void Main()
{
int []arr = { 3, 6, 2, 7, 3, 8, 4, 9, 1, 5, 0, 4 };
int m = 5;
int n = arr.Length;
calcSum_array(arr, n, m);
}
} // This code is contributed by vt_m. |
<?php // PHP Program to find sum array // for a given array. // function to calc sum_array and print function calcSum_array(& $arr , $n , $m )
{ $sum = 0;
$sum_array = array ();
// calc 1st m/2 + 1 element
// for 1st window
for ( $i = 0;
$i < (int)( $m / 2) + 1; $i ++)
$sum = $sum + $arr [ $i ];
$sum_array [0] = $sum ;
// use sliding window to
// calculate rest of sum_array
for ( $i = 1; $i < $n ; $i ++)
{
if ( $i - (int)( $m / 2) - 1 >= 0)
$sum = $sum - $arr [ $i -
(int)( $m / 2) - 1];
if ( $i + (int)( $m / 2) < $n )
$sum = $sum + $arr [ $i +
(int)( $m / 2)];
$sum_array [ $i ] = $sum ;
}
// print sum_array
for ( $i = 0; $i < $n ; $i ++)
echo $sum_array [ $i ] . " " ;
} // Driver Code $arr = array (3, 6, 2, 7, 3, 8,
4, 9, 1, 5, 0, 4 );
$m = 5;
$n = sizeof( $arr );
calcSum_array( $arr , $n , $m );
// This code is contributed by Mukul Singh ?> |
<script> // JavaScript Program to find sum array for a given // array. // function to calc sum_array and print function calcSum_array(arr, n, m)
{ let sum = 0;
let sum_array = new Array(n);
// calc 1st m/2 + 1 element for 1st window
for (let i = 0; i < Math.floor(m / 2) + 1; i++)
sum += arr[i];
sum_array[0] = sum;
// use sliding window to
// calculate rest of sum_array
for (let i = 1; i < n; i++)
{
if (i - Math.floor(m / 2) - 1 >= 0)
sum -= arr[i - Math.floor(m / 2) - 1];
if (i + Math.floor(m / 2) < n)
sum += arr[i + Math.floor(m / 2)];
sum_array[i] = sum;
}
// print sum_array
for (let i = 0; i < n; i++)
document.write(sum_array[i] + " " );
} // Driver program let arr = [ 3, 6, 2, 7, 3, 8, 4,
9, 1, 5, 0, 4 ];
let m = 5;
let n = arr.length;
calcSum_array(arr, n, m);
// This code is contributed by Surbhi Tyagi. </script> |
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