Mean of range in array
Last Updated :
09 Dec, 2022
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
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++
#include <bits/stdc++.h>
using namespace std;
int findMean(int arr[], int l, int r)
{
int sum = 0, count = 0;
for (int i = l; i <= r; i++) {
sum += arr[i];
count++;
}
int mean = floor(sum / count);
return mean;
}
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;
}
|
C
#include <stdio.h>
#include <math.h>
int findMean(int arr[], int l, int r)
{
int sum = 0, count = 0;
for (int i = l; i <= r; i++) {
sum += arr[i];
count++;
}
int mean = floor(sum / count);
return mean;
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
printf("%d\n",findMean(arr, 0, 2));
printf("%d\n",findMean(arr, 1, 3));
printf("%d\n",findMean(arr, 0, 4));
return 0;
}
|
Java
import java.io.*;
public class Main {
static int findMean(int arr[], int l, int r)
{
int sum = 0, count = 0;
for (int i = l; i <= r; i++) {
sum += arr[i];
count++;
}
int mean = (int)Math.floor(sum / count);
return mean;
}
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
import math
def findMean(arr, l, r):
sum, count = 0, 0
for i in range(l, r + 1):
sum += arr[i]
count += 1
mean = math.floor(sum / count)
return mean
arr = [ 1, 2, 3, 4, 5 ]
print(findMean(arr, 0, 2))
print(findMean(arr, 1, 3))
print(findMean(arr, 0, 4))
|
C#
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|
PHP
<?php
function findMean($arr, $l, $r)
{
$sum = 0;
$count = 0;
for ($i = $l; $i <= $r; $i++)
{
$sum += $arr[$i];
$count++;
}
$mean = floor($sum / $count);
return $mean;
}
$arr = array(1, 2, 3, 4, 5);
echo findMean($arr, 0, 2), "\n";
echo findMean($arr, 1, 3), "\n";
echo findMean($arr, 0, 4), "\n";
?>
|
Javascript
<script>
function findMean(arr, l, r)
{
let sum = 0, count = 0;
for (let i = l; i <= r; i++) {
sum += arr[i];
count++;
}
let mean = Math.floor(sum / count);
return mean;
}
let arr = [ 1, 2, 3, 4, 5 ];
document.write(findMean(arr, 0, 2) + "</br>");
document.write(findMean(arr, 1, 3) + "</br>");
document.write(findMean(arr, 0, 4) + "</br>");
</script>
|
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 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++
Java
import java.io.*;
public class Main {
public static final int MAX = 1000005;
static int prefixSum[] = new int[MAX];
static void calculatePrefixSum(int arr[], int n)
{
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
static int findMean(int l, int r)
{
if (l == 0)
return (int)Math.floor(prefixSum[r] / (r + 1));
return (int)Math.floor((prefixSum[r] -
prefixSum[l - 1]) / (r - l + 1));
}
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
import math as mt
MAX = 1000005
prefixSum = [0 for i in range(MAX)]
def calculatePrefixSum(arr, n):
prefixSum[0] = arr[0]
for i in range(1,n):
prefixSum[i] = prefixSum[i - 1] + arr[i]
def findMean(l, r):
if (l == 0):
return mt.floor(prefixSum[r] / (r + 1))
return (mt.floor((prefixSum[r] -
prefixSum[l - 1]) /
(r - l + 1)))
arr = [1, 2, 3, 4, 5]
n = len(arr)
calculatePrefixSum(arr, n)
print(findMean(0, 2))
print(findMean(1, 3))
print(findMean(0, 4))
|
C#
using System;
public class GFG {
public static readonly int MAX = 1000005;
static int []prefixSum = new int[MAX];
static void calculatePrefixSum(int []arr, int n)
{
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
static int findMean(int l, int r)
{
if (l == 0)
return (int)Math.Floor((double)(prefixSum[r] / (r + 1)));
return (int)Math.Floor((double)(prefixSum[r] -
prefixSum[l - 1]) / (r - l + 1));
}
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));
}
}
|
Javascript
<script>
let MAX = 1000005;
let prefixSum = new Array(MAX);
prefixSum.fill(0);
function calculatePrefixSum(arr, n)
{
prefixSum[0] = arr[0];
for(let i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
function findMean(l, r)
{
if (l == 0)
return parseInt(Math.floor(prefixSum[r] /
(r + 1)), 10);
return parseInt(Math.floor((prefixSum[r] -
prefixSum[l - 1]) /
(r - l + 1)), 10);
}
let arr = [ 1, 2, 3, 4, 5 ];
let n = arr.length;
calculatePrefixSum(arr, n);
document.write(findMean(1, 2) + "</br>");
document.write(findMean(1, 3) + "</br>");
document.write(findMean(1, 4) + "</br>");
</script>
|
C
#include <stdio.h>
#include <math.h>
#define MAX 1000005
int prefixSum[MAX];
void calculatePrefixSum(int arr[], int n)
{
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
int findMean(int l, int r)
{
if (l == 0)
return floor(prefixSum[r]/(r+1));
return floor((prefixSum[r] -
prefixSum[l - 1]) / (r - l + 1));
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
calculatePrefixSum(arr, n);
printf("%d\n",findMean(0, 2));
printf("%d\n",findMean(1, 3));
printf("%d\n",findMean(0, 4));
return 0;
}
|
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.
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