Given an array arr[] of positive integers. The task is to reverse a subarray to minimize the sum of elements at even places and print the minimum sum.
Note: Perform the move only one time. Subarray might not be reversed.
Example:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 7
Explanation:
Sum of elements at even positions initially = arr[0] + arr[2] + arr[4] = 1 + 3 + 5 = 9
On reversing the subarray from position [1, 4], the array becomes: {1, 5, 4, 3, 2}
Now the sum of elements at even positions = arr[0] + arr[2] + arr[4] = 1 + 4 + 2 = 7, which is the minimum sum.Input: arr[] = {0, 1, 4, 3}
Output: 1
Explanation:
Sum of elements at even positions initially = arr[0] + arr[2] = 0 + 4 = 4
On reversing the subarray from position [1, 2], the array becomes: {0, 4, 1, 3}
Now the sum of elements at even positions = arr[0] + arr[2] = 0 + 1 = 1, which is the minimum sum.
Naive Approach: The idea is to apply the Brute Force method and generate all the subarrays and check the sum of elements at the even position. Print the sum which is the minimum among all.
Below is the code for the above approach :
C++
// C++ implementation to reverse a subarray// of the given array to minimize the// sum of elements at even position#include <bits/stdc++.h>#define N 5using namespace std;// Function that will give// the max negative valueint after_rev(vector<int> v){ int mini = 0, count = 0; for (int i = 0; i < v.size(); i++) { count += v[i]; // Check for count // greater than 0 // as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}// Function to print the minimum sumvoid print(int arr[N]){ int sum = 0; // Taking sum of only // even index elements for (int i = 0; i < N; i += 2) sum += arr[i]; // Naive approach to generate all subarrays // and check their sums at even positions for (int i = 0; i < N; i++) { for (int j = i; j < N; j++) { // Reverse the subarray from i to j reverse(arr + i, arr + j + 1); // Update the sum if the current subarray // gives a smaller sum at even positions int current_sum = 0; for (int k = 0; k < N; k += 2) current_sum += arr[k]; sum = min(sum, current_sum); // Reverse the subarray back to original reverse(arr + i, arr + j + 1); } } cout << sum << endl;}// Driver codeint main(){ int arr[N] = { 0, 1, 4, 3 }; print(arr); return 0;} |
Java
import java.util.Arrays;public class Main { static int afterRev(int[] arr) { int mini = 0, count = 0; for (int i = 0; i < arr.length; i++) { count += arr[i]; // Check for count greater than 0 // as we require only negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini; } static void print(int[] arr) { int sum = 0; // Taking sum of only even index elements for (int i = 0; i < arr.length; i += 2) sum += arr[i]; // Naive approach to generate all subarrays // and check their sums at even positions for (int i = 0; i < arr.length; i++) { for (int j = i; j < arr.length; j++) { // Reverse the subarray from i to j reverse(arr, i, j); // Update the sum if the current subarray // gives a smaller sum at even positions int currentSum = 0; for (int k = 0; k < arr.length; k += 2) currentSum += arr[k]; sum = Math.min(sum, currentSum); // Reverse the subarray back to original reverse(arr, i, j); } } System.out.println(sum); } static void reverse(int[] arr, int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } } public static void main(String[] args) { int[] arr = { 0, 1, 4, 3 }; print(arr); }} |
Python3
# Function that will give the max negative valuedef after_rev(v): mini = 0 count = 0 for i in range(len(v)): count += v[i] # Check for count greater than 0 # as we require only negative solution if count > 0: count = 0 if mini > count: mini = count return mini# Function to print the minimum sumdef print_sum(arr): sum_val = 0 # Taking sum of only even index elements for i in range(0, len(arr), 2): sum_val += arr[i] # Naive approach to generate all subarrays # and check their sums at even positions for i in range(len(arr)): for j in range(i, len(arr)): # Reverse the subarray from i to j arr[i:j+1] = arr[i:j+1][::-1] # Update the sum if the current subarray # gives a smaller sum at even positions current_sum = 0 for k in range(0, len(arr), 2): current_sum += arr[k] sum_val = min(sum_val, current_sum) # Reverse the subarray back to the original arr[i:j+1] = arr[i:j+1][::-1] print(sum_val)# Driver codeif __name__ == "__main__": arr = [0, 1, 4, 3] print_sum(arr) |
C#
