Given an array, arr[] and a function F(i, j). The task is to compute max{F(i, j)} over all sub-arrays [i..j].
The function F() is defined as:
![F(l, r) = \sum_{i = l}^{r - 1} |arr[i] - arr[i+1]|.(-1)^{i-l}](https://neveropen.co.uk/wp-content/uploads/2023/11/wardslaus_quicklatex.com-7f69357a9231eaf8341997dd54b6d3e8_l3.png "Rendered by QuickLaTeX.com")
Examples:
Input : arr[] = { 1, 5, 4, 7 }
Output : 6
Values of F(i, j) for all the sub-arrays:
{ 1, 5 } = |1 – 5| * (1) = 4
{ 1, 5, 4 } = |1 – 5| * (1) + |5 – 4| * (-1) = 3
{ 1, 5, 4, 7 } = |1 – 5| * (1) + |5 – 4| * (-1) + |4 – 7| * (1) = 6
{ 5, 4 } = |5 – 4| * (1) = 1
{ 5, 4, 7 } = |5 – 4| * (1) + |4 – 7| * (-1) = -2
{ 4, 7 } = |4 – 7| * (1) = 3
Max of all the above values = 6.Input : arr[] = { 1, 4, 2, 3, 1 }
Output : 3
Naive Approach: A naive approach is to traverse over all sub-arrays and calculate the maximum of function F over all the sub-arrays.
Efficient Approach: A better approach is to consider segments in F(l, r) with odd and even l separately. Two different arrays B[] and C[] can be constructed for this purpose such that:
B[i] = |arr[i] - arr[i + 1]| * (-1)i C[i] = |arr[i] - arr[i + 1]| * (-1)i + 1
Now if we observe closely, we just need to find the maximum sum subarray of the arrays B[] and C[] and final answer of the function will be the maximum among both the arrays.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define MAX 100005using namespace std;// Function to return maximum sum of a sub-arrayint kadaneAlgorithm(const int* ar, int n){ int sum = 0, maxSum = 0; for (int i = 0; i < n; i++) { sum += ar[i]; if (sum < 0) sum = 0; maxSum = max(maxSum, sum); } return maxSum;}// Function to return maximum value of function Fint maxFunction(const int* arr, int n){ int b[MAX], c[MAX]; // Compute arrays B[] and C[] for (int i = 0; i < n - 1; i++) { if (i & 1) { b[i] = abs(arr[i + 1] - arr[i]); c[i] = -b[i]; } else { c[i] = abs(arr[i + 1] - arr[i]); b[i] = -c[i]; } } // Find maximum sum sub-array of both of the // arrays and take maximum among them int ans = kadaneAlgorithm(b, n - 1); ans = max(ans, kadaneAlgorithm(c, n - 1)); return ans;}// Driver codeint main(){ int arr[] = { 1, 5, 4, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxFunction(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{static int MAX = 100005;// Function to return maximum sum of a sub-array static int kadaneAlgorithm(int[] ar, int n) { int sum = 0, maxSum = 0; for (int i = 0; i < n; i++) { sum += ar[i]; if (sum < 0) sum = 0; maxSum = Math.max(maxSum, sum); } return maxSum; } // Function to return maximum value // of function F static int maxFunction(int[] arr, int n) { int []b = new int[MAX]; int []c = new int[MAX]; // Compute arrays B[] and C[] for (int i = 0; i < n - 1; i++) { if (i % 2 == 1) { b[i] = Math.abs(arr[i + 1] - arr[i]); c[i] = -b[i]; } else { c[i] = Math.abs(arr[i + 1] - arr[i]); b[i] = -c[i]; } } // Find maximum sum sub-array of both of the // arrays and take maximum among them int ans = kadaneAlgorithm(b, n - 1); ans = Math.max(ans, kadaneAlgorithm(c, n - 1)); return ans; } // Driver code public static void main(String[] args) { int arr[] = { 1, 5, 4, 7 }; int n = arr.length; System.out.println(maxFunction(arr, n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the above approach MAX = 100005;# Function to return maximum # sum of a sub-array def kadaneAlgorithm(ar, n) : sum = 0; maxSum = 0; for i in range(n) : sum += ar[i]; if (sum < 0) : sum = 0; maxSum = max(maxSum, sum); return maxSum; # Function to return maximum # value of function F def maxFunction(arr, n) : b = [0] * MAX; c = [0] * MAX; # Compute arrays B[] and C[] for i in range(n - 1) : if (i & 1) : b[i] = abs(arr[i + 1] - arr[i]); c[i] = -b[i]; else : c[i] = abs(arr[i + 1] - arr[i]); b[i] = -c[i]; # Find maximum sum sub-array of both of the # arrays and take maximum among them ans = kadaneAlgorithm(b, n - 1); ans = max(ans, kadaneAlgorithm(c, n - 1)); return ans; # Driver code if __name__ == "__main__" : arr = [ 1, 5, 4, 7 ]; n = len(arr) print(maxFunction(arr, n)); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System; class GFG{static int MAX = 100005;// Function to return maximum sum of a sub-array static int kadaneAlgorithm(int[] ar, int n) { int sum = 0, maxSum = 0; for (int i = 0; i < n; i++) { sum += ar[i]; if (sum < 0) sum = 0; maxSum = Math.Max(maxSum, sum); } return maxSum; } // Function to return maximum value // of function F static int maxFunction(int[] arr, int n) { int []b = new int[MAX]; int []c = new int[MAX]; // Compute arrays B[] and C[] for (int i = 0; i < n - 1; i++) { if (i % 2 == 1) { b[i] = Math.Abs(arr[i + 1] - arr[i]); c[i] = -b[i]; } else { c[i] = Math.Abs(arr[i + 1] - arr[i]); b[i] = -c[i]; } } // Find maximum sum sub-array of both of the // arrays and take maximum among them int ans = kadaneAlgorithm(b, n - 1); ans = Math.Max(ans, kadaneAlgorithm(c, n - 1)); return ans; } // Driver code public static void Main(String[] args) { int []arr = { 1, 5, 4, 7 }; int n = arr.Length; Console.WriteLine(maxFunction(arr, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation of the above approachconst MAX = 100005;// Function to return maximum sum of a sub-arrayfunction kadaneAlgorithm(ar, n){ let sum = 0, maxSum = 0; for(let i = 0; i < n; i++) { sum += ar[i]; if (sum < 0) sum = 0; maxSum = Math.max(maxSum, sum); } return maxSum;}// Function to return maximum // value of function Ffunction maxFunction(arr, n){ let b = new Array(MAX), c = new Array(MAX); // Compute arrays B[] and C[] for(let i = 0; i < n - 1; i++) { if (i & 1) { b[i] = Math.abs(arr[i + 1] - arr[i]); c[i] = -b[i]; } else { c[i] = Math.abs(arr[i + 1] - arr[i]); b[i] = -c[i]; } } // Find maximum sum sub-array of both of the // arrays and take maximum among them let ans = kadaneAlgorithm(b, n - 1); ans = Math.max(ans, kadaneAlgorithm(c, n - 1)); return ans;}// Driver codelet arr = [ 1, 5, 4, 7 ];let n = arr.length;document.write(maxFunction(arr, n));// This code is contributed by Manoj.</script> |
Output:
6
Time Complexity: O(n)
Auxiliary Space: O(MAX)
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