Given an array of n integers. You are given q queries. Write a program to print the floor value of mean in range l to r for each query in a new line.
Examples :
Input : arr[] = {1, 2, 3, 4, 5} q = 3 0 2 1 3 0 4 Output : 2 3 3 Here for 0 to 2 (1 + 2 + 3) / 3 = 2 Input : arr[] = {6, 7, 8, 10} q = 2 0 3 1 2 Output : 7 7
Naive Approach: We can run loop for each query l to r and find sum and number of elements in range. After this we can print floor of mean for each query.
// Java program to find floor value // of mean in range l to r public class Main {
// To find mean of range in l to r
static int findMean( int arr[], int l, int r)
{
// Both sum and count are
// initialize to 0
int sum = 0 , count = 0 ;
// To calculate sum and number
// of elements in range l to r
for ( int i = l; i <= r; i++) {
sum += arr[i];
count++;
}
// Calculate floor value of mean
int mean = ( int )Math.floor(sum / count);
// Returns mean of array
// in range l to r
return mean;
}
// Driver program to test findMean()
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 };
System.out.println(findMean(arr, 0 , 2 ));
System.out.println(findMean(arr, 1 , 3 ));
System.out.println(findMean(arr, 0 , 4 ));
}
} |
Output :
2 3 3
Time complexity: O(n*q) where q is the number of queries and n is the size of the array. Here in the above code q is 3 as the findMean function is used 3 times.
Auxiliary Space: O(1)
Efficient Approach: We can find sum of numbers using numbers using prefix sum. The prefixSum[i] denotes the sum of first i elements. So sum of numbers in range l to r will be prefixSum[r] – prefixSum[l-1]. Number of elements in range l to r will be r – l + 1. So we can now print mean of range l to r in O(1).
// Java program to find floor value // of mean in range l to r public class Main {
public static final int MAX = 1000005 ;
static int prefixSum[] = new int [MAX];
// To calculate prefixSum of array
static void calculatePrefixSum( int arr[], int n)
{
// Calculate prefix sum of array
prefixSum[ 0 ] = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
prefixSum[i] = prefixSum[i - 1 ] + arr[i];
}
// To return floor of mean
// in range l to r
static int findMean( int l, int r)
{
if (l == 0 )
return ( int )Math.floor(prefixSum[r] / (r + 1 ));
// Sum of elements in range l to
// r is prefixSum[r] - prefixSum[l-1]
// Number of elements in range
// l to r is r - l + 1
return ( int )Math.floor((prefixSum[r] -
prefixSum[l - 1 ]) / (r - l + 1 ));
}
// Driver program to test above functions
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 };
int n = arr.length;
calculatePrefixSum(arr, n);
System.out.println(findMean( 1 , 2 ));
System.out.println(findMean( 1 , 3 ));
System.out.println(findMean( 1 , 4 ));
}
} |
Output:
2 3 3
Time complexity: O(n+q) where q is the number of queries and n is the size of the array. Here in the above code q is 3 as the findMean function is used 3 times.
Auxiliary Space: O(k) where k=1000005.
Please refer complete article on Mean of range in array for more details!