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.
// CPP program to find floor value // of mean in range l to r #include <bits/stdc++.h> 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;
} |
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).
// CPP program to find floor value // of mean in range l to r #include <bits/stdc++.h> #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[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 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[0]);
calculatePrefixSum(arr, n);
cout << findMean(0, 2) << endl;
cout << findMean(1, 3) << endl;
cout << findMean(0, 4) << endl;
return 0;
} |
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!