Given an array arr[], we need to find the maximum sum of the even indexed elements that can be obtained by performing right shift operation on any sub-array of even length by 1.
Examples:
Input: arr[] = {5, 1, 3, 4, 5, 6}
Output: 15
Explanation:
We can perform a right shift on index 2 to 5 then resulting array is:
arr[] = {5, 1, 6, 3, 4, 5}
Sum of elements at even indexes = 5 + 6 + 4 = 15
Input: arr[] = {7, 9, 1, 8, 3, 10, 4, 12}
Output: 39
Explanation:
We can perform a right shift on index 0 to 7 then resulting array is:
arr[] = {12, 7, 9, 1, 8, 3, 10, 4}
Sum of elements at even indexes = 12 + 9 + 8 + 10 = 39
Naive Approach: The naive approach is to right shift every possible subarray of even length by one and find the sum of elements at even index for all the array formed after every possible subarray shifts. The maximum sum in all the array is the required result.
Below is the implementation of the above approach:
C++
#include<iostream>#include<algorithm>using namespace std;int main(){ int arr[] = {5, 1, 3, 4, 5, 6}; int n = sizeof(arr)/sizeof(arr[0]); int max_sum = 0; for(int i=0; i<n; i+=2){ // iterate over all even indices as starting point for(int j=i; j<n; j+=2){ // iterate over all even indices as ending point for(int k=i; k<=j; k+=2){ // right shift every even indexed element of subarray by one swap(arr[k], arr[k+1]); } int sum = 0; for(int x=0; x<n; x+=2){ // calculate sum of all even indexed elements in resulting array sum += arr[x]; } max_sum = max(max_sum, sum); // update max_sum if current sum is greater for(int k=i; k<=j; k+=2){ // restore the original array by left shifting the subarray swap(arr[k], arr[k+1]); } } } cout<<max_sum<<endl; return 0;}// This code is contributed by rudra1807raj |
Java
import java.util.*;public class Main { public static void main(String[] args) { int[] arr = { 5, 1, 3, 4, 5, 6 }; int n = arr.length; int max_sum = 0; // iterate over all even indices as starting point for (int i = 0; i < n; i += 2) { // iterate over all even indices as ending point for (int j = i; j < n; j += 2) { // right shift every even indexed element of // subarray by one for (int k = i; k <= j; k += 2) { int temp = arr[k]; arr[k] = arr[k + 1]; arr[k + 1] = temp; } int sum = 0; // calculate sum of all even indexed // elements in resulting array for (int x = 0; x < n; x += 2) { sum += arr[x]; } // update max_sum if current sum is greater max_sum = Math.max(max_sum, sum); // restore the original array by left // shifting the subarray for (int k = i; k <= j; k += 2) { int temp = arr[k]; arr[k] = arr[k + 1]; arr[k + 1] = temp; } } } System.out.println(max_sum); }}// This code is contributed by Samim Hossain Mondal. |
Python
def main(): arr = [5, 1, 3, 4, 5, 6] n = len(arr) max_sum = 0 for i in range(0, n, 2): # iterate over all even indices as starting point for j in range(i, n, 2): # iterate over all even indices as ending point for k in range(i, j+1, 2): # right shift every even indexed # element of subarray by one arr[k], arr[k+1] = arr[k+1], arr[k] sum_even = sum(arr[0::2]) # calculate sum of all even # indexed elements in resulting array max_sum = max(max_sum, sum_even) # update max_sum if current sum is greater for k in range(i, j+1, 2): # restore the original array by left shifting the subarray arr[k], arr[k+1] = arr[k+1], arr[k] print(max_sum)if __name__ == "__main__": main() |
C#
using System;class GFG { public static void Main(string[] args) { int[] arr = { 5, 1, 3, 4, 5, 6 }; int n = arr.Length; int max_sum = 0; // Iterate over all even indices as the starting // point for (int i = 0; i < n; i += 2) { // Iterate over all even indices as the ending // point for (int j = i; j < n; j += 2) { // Right shift every even indexed element of // the subarray by one for (int k = i; k <= j; k += 2) { int temp = arr[k]; arr[k] = arr[k + 1]; arr[k + 1] = temp; } int sum = 0; // Calculate the sum of all even indexed // elements in the resulting array for (int x = 0; x < n; x += 2) { sum += arr[x]; } // Update max_sum if the current sum is // greater max_sum = Math.Max(max_sum, sum); // Restore the original array by left // shifting the subarray for (int k = i; k <= j; k += 2) { int temp = arr[k]; arr[k] = arr[k + 1]; arr[k + 1] = temp; } } } Console.WriteLine(max_sum); }}// This code is contributed by Samim Hossain Mondal. |
Output
15
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimized the naive above approach we can observe that after performing the right shift on any even subarray by 1 the odd index value gets replaced by even index value and vice-versa. If we find the sum of element at even indexes(say sum) before shifting, then after shifting the sum gets changed by the sum of the consecutive difference between elements at even and odd index.