using System;class Program{ static int AfterRev(int[] arr) { int mini = 0, count = 0; for (int i = 0; i < arr.Length; i++) { count += arr[i]; // Check for count // greater than 0 // as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini; } static void Print(int[] arr) { int sum = 0; // Taking sum of only // even index elements for (int i = 0; i < arr.Length; i += 2) sum += arr[i]; // Naive approach to generate all subarrays // and check their sums at even positions for (int i = 0; i < arr.Length; i++) { for (int j = i; j < arr.Length; j++) { // Reverse the subarray from i to j Array.Reverse(arr, i, j - i + 1); // Update the sum if the current subarray // gives a smaller sum at even positions int currentSum = 0; for (int k = 0; k < arr.Length; k += 2) currentSum += arr[k]; sum = Math.Min(sum, currentSum); // Reverse the subarray back to original Array.Reverse(arr, i, j - i + 1); } } Console.WriteLine(sum); } static void Main(string[] args) { int[] arr = { 0, 1, 4, 3 }; Print(arr); }} |
Javascript
function afterRev(arr) { let mini = 0; let count = 0; for (let i = 0; i < arr.length; i++) { count += arr[i]; // Check for count // greater than 0 // as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}function print(arr) { let sum = 0; for (let i = 0; i < arr.length; i += 2) sum += arr[i]; let originalArr = [...arr]; // Create a copy of the original array for (let i = 0; i < arr.length; i++) { for (let j = i; j < arr.length; j++) { // Reverse the subarray from i to j arr = reverseSubarray(arr, i, j); let currentSum = 0; for (let k = 0; k < arr.length; k += 2) currentSum += arr[k]; sum = Math.min(sum, currentSum); // Restore the original subarray arr = originalArr.slice(); } } console.log(sum);}function reverseSubarray(arr, start, end) { let subarray = arr.slice(start, end + 1); subarray.reverse(); for (let i = start, j = 0; i <= end; i++, j++) { arr[i] = subarray[j]; } return arr;}const arr = [0, 1, 4, 3];print(arr); |
Output
1
Time Complexity: O(N3)
Auxiliary Space: O(N)
Efficient Approach: The idea is to observe the following important points for array arr[]:
- Reversing the subarray of odd length won’t change the sum because all elements at even index will again come to the even index.
- Reversing an even length subarray will make all elements at index position come to the even index and elements at even index goes to the odd index.
- Sum of elements will change only when an odd index element is put at even indexed elements. Let’s say element at index 1 can go to index 0 or index 2.
- On reversing from an even index to another even index example from index 2 to 4, it will be covering in the first case or if from odd index to even index example from index 1 to 4, it will be covering the second case.
Below are the steps of the approach based on the above observations:
- So the idea is to find the sum of only even index elements and initialize two arrays lets say v1 and v2 such that v1 will keep account of change if first index element goes to 0 whereas v2 will keep the account of change if first index element goes to 2.
- Get the minimum of the two values and check if it’s lesser than 0. If it is, then add it to the answer and finally return the answer.
Below is the implementation of the above approach :
C++
// C++ implementation to reverse a subarray// of the given array to minimize the// sum of elements at even position#include <bits/stdc++.h>#define N 5using namespace std;// Function that will give// the max negative valueint after_rev(vector<int> v){ int mini = 0, count = 0; for (int i = 0; i < v.size(); i++) { count += v[i]; // Check for count // greater than 0 // as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}// Function to print the minimum sumvoid print(int arr[N]){ int sum = 0; // Taking sum of only // even index elements for (int i = 0; i < N; i += 2) sum += arr[i]; // Initialize two vectors v1, v2 vector<int> v1, v2; // v1 will keep account for change // if 1th index element goes to 0 for (int i = 0; i + 1 < N; i += 2) v1.push_back(arr[i + 1] - arr[i]); // v2 will keep account for change // if 1th index element goes to 2 for (int i = 1; i + 1 < N; i += 2) v2.push_back(arr[i] - arr[i + 1]); // Get the max negative value int change = min(after_rev(v1), after_rev(v2)); if (change < 0) sum += change; cout << sum << endl;}// Driver codeint main(){ int arr[N] = { 0, 1, 4, 3 }; print(arr); return 0;} |
Java
