Minimizing Maximum Absolute Subarray Sums

Given an array arr[] of size N, we can choose any real number X which when subtracted from all the elements of the array then the maximum absolute subarray sum among all the subarrays is minimum. The task is to return the minimum of maximum absolute sum among all the subarrays.

Note: The answer should be correct upto 6 decimal places.

Constraints:

• 1 <= N <= 10⁵
• abs(arr[i]) <= 10⁴

Examples:

Input: N = 3, arr[] = {50, -50, 50}
Output: 50.000000
Explanation: We can subtract X = 0 from all the elements of arr[], so the new array becomes {50, -50, 50}. Now, for all the subarrays:

• Subarray arr[0…0] = {50}, absolute sum = 50
• Subarray arr[1…1] = {-50}, absolute sum = 50
• Subarray arr[2…2] = {50}, absolute sum = 50
• Subarray arr[0…1] = {50, -50}, absolute sum = 0
• Subarray arr[1…2] = {-50, 50}, absolute sum = 0
• Subarray arr[0…2] = {50, -50, 50}, absolute sum = 50

So the maximum of all subarrays is 50. Subtracting any other number from arr[] will result to a greater value of maximum of all subarrays.

Input: N = 3, arr[] = {1, 2, 3}
Output: 1.000000
Explanation: We can subtract X = 2 from all the elements of arr[] so the new array becomes {-1, 0, 1}. Now, for all the subarrays:

• Subarray arr[0…0] = {-1}, absolute sum = 1
• Subarray arr[1…1] = {0}, absolute sum = 0
• Subarray arr[2…2] = {1}, absolute sum = 1
• Subarray arr[0…1] = {-1, 0}, absolute sum = 1
• Subarray arr[1…2] = {0, 1}, absolute sum = 1
• Subarray arr[0…2] = {-1, 0, 1}, absolute sum = 0

So the maximum of all subarrays is 1. Subtracting any other number from arr[] will result to a greater value of maximum of all subarrays.

Approach: The problem can be solved using the following approach:

If we observe carefully, then there will be an optimal value of X for which the maximum absolute sum among all subarrays will be minimum. Now, if we increase or decrease the value of then the maximum absolute sum among all subarrays will keep on increasing. Since this is same as Unimodal functions, we can apply Ternary Search to find the optimal value of X. As we are only concerned with the maximum absolute value of the subarray, we can calculate it by returning the maximum of largest subarray sum and abs(smallest subarray sum) using Kadane’s Algorithm. We will consider 2 points mid1 and mid2 in our search space [-10⁴, 10⁴] and store the corresponding values of maximum absolute sum of subarray in val1 and val2 respectively. Now, we can have 3 cases:

• If val1 == val2, then we can reduce our search space to [mid1, mid2]
• Else if val1 > val2, then we can reduce our search space to [mid1, hi]
• Else if val1 < val2, then we can reduce our search space to [lo, mid2]

Since the error limit is 1e-6 and in each iteration we are reducing our search space by a factor of at least 2/3. Therefore, 200 iterations are sufficient to get our answer.

Step-by-step approach:

• Declare a method getMinSubarraySum() to calculate the minimum subarray sum
• Declare a method getMaxSubarraySum() to calculate the maximum subarray sum
• Initialize lo = -1e4 and hi = 1e4
• Find two points mid1 and mid2 and store maximum absolute sum of subarray corresponding to mid1 and mid2 to val1 and val2 respectively.
• Reduce our search space using val1 and val2 for 200 iterations.
• Return the final answer as solve(lo) or solve(hi)

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
#define ll long long
using namespace std;

// Method to calculate the minimum subarray sum using
// Kadane's Algorithm
double getMinSubarraySum(double* arr, ll N)
{
    double sum = 0, ans = arr[0];
    for (int i = 0; i < N; i++) {
        sum += arr[i];
        ans = min(ans, sum);
        if (sum > 0)
            sum = 0;
    }
    return ans;
}

// Method to calculate the maximum subarray sum using
// Kadane's Algorithm
double getMaxSubarraySum(double* arr, ll N)
{
    double sum = 0, ans = arr[0];
    for (int i = 0; i < N; i++) {
        sum += arr[i];
        ans = max(ans, sum);
        if (sum < 0)
            sum = 0;
    }
    return ans;
}

// Minimize the maximum absolute value of sum of any
// subarray
double solve(double* arr, ll N, double X)
{
    // Update the array by subtracting X from each element
    for (int i = 0; i < N; i++)
        arr[i] -= X;

    // Calculate the maximum absolute sum among all
    // subarrays using minimum subarray sum and maximum
    // subarray sum
    double ans = max(abs(getMaxSubarraySum(arr, N)),
                    abs(getMinSubarraySum(arr, N)));

    // Retrieve the original values of array by adding X to
    // each element
    for (int i = 0; i < N; i++)
        arr[i] += X;

    return ans;
}

int main()
{

    ll N = 3;
    double arr[] = { 1, 2, 3 };

