Mean of Range in Array

You are given an array arr[] consisting of n positive integers and a set of q queries represented by a 2D array queries queries[][]. Each query contains two integers l and r, indicating a range of indices in the array.

For every query, calculate the mean of the elements in the subarray from index l to index r (inclusive). After computing the mean, return its floor value.

Examples:

Input: arr[] = [3, 7, 2, 8, 2] , queries[][] = [[0,1], [1, 3], [2, 4]]
Output: 5 5 4
Explanation:
For query [0, 1] - Elements in the range are [3, 7], Mean is (3+7)/2 = 5, Floor value is 5.
For query [1, 3] - Elements in the range are [7, 2, 8], Mean is (7+2+8)/3 = 5.6, Floor value is 5.
For query [2, 4] - Elements in the range are [2, 8, 2], Mean is (2+8+2)/3 = 4, Floor value is 4.

Input: arr[] = [10, 20, 30, 40, 50, 60], queries[][] = [[3, 5]]
Output: 50
Explanation: For query [3, 5] - Elements in the range are [40, 50, 60], Mean is (40+50+60)/3 = 50, Floor value is 50.

[Naive Approach] Direct Range Traversal

The idea is to process each query separately by iterating over the given range l to r and computing the sum and count of elements. Since the mean is sum/count, we can directly compute its floor value using integer division. The approach ensures that each query is handled independently.

Dry run for arr[] = [3, 7, 2, 8, 2] , queries[][] = [[0,1], [1, 3], [2, 4]]:

• Query [0,1]: take elements from index 0 to 1 (3,7), sum = 10, count = 2, mean = 10/2 = 5, result = 5
• Query [1,3]: take elements from index 1 to 3 (7,2,8), sum = 17, count = 3, mean = 17/3 = 5 (floor), result = 5
• Query [2,4]: take elements from index 2 to 4 (2,8,2), sum = 12, count = 3, mean = 12/3 = 4, result = 4

Final answer: 5 5 4

5 5 4 

Time Complexity: O(n * q), for each of the q queries, we iterate over the range l to r, taking O(n) time.
Auxiliary Space: O(1)

Expected Approach] Using Prefix Sum

Use a prefix sum array to calculate the sum from l to r in O(1) time. For each query, compute the range sum using the prefix array and divide it by the number of elements to obtain the mean.

Dry run for arr[] = [3, 7, 2, 8, 2] , queries[][] = [[0,1], [1, 3], [2, 4]]:

• Prefix array formation: prefixSum = [0, 3, 10, 12, 20, 22].
• Query [0,1]: l = 0, r = 1, sum = prefixSum[2] - prefixSum[0] = 10 - 0 = 10, count = 2, mean = 10/2 = 5, result = 5.
• Query [1,3]: l = 1, r = 3, sum = prefixSum[4] - prefixSum[1] = 20 - 3 = 17, count = 3, mean = 17/3 = 5 (floor), result = 5.
• Query [2,4]: l = 2, r = 4, sum = prefixSum[5] - prefixSum[2] = 22 - 10 = 12, count = 3, mean = 12/3 = 4, result = 4.

Final answer: 5 5 4

5 5 4 

Time Complexity: O(n + q), as the method efficiently computes prefix sum in O(n) and processes each query in O(1).
Auxiliary Space: O(n)