Length of longest sub-array with maximum arithmetic mean.
Given an array of n-elements find the longest sub-array with the greatest arithmetic mean. The length of the sub-array must be greater than 1 and the mean should be calculated as an integer only.
Examples:
Input : arr[] = {3, 2, 1, 2} Output : 2 sub-array 3, 2 has greatest arithmetic mean Input :arr[] = {3, 3, 3, 2} Output : 3
The idea is to first find the greatest mean of two consecutive elements from the array. Again iterate over the array and try to find the longest sequence in which each element must be greater or equal to the greatest mean calculated.
Above approach works because of these key points:
- Minimum possible sequence length is 2 and hence the greatest mean of two consecutive elements will always be part of the result.
- Any element which is equal or greater than the calculated mean may be the part of the longest sequence.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find maximum distance // between unequal elements int longestSubarray( int arr[], int n) { // Calculate maxMean int maxMean = 0; for ( int i = 1; i < n; i++) maxMean = max(maxMean, (arr[i] + arr[i - 1]) / 2); // Iterate over array and calculate largest subarray // with all elements greater or equal to maxMean int ans = 0; int subarrayLength = 0; for ( int i = 0; i < n; i++) if (arr[i] >= maxMean) ans = max(ans, ++subarrayLength); else subarrayLength = 0; return ans; } // Driver code int main() { int arr[] = { 4, 3, 3, 2, 1, 4 }; int n = sizeof (arr) / sizeof (arr[0]); cout << longestSubarray(arr, n); return 0; } |
Java
// Java implementation of the above approach import java.io.*; class GFG { // Function to find maximum distance // between unequal elements static int longestSubarray( int arr[], int n) { // Calculate maxMean int maxMean = 0 ; for ( int i = 1 ; i < n; i++) maxMean = Math.max(maxMean, (arr[i] + arr[i - 1 ]) / 2 ); // Iterate over array and calculate largest subarray // with all elements greater or equal to maxMean int ans = 0 ; int subarrayLength = 0 ; for ( int i = 0 ; i < n; i++) if (arr[i] >= maxMean) ans = Math.max(ans, ++subarrayLength); else subarrayLength = 0 ; return ans; } // Driver code public static void main (String[] args) { int arr[] = { 4 , 3 , 3 , 2 , 1 , 4 }; int n = arr.length; System.out.println (longestSubarray(arr, n)); } } // This code is contributed by ajit_00023 |
Python3
# Python implementation of the above approach # Function to find maximum distance # between unequal elements def longestSubarray(arr, n): # Calculate maxMean maxMean = 0 ; for i in range ( 1 , n): maxMean = max (maxMean, (arr[i] + arr[i - 1 ]) / / 2 ); # Iterate over array and calculate largest subarray # with all elements greater or equal to maxMean ans = 0 ; subarrayLength = 0 ; for i in range (n): if (arr[i] > = maxMean): subarrayLength + = 1 ; ans = max (ans, subarrayLength); else : subarrayLength = 0 ; return ans; # Driver code arr = [ 4 , 3 , 3 , 2 , 1 , 4 ]; n = len (arr); print (longestSubarray(arr, n)); # This code contributed by PrinciRaj1992 |
C#
// C# program for the above approach using System; class GFG { // Function to find maximum distance // between unequal elements static int longestSubarray( int []arr, int n) { // Calculate maxMean int maxMean = 0; for ( int i = 1; i < n; i++) maxMean = Math.Max(maxMean, (arr[i] + arr[i - 1]) / 2); // Iterate over array and calculate // largest subarray with all elements // greater or equal to maxMean int ans = 0; int subarrayLength = 0; for ( int i = 0; i < n; i++) if (arr[i] >= maxMean) ans = Math.Max(ans, ++subarrayLength); else subarrayLength = 0; return ans; } // Driver code public static void Main () { int []arr = { 4, 3, 3, 2, 1, 4 }; int n = arr.Length; Console.WriteLine(longestSubarray(arr, n)); } } // This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript implementation of the above approach // Function to find maximum distance // between unequal elements function longestSubarray(arr, n) { // Calculate maxMean var maxMean = 0; for ( var i = 1; i < n; i++) maxMean = Math.max(maxMean, parseInt((arr[i] + arr[i - 1]) / 2)); // Iterate over array and calculate largest subarray // with all elements greater or equal to maxMean var ans = 0; var subarrayLength = 0; for ( var i = 0; i < n; i++) if (arr[i] >= maxMean) ans = Math.max(ans, ++subarrayLength); else subarrayLength = 0; return ans; } // Driver code var arr = [ 4, 3, 3, 2, 1, 4 ]; var n = arr.length; document.write( longestSubarray(arr, n)); </script> |
Output:
3
Time Complexity : O(N)
Auxiliary Space: O(1)
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