Minimize the maximum subarray sum with 1s and -2s

Last Updated : 05 Mar, 2024

Given two integers X and Y. X and Y represent the frequency of elements 1 and -2 you have. You have to arrange all elements such that the maximum sum over all subarrays is minimized, and then find the maximum subarray sum of such rearrangement.

Examples:

Input: X = 1, Y = 1
Output: 1
Explanation: X = 1 and Y = 1 means we have one 1 and one -2. There can be only two possible arrangements with these elements as {1, -2} or {-2, 1}. In both of the cases the maximum subarray sum will be 1.

Input: X = 5, Y = 1
Output: 3
Explanation: The optimal arrangement of elements will be {1, 1, -2, 1, 1, 1}. Then, the maximum subarray sum will be 3.

Approach: To solve the problem, follow the below idea:

In case the number of 1's are too great, the -2s won't affect the subarray sum too much. Hence, we also calculate the sum of the whole array, which is (X - 2*Y) (Since the whole array is also its own subarray).
If there are Y number of -2s then there will be (Y+1) groups of 1s in arrangement. For example, X = 5 and Y = 2, then the arrangement will be as follows: {}, -2, {}, -2, {}. Where "{}" shows the collective group of ones (equal to (Y+1) = 3). We need to distribute X number of 1s in (Y+1) spaces equally. So that the distribution will be as Ceil (X/ (Y+1)).
Then, the maximum subarray sum will be max of both of these as max((X - 2*Y), Ceil (X/ (Y+1))).
The problem is observation based. The main idea is, for any subarray to have its sum minimized, we need to distribute the 1s among the -2s such that the count of 1s in between -2s are smallest. This can only be done by optimally dividing the number of 1's between the 2's.

Illustration:

Let us understand it with some examples and find some observations:

Example 1: X = 5, Y = 2
We have five 1s and two -2s.
We need to distribute 1s in between -2s, then the optimal arrangement will be as follows: {1, 1, -2, 1, -2, 1, 1}. The maximum subarray sum will be 2 in this case. Max((X - 2*Y), Ciel (X/ (Y+1))) = 2.
Example 2: X = 5, Y = 1
We have five 1s and one -2s.
The optimal arrangement will be as follows: {1, 1, -2, 1, 1, 1}. The maximum subarray sum will be 3 in this case. Ceil(5/(1+1)) = 3. Max((X - 2*Y), Ciel (X/ (Y+1))) = 3.
Example 3: X = 1, Y = 3
We have one 1 and three -2s.
The optimal arrangement will be as follows: {-2, -2, -2, 1}. The maximum subarray sum will be 1. Max((X - 2*Y), Ciel (X/ (Y+1))) = 1.

Step-by-step algorithm:

Declare two variables let say A and B.
Initialize A with Ciel (X/ (Y+1)).
Initialize B with sum of all elements as (X - 2*Y).
Output Max (A, B).

Below is the implementation of the algorithm:

C++

#include <iostream> #include <cmath>  // Function to output the maximum subarray sum after // optimal arrangement void FindMaxSum(double X, double Y) {     double A = std::ceil(X / (Y + 1.0));     double B = (X - 2 * Y);     std::cout << std::max(A, B) << std::endl; }  // Driver Function int main() {     // Inputs     double X = 5, Y = 2;      // Function_call     FindMaxSum(X, Y);      return 0; }   // This code is contributed by akshitaguprzj3

Java

// Java code to implement the approach  import java.util.*;  // Driver Class public class Main {      // Driver Function     public static void main(String[] args)     {         // Inputs         double X = 5, Y = 2;          // Function_call         FindMaxSum(X, Y);     }      // Function to output the maximum subarray sum after     // optimal arrangement     public static void FindMaxSum(Double X, Double Y)     {         double A = Math.ceil(X / (Y + 1.0));         double B = (X - 2 * Y);         System.out.println(Math.max(A, B));     } }

Python3

# Python Implementation  import math  def find_max_sum(X, Y):     A = math.ceil(X / (Y + 1.0))     B = X - 2 * Y     return max(A, B)  X = 5 Y = 2  max_sum = find_max_sum(X, Y) print(max_sum)  # This code is contributed by Sakshi

// C# code to implement the approach using System;  // Driver Class public class GFG {     // Driver Function     public static void Main(string[] args)     {         // Inputs         double X = 5, Y = 2;          // Function_call         FindMaxSum(X, Y);     }      // Function to output the maximum subarray sum after     // optimal arrangement     public static void FindMaxSum(double X, double Y)     {         double A = Math.Ceiling(X / (Y + 1.0));         double B = (X - 2 * Y);         Console.WriteLine(Math.Max(A, B));     } }

JavaScript

// Function to output the maximum subarray sum after optimal arrangement function findMaxSum(X, Y) {     let A = Math.ceil(X / (Y + 1));     let B = X - 2 * Y;     console.log(Math.max(A, B)); }  // Driver Function function main() {     // Inputs     let X = 5, Y = 2;      // Function call     findMaxSum(X, Y); }  // Invoke main function main();

Output

2.0

Time Complexity: O(1)
Auxiliary Space: O(1)

Maximum sum subarray of size K with sum less than X

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Minimize the maximum subarray sum with 1s and -2s

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