Mean of range in array

Given an array of n integers. You are given q queries. Write a program to print 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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.

C++

 // CPP program to find floor value // of mean in range l to r #include using namespace std;    // To find mean of range in l to r 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 = floor(sum / count);        // Returns mean of array     // in range l to r     return mean; }    // Driver program to test findMean() int main() {     int arr[] = { 1, 2, 3, 4, 5 };     cout << findMean(arr, 0, 2) << endl;     cout << findMean(arr, 1, 3) << endl;     cout << findMean(arr, 0, 4) << endl;     return 0; }

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));     } }

Python3

 # Python 3 program to find floor value # of mean in range l to r import math    # To find mean of range in l to r def findMean(arr, l, r):            # Both sum and count are     # initialize to 0     sum, count = 0, 0            # To calculate sum and number     # of elements in range l to r     for i in range(l, r + 1):         sum += arr[i]         count += 1        # Calculate floor value of mean     mean = math.floor(sum / count)        # Returns mean of array     # in range l to r     return mean    # Driver Code arr = [ 1, 2, 3, 4, 5 ]        print(findMean(arr, 0, 2)) print(findMean(arr, 1, 3)) print(findMean(arr, 0, 4))    # This code is contributed  # by PrinciRaj1992

C#

 //C# program to find floor value // of mean in range l to r using System;    public class GFG {         // 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((double)sum / count);             // Returns mean of array         // in range l to r         return mean;     }         // Driver program to test findMean()     public static void Main()     {         int []arr = { 1, 2, 3, 4, 5 };         Console.WriteLine(findMean(arr, 0, 2));         Console.WriteLine(findMean(arr, 1, 3));         Console.WriteLine(findMean(arr, 0, 4));     } }    /*This code is contributed by PrinciRaj1992*/

PHP



Output :

2
3
3

Time Complexity: O(n)

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).

C++

 // CPP program to find floor value // of mean in range l to r #include #define MAX 1000005 using namespace std;    int prefixSum[MAX];    // To calculate prefixSum of array void calculatePrefixSum(int arr[], int n) {     // Calculate prefix sum of array     prefixSum = arr;     for (int i = 1; i < n; i++)         prefixSum[i] = prefixSum[i - 1] + arr[i]; }    // To return floor of mean // in range l to r int findMean(int l, int r) {     if (l == 0)       return 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 floor((prefixSum[r] -            prefixSum[l - 1]) / (r - l + 1)); }    // Driver program to test above functions int main() {     int arr[] = { 1, 2, 3, 4, 5 };     int n = sizeof(arr) / sizeof(arr);     calculatePrefixSum(arr, n);     cout << findMean(0, 2) << endl;     cout << findMean(1, 3) << endl;     cout << findMean(0, 4) << endl;     return 0; }

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 = arr;         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));     } }

Python3

 # Python3 program to find floor value # of mean in range l to r import math as mt    MAX = 1000005 prefixSum = [0 for i in range(MAX)]    # To calculate prefixSum of array def calculatePrefixSum(arr, n):        # Calculate prefix sum of array     prefixSum = arr        for i in range(1,n):         prefixSum[i] = prefixSum[i - 1] + arr[i]    # To return floor of mean # in range l to r def findMean(l, r):        if (l == 0):         return mt.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 (mt.floor((prefixSum[r] -                        prefixSum[l - 1]) /                           (r - l + 1)))    # Driver Code arr = [1, 2, 3, 4, 5]    n = len(arr)    calculatePrefixSum(arr, n) print(findMean(0, 2)) print(findMean(1, 3)) print(findMean(0, 4))    # This code is contributed by Mohit Kumar

C#

 // C# program to find floor value // of mean in range l to r using System;                        public class GFG { public static readonly 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 = arr;         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((double)(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((double)(prefixSum[r] -                 prefixSum[l - 1]) / (r - l + 1));     }         // Driver program to test above functions     public static void Main()     {         int []arr = { 1, 2, 3, 4, 5 };         int n = arr.Length;         calculatePrefixSum(arr, n);         Console.WriteLine(findMean(1, 2));         Console.WriteLine(findMean(1, 3));         Console.WriteLine(findMean(1, 4));     } }    //This code is contributed by PrinciRaj1992

Output:

2
3
3

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