For Example:
arr[] = {1, 2, 3, 4}
Sum element at even index in the above array = 1 + 3 = 4
Right shift array by 1, we have
arr1[] = {4, 1, 2, 3}
Sum element at even index in the above array = 4 + 2 = 6
therefore the sum get differ by 2 in the above two array which is equals the sum of consecutive difference in arr[] as ( (2 – 1) + (4 – 3) = 2.
We will use the above concepts to solve this problem. Below are the steps:
- Create two arrays(say arr1[] and arr2[]) such that arr1[] will store the consecutive difference of the element in the array arr[] as:
{(a[1] – a[0]), (a[3] – a[2]), . . ., (a[n]-a[n-1])}
- And arr2[] will store the consecutive difference of the element in the array arr[] as:
{(a[1] – a[2]), (a[3] – a[4]), . . ., (a[n-1]-a[n])}
- Then find the maximum subarray sum using Kadane’s Algorithm in the above two array formed.
- Now the maximum sum is the sum of element at even indexes in the original array(arr[]) + maximum subarray sum of the two arrays arr1[] and arr2[].
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Kadane's Algorithm to find // the maximum sum sub array int kadane(vector<int> v) { int maxSoFar = 0; int maxEndingHere = 0; // Iterate array v for (int i = 0; i < v.size(); i++) { maxEndingHere += v[i]; // Update the maximum maxSoFar = max(maxEndingHere, maxSoFar); // Update maxEndingHere to 0 if it // is less than 0 maxEndingHere = max(maxEndingHere, 0); } // Return maximum sum return maxSoFar; } // Function to find the sum // of even indexed elements // after modification in array. int sumOfElements(int* arr, int n) { int sum = 0; // Find initial sum of // even indexed elements for (int i = 0; i < n; i++) { if (i % 2 == 0) sum += arr[i]; } // Create two vectors to store // the consecutive differences // of elements vector<int> v1; vector<int> v2; for (int i = 1; i < n; i += 2) { v1.push_back(arr[i] - arr[i - 1]); if (i + 1 < n) { v2.push_back(arr[i] - arr[i + 1]); } } // Find the maximum sum subarray int option1 = kadane(v1); int option2 = kadane(v2); // Add the maximum value // to initial sum int ans = sum + max(option1, option2); // Return the answer return ans; } // Driver Code int main() { // Given array arr[] int arr[] = { 5, 1, 3, 4, 5, 6 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call cout << sumOfElements(arr, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Kadane's Algorithm to find // the maximum sum sub array static int kadane(Vector<Integer> v) { int maxSoFar = 0; int maxEndingHere = 0; // Iterate array v for(int i = 0; i < v.size(); i++) { maxEndingHere += v.get(i); // Update the maximum maxSoFar = Math.max(maxEndingHere, maxSoFar); // Update maxEndingHere to 0 if it // is less than 0 maxEndingHere = Math.max(maxEndingHere, 0); } // Return maximum sum return maxSoFar; } // Function to find the sum // of even indexed elements // after modification in array. static int sumOfElements(int []arr, int n) { int sum = 0; // Find initial sum of // even indexed elements for(int i = 0; i < n; i++) { if (i % 2 == 0) sum += arr[i]; } // Create two vectors to store // the consecutive differences // of elements Vector<Integer> v1 = new Vector<Integer>(); Vector<Integer> v2 = new Vector<Integer>(); for(int i = 1; i < n; i += 2) { v1.add(arr[i] - arr[i - 1]); if (i + 1 < n) { v2.add(arr[i] - arr[i + 1]); } } // Find the maximum sum subarray int option1 = kadane(v1); int option2 = kadane(v2); // Add the maximum value // to initial sum int ans = sum + Math.max(option1, option2); // Return the answer return ans; } // Driver Code public static void main(String[] args) { // Given array arr[] int arr[] = { 5, 1, 3, 4, 5, 6 }; int N = arr.length; // Function Call System.out.print(sumOfElements(arr, N)); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach # Kadane's Algorithm to find # the maximum sum sub array def kadane(v): maxSoFar = 0; maxEndingHere = 0; # Iterate array v for i in range(len(v)): maxEndingHere += v[i]; # Update the maximum maxSoFar = max(maxEndingHere, maxSoFar); # Update maxEndingHere to 0 # if it is less than 0 maxEndingHere = max(maxEndingHere, 0); # Return maximum sum return maxSoFar; # Function to find the sum # of even indexed elements # after modification in array. def sumOfElements(arr, n): sum = 0; # Find initial sum of # even indexed elements for i in range(n): if (i % 2 == 0): sum += arr[i]; # Create two vectors to store # the consecutive differences # of elements v1 = []; v2 = []; for i in range(1, n, 2): v1.append(arr[i] - arr[i - 1]); if (i + 1 < n): v2.append(arr[i] - arr[i + 1]); # Find the maximum sum subarray option1 = kadane(v1); option2 = kadane(v2); # Add the maximum value # to initial sum ans = sum + max(option1, option2); # Return the answer return ans; # Driver Code if __name__ == "__main__" : # Given array arr[] arr = [ 5, 1, 3, 4, 5, 6 ]; N = len(arr); # Function call print(sumOfElements(arr, N)); # This code is contributed by AnkitRai01 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Kadane's Algorithm to find // the maximum sum sub array static int kadane(List<int> v){ int maxSoFar = 0; int maxEndingHere = 0; // Iterate array v for(int i = 0; i < v.Count; i++) { maxEndingHere += v[i]; // Update the maximum maxSoFar = Math.Max(maxEndingHere, maxSoFar); // Update maxEndingHere to 0 if it // is less than 0 maxEndingHere = Math.Max(maxEndingHere, 0); } // Return maximum sum return maxSoFar;}// Function to find the sum// of even indexed elements// after modification in array.static int sumOfElements(int []arr, int n){ int sum = 0; // Find initial sum of // even indexed elements for(int i = 0; i < n; i++) { if (i % 2 == 0) sum += arr[i]; } // Create two vectors to store // the consecutive differences // of elements List<int> v1 = new List<int>(); List<int> v2 = new List<int>(); for(int i = 1; i < n; i += 2) { v1.Add(arr[i] - arr[i - 1]); if (i + 1 < n) { v2.Add(arr[i] - arr[i + 1]); } } // Find the maximum sum subarray int option1 = kadane(v1); int option2 = kadane(v2); // Add the maximum value // to initial sum int ans = sum + Math.Max(option1, option2); // Return the answer return ans;}// Driver Codepublic static void Main(String[] args){ // Given array []arr int []arr = { 5, 1, 3, 4, 5, 6 }; int N = arr.Length; // Function call Console.Write(sumOfElements(arr, N));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Kadane's Algorithm to find // the maximum sum sub array function kadane(v) { var maxSoFar = 0; var maxEndingHere = 0; // Iterate array v for (var i = 0; i < v.length; i++) { maxEndingHere += v[i]; // Update the maximum maxSoFar = Math.max(maxEndingHere,maxSoFar); // Update maxEndingHere to 0 if it // is less than 0 maxEndingHere = Math.max(maxEndingHere, 0); } // Return maximum sum return maxSoFar; } // Function to find the sum // of even indexed elements // after modification in array. function sumOfElements(arr, n) { var sum = 0; // Find initial sum of // even indexed elements for (var i = 0; i < n; i++) { if (i % 2 == 0) sum += arr[i]; } // Create two vectors to store // the consecutive differences // of elements var v1 = []; var v2 = []; for (var i = 1; i < n; i += 2) { v1.push(arr[i] - arr[i - 1]); if (i + 1 < n) { v2.push(arr[i] - arr[i + 1]); } } // Find the maximum sum subarray var option1 = kadane(v1); var option2 = kadane(v2); // Add the maximum value // to initial sum var ans = sum + Math.max(option1, option2); // Return the answer return ans; } var arr = [ 5, 1, 3, 4, 5, 6 ]; var N = arr.length; // Function Call document.write(sumOfElements(arr, N));// This code is contributed by SoumikMondal</script> |
Output
15
Time Complexity: O(N), since traversal on the array is required to complete all operations hence the overall time required by the algorithm is linear.
Auxiliary Space: O(N), extra vectors are used hence algorithm takes up linear space.
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