    // Run 200 iterations to get our answer
    ll cnt = 200;

    // Initialize the lower and upper limit of our search
    // space
    double lo = -1e4, hi = 1e4;

    while (cnt--) {
        double mid1 = lo + (hi - lo) / 3.0;
        double mid2 = hi - (hi - lo) / 3.0;

        // Calculate the maximum sum when X = mid1
        double val1 = solve(arr, N, mid1);

        // Calculate the maximum sum when X = mid2
        double val2 = solve(arr, N, mid2);

        // Minimize the search space
        if (val1 == val2) {
            lo = mid1;
            hi = mid2;
        }
        else if (val1 < val2) {
            hi = mid2;
        }
        else {
            lo = mid1;
        }
    }
    cout << fixed << setprecision(6) << solve(arr, N, lo)
        << "\n";
    return 0;
}

Java

import java.util.Arrays;

public class Main {
    // Method to calculate the minimum subarray sum using
    // Kadane's Algorithm
    static double getMinSubarraySum(double[] arr, int N) {
        double sum = 0, ans = arr[0];
        for (int i = 0; i < N; i++) {
            sum += arr[i];
            ans = Math.min(ans, sum);
            if (sum > 0)
                sum = 0;
        }
        return ans;
    }

    // Method to calculate the maximum subarray sum using
    // Kadane's Algorithm
    static double getMaxSubarraySum(double[] arr, int N) {
        double sum = 0, ans = arr[0];
        for (int i = 0; i < N; i++) {
            sum += arr[i];
            ans = Math.max(ans, sum);
            if (sum < 0)
                sum = 0;
        }
        return ans;
    }

    // Minimize the maximum absolute value of the sum of any
    // subarray
    static double solve(double[] arr, int N, double X) {
        // Update the array by subtracting X from each element
        for (int i = 0; i < N; i++)
            arr[i] -= X;

        // Calculate the maximum absolute sum among all
        // subarrays using the minimum subarray sum and maximum
        // subarray sum
        double ans = Math.max(Math.abs(getMaxSubarraySum(arr, N)),
                Math.abs(getMinSubarraySum(arr, N)));

        // Retrieve the original values of the array by adding X to
        // each element
        for (int i = 0; i < N; i++)
            arr[i] += X;

        return ans;
    }

    public static void main(String[] args) {
        int N = 3;
        double[] arr = {1, 2, 3};

        // Run 200 iterations to get our answer
        int cnt = 200;

        // Initialize the lower and upper limit of our search space
        double lo = -1e4, hi = 1e4;

        while (cnt-- > 0) {
            double mid1 = lo + (hi - lo) / 3.0;
            double mid2 = hi - (hi - lo) / 3.0;

            // Calculate the maximum sum when X = mid1
            double val1 = solve(Arrays.copyOf(arr, N), N, mid1);

            // Calculate the maximum sum when X = mid2
            double val2 = solve(Arrays.copyOf(arr, N), N, mid2);

            // Minimize the search space
            if (val1 == val2) {
                lo = mid1;
                hi = mid2;
            } else if (val1 < val2) {
                hi = mid2;
            } else {
                lo = mid1;
            }
        }
        System.out.printf("%.6f%n", solve(arr, N, lo));
    }
}

Python3

def get_min_subarray_sum(arr, N):
    """
    Function to find the minimum subarray sum using Kadane's algorithm.
    """
    sum_val = 0
    ans = arr[0]
    for i in range(N):
        sum_val += arr[i]
        ans = min(ans, sum_val)
        if sum_val > 0:
            sum_val = 0
    return ans


def get_max_subarray_sum(arr, N):
    """
    Function to find the maximum subarray sum using Kadane's algorithm.
    """
    sum_val = 0
    ans = arr[0]
    for i in range(N):
        sum_val += arr[i]
        ans = max(ans, sum_val)
        if sum_val < 0:
            sum_val = 0
    return ans


def solve(arr, N, X):
    """
    Function to solve the problem by modifying the array and finding the maximum absolute subarray sum.
    """
    # Subtract X from each element in the array
    for i in range(N):
        arr[i] -= X

    # Find the maximum absolute subarray sum after modification
    max_abs_sum = max(abs(get_max_subarray_sum(arr, N)), abs(get_min_subarray_sum(arr, N)))

    # Add X back to each element in the array to restore the original array
    for i in range(N):
        arr[i] += X

    return max_abs_sum


def main():
    N = 3
    arr = [1, 2, 3]
    cnt = 200
    lo, hi = -1e4, 1e4

    # Perform ternary search to find the optimal value of X
    while cnt > 0:
        mid1 = lo + (hi - lo) / 3.0
        mid2 = hi - (hi - lo) / 3.0

        val1 = solve(arr, N, mid1)
        val2 = solve(arr, N, mid2)

        # Adjust the search range based on the comparison of values
        if val1 == val2:
            lo = mid1
            hi = mid2
        elif val1 < val2:
            hi = mid2
        else:
            lo = mid1

        cnt -= 1

    # Print the result with 6 decimal places
    print(f"{solve(arr, N, lo):.6f}")


if __name__ == "__main__":
    main()

