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 rpublic 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 rpublic 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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