// Java implementation to reverse a subarray// of the given array to minimize the// sum of elements at even positionimport java.util.*;class GFG{static final int N = 5;// Function that will give// the max negative valuestatic int after_rev(Vector<Integer> v){ int mini = 0, count = 0; for(int i = 0; i < v.size(); i++) { count += v.get(i); // Check for count greater // than 0 as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}// Function to print the minimum sumstatic void print(int arr[]){ int sum = 0; // Taking sum of only // even index elements for(int i = 0; i < N; i += 2) sum += arr[i]; // Initialize two vectors v1, v2 Vector<Integer> v1, v2; v1 = new Vector<Integer>(); v2 = new Vector<Integer>(); // v1 will keep account for change // if 1th index element goes to 0 for(int i = 0; i + 1 < N; i += 2) v1.add(arr[i + 1] - arr[i]); // v2 will keep account for change // if 1th index element goes to 2 for(int i = 1; i + 1 < N; i += 2) v2.add(arr[i] - arr[i + 1]); // Get the max negative value int change = Math.min(after_rev(v1), after_rev(v2)); if (change < 0) sum += change; System.out.print(sum + "\n");}// Driver codepublic static void main(String[] args){ int arr[] = { 0, 1, 4, 3, 0 }; print(arr);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to reverse # a subarray of the given array to # minimize the sum of elements at # even position # Function that will give# the max negative valuedef after_rev(v): mini = 0 count = 0 for i in range(len(v)): count += v[i] # Check for count greater # than 0 as we require only # negative solution if(count > 0): count = 0 if(mini > count): mini = count return mini# Function to print the# minimum sumdef print_f(arr): sum = 0 # Taking sum of only # even index elements for i in range(0, len(arr), 2): sum += arr[i] # Initialize two vectors v1, v2 v1, v2 = [], [] # v1 will keep account for change # if 1th index element goes to 0 i = 1 while i + 1 < len(arr): v1.append(arr[i + 1] - arr[i]) i += 2 # v2 will keep account for change # if 1th index element goes to 2 i = 1 while i + 1 < len(arr): v2.append(arr[i] - arr[i + 1]) i += 2 # Get the max negative value change = min(after_rev(v1), after_rev(v2)) if(change < 0): sum += change print(sum)# Driver codeif __name__ == '__main__': arr = [ 0, 1, 4, 3 ] print_f(arr)# This code is contributed by Shivam Singh |
C#
// C# implementation to reverse a subarray// of the given array to minimize the// sum of elements at even positionusing System;using System.Collections.Generic;class GFG{static readonly int N = 5;// Function that will give// the max negative valuestatic int after_rev(List<int> v){ int mini = 0, count = 0; for(int i = 0; i < v.Count; i++) { count += v[i]; // Check for count greater // than 0 as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}// Function to print the minimum sumstatic void print(int []arr){ int sum = 0; // Taking sum of only // even index elements for(int i = 0; i < N; i += 2) sum += arr[i]; // Initialize two vectors v1, v2 List<int> v1, v2; v1 = new List<int>(); v2 = new List<int>(); // v1 will keep account for change // if 1th index element goes to 0 for(int i = 0; i + 1 < N; i += 2) v1.Add(arr[i + 1] - arr[i]); // v2 will keep account for change // if 1th index element goes to 2 for(int i = 1; i + 1 < N; i += 2) v2.Add(arr[i] - arr[i + 1]); // Get the max negative value int change = Math.Min(after_rev(v1), after_rev(v2)); if (change < 0) sum += change; Console.Write(sum + "\n");}// Driver codepublic static void Main(String[] args){ int []arr = { 0, 1, 4, 3, 0 }; print(arr);}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript implementation to reverse a subarray// of the given array to minimize the// sum of elements at even positionvar N = 3;// Function that will give// the max negative valuefunction after_rev(v){ var mini = 0, count = 0; for (var i = 0; i < v.length; i++) { count += v[i]; // Check for count // greater than 0 // as we require only // negative solution if (count > 0) count = 0; if (mini > count) mini = count; } return mini;}// Function to print the minimum sumfunction print(arr){ var sum = 0; // Taking sum of only // even index elements for (var i = 0; i < N; i += 2) sum += arr[i]; // Initialize two vectors v1, v2 var v1 = [], v2 = []; // v1 will keep account for change // if 1th index element goes to 0 for (var i = 0; i + 1 < N; i += 2) v1.push(arr[i + 1] - arr[i]); // v2 will keep account for change // if 1th index element goes to 2 for (var i = 1; i + 1 < N; i += 2) v2.push(arr[i] - arr[i + 1]); // Get the max negative value var change = Math.min(after_rev(v1), after_rev(v2)); if (change < 0) sum += change; document.write( sum );}// Driver codevar arr = [0, 1, 4, 3];print(arr);// This code is contributed by importantly.</script> |
Output
1
Time Complexity: O(N)
Auxiliary Space: O(N)
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