C#

using System;

class MainClass
{
    // Method to calculate the minimum subarray sum using
    // Kadane's Algorithm
    static double GetMinSubarraySum(double[] arr, int N)
    {
        double sum = 0, ans = arr[0];
        for (int i = 0; i < N; i++)
        {
            sum += arr[i];
            ans = Math.Min(ans, sum);
            if (sum > 0)
                sum = 0;
        }
        return ans;
    }

    // Method to calculate the maximum subarray sum using
    // Kadane's Algorithm
    static double GetMaxSubarraySum(double[] arr, int N)
    {
        double sum = 0, ans = arr[0];
        for (int i = 0; i < N; i++)
        {
            sum += arr[i];
            ans = Math.Max(ans, sum);
            if (sum < 0)
                sum = 0;
        }
        return ans;
    }

    // Minimize the maximum absolute value of the sum of any
    // subarray
    static double Solve(double[] arr, int N, double X)
    {
        // Update the array by subtracting X from each element
        for (int i = 0; i < N; i++)
            arr[i] -= X;

        // Calculate the maximum absolute sum among all
        // subarrays using the minimum subarray sum and maximum
        // subarray sum
        double ans = Math.Max(Math.Abs(GetMaxSubarraySum(arr, N)),
                Math.Abs(GetMinSubarraySum(arr, N)));

        // Retrieve the original values of the array by adding X to
        // each element
        for (int i = 0; i < N; i++)
            arr[i] += X;

        return ans;
    }

    public static void Main(string[] args)
    {
        int N = 3;
        double[] arr = { 1, 2, 3 };

        // Run 200 iterations to get our answer
        int cnt = 200;

        // Initialize the lower and upper limit of our search space
        double lo = -1e4, hi = 1e4;

        while (cnt-- > 0)
        {
            double mid1 = lo + (hi - lo) / 3.0;
            double mid2 = hi - (hi - lo) / 3.0;

            // Calculate the maximum sum when X = mid1
            double val1 = Solve((double[])arr.Clone(), N, mid1);

            // Calculate the maximum sum when X = mid2
            double val2 = Solve((double[])arr.Clone(), N, mid2);

            // Minimize the search space
            if (val1 == val2)
            {
                lo = mid1;
                hi = mid2;
            }
            else if (val1 < val2)
            {
                hi = mid2;
            }
            else
            {
                lo = mid1;
            }
        }
        Console.WriteLine($"{Solve(arr, N, lo):F6}");
    }
}

Javascript

// Method to calculate the minimum subarray sum using Kadane's Algorithm
function getMinSubarraySum(arr) {
    let sum = 0, ans = arr[0];
    for (let i = 0; i < arr.length; i++) {
        sum += arr[i];
        ans = Math.min(ans, sum);
        if (sum > 0)
            sum = 0;
    }
    return ans;
}

// Method to calculate the maximum subarray sum using Kadane's Algorithm
function getMaxSubarraySum(arr) {
    let sum = 0, ans = arr[0];
    for (let i = 0; i < arr.length; i++) {
        sum += arr[i];
        ans = Math.max(ans, sum);
        if (sum < 0)
            sum = 0;
    }
    return ans;
}

// Minimize the maximum absolute value of sum of any subarray
function solve(arr, X) {
    // Update the array by subtracting X from each element
    for (let i = 0; i < arr.length; i++)
        arr[i] -= X;

    // Calculate the maximum absolute sum among all subarrays using minimum subarray sum and maximum subarray sum
    let ans = Math.max(Math.abs(getMaxSubarraySum(arr)), Math.abs(getMinSubarraySum(arr)));

    // Retrieve the original values of array by adding X to each element
    for (let i = 0; i < arr.length; i++)
        arr[i] += X;

    return ans;
}

// Main function
function main() {
    const N = 3;
    const arr = [1, 2, 3];

    // Run 200 iterations to get our answer
    let cnt = 200;

    // Initialize the lower and upper limit of our search space
    let lo = -1e4, hi = 1e4;

    while (cnt--) {
        let mid1 = lo + (hi - lo) / 3.0;
        let mid2 = hi - (hi - lo) / 3.0;

        // Calculate the maximum sum when X = mid1
        let val1 = solve(arr.slice(), mid1);

        // Calculate the maximum sum when X = mid2
        let val2 = solve(arr.slice(), mid2);

        // Minimize the search space
        if (val1 === val2) {
            lo = mid1;
            hi = mid2;
        } else if (val1 < val2) {
            hi = mid2;
        } else {
            lo = mid1;
        }
    }
    console.log(solve(arr, lo).toFixed(6));
}

// Run the main function
main();
//This code is contributed by Utkarsh

Output

1.000000

Time Complexity: O(N * 2 * log₃(M)) = O(N * log₃(M)), where N is the size of arr[] and M is the maximum element in arr[].
Auxiliary Space: O(1)