Java Program for Mean of range in array

Last Updated : 31 May, 2022
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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 
// 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 
// 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!


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