Minimize the maximum value of the absolute difference

Last Updated : 28 Dec, 2023

Given array A[] of size N (1 <= N <= 10⁵). The following operation should be performed until array A[] is not empty. choose three integers X, Y, and Z, Take out an element from array A[] (1 <= A[i] <= 10⁹), and choose one integer out of the chosen three integers X, Y, and Z record absolute difference between them, then delete the element. The task for this problem is to minimize the maximum value of this absolute difference.

Examples:

Input: A[] = {1, 7, 7, 9, 9, 9}
Output: 0
Explanation: If X = 1, Y = 7 and Z = 9

for A[1] = 1 choose X = 1 the absolute difference is 0
for A[2] = 7 choose Y = 7 the absolute difference is 0
for A[3] = 7 choose Y = 7 the absolute difference is 0
for A[4] = 9 choose Z = 9 the absolute difference is 0
for A[5] = 9 choose Z = 9 the absolute difference is 0
for A[6] = 9 choose Z = 9 the absolute difference is 0
the maximum absolute difference is 0

Input: A[] = {5, 4, 2, 1, 30, 60}
Output: 2
Explanation: If X = 3, Y = 30 and Z = 60

for A[1] = 5 choose X = 3 the absolute difference is 2
for A[2] = 4 choose X = 3 the absolute difference is 1
for A[3] = 2 choose X = 3 the absolute difference is 1
for A[4] = 1 choose X = 3 the absolute difference is 2
for A[5] = 30 choose Y = 30 the absolute difference is 0
for A[6] = 60 choose Z = 60 the absolute difference is 0
maximum absolute difference is 2

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

Binary Search can be used to solve this problem and the range of binary search will be 1 to maxArrayElement(A[]) is monotonic function represents whether absolute difference of mid can be achieved or not. it is of the form FFFFFFFTTTTTTT. we have to find when the first time function becomes true using Binary Search.

Below are the steps for the above approach:

Set low and high range of binary serach.
ispos(mid) function used to check whether mid can be the maximum absolute difference.
Run while loop till high and low are not equal.
In while loop find middle element and store it in mid variable.
Check if that mid can be maximum value of array using ispos() function. If it is true then set high = mid else low = mid + 1.
After loop ends if ispos() true for low then return low else return high.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to check given absolute difference
// is possible or not
bool ispos(int mid, int A[], int N)
{
    // declaring array containing X
    vector<int> sze{ A[0] + mid };

    // Iterating over N elements starting
    // from first index
    for (int i = 1; i < N; i++) {

        // Declaring answer variable for tracking
        int ans = INT_MAX;

        // Iterating on chosen elements
        for (int j = 0; j < sze.size(); j++) {
            ans = min(ans, abs(A[i] - sze[j]));
        }

        // if answer is greater than mid
        if (ans > mid) {

            // Return false if elements required to
            // have maximum absolute difference mid
            // is greater than 3
            if (sze.size() == 3)
                return false;

            // push A[i] + mid to our array
            sze.push_back(A[i] + mid);
        }
    }

    // Return true if it is possible
    return true;
}

// Function to find Minimize the
// maximum value of absolute difference
int minimizeMaximumAbsDiff(int A[], int N)
{
    // sort the array
    sort(A, A + N);

    // Range of binary search
    int low = 0, high = A[N - 1];

    // Running loop till high
    // is not equal to low
    while (high - low > 1) {

        // mid is average of low and high
        int mid = (low + high) / 2;

        // Checking test function
        if (ispos(mid, A, N)) {
            high = mid;
        }
        else {
            low = mid + 1;
        }
    }

    // Checking whether low can be answer
    if (ispos(low, A, N))
        return low;

    // If not then it is high
    else
        return high;
}

// Driver Code
int32_t main()
{

    // Input 1
    int N = 6, A[] = { 1, 7, 7, 9, 9, 9 };

    // Function Call
    cout << minimizeMaximumAbsDiff(A, N) << endl;

    // Input 2
    int N1 = 6, A1[] = { 5, 4, 2, 1, 30, 60 };

    // Function Call
    cout << minimizeMaximumAbsDiff(A1, N1) << endl;

    return 0;
}

Java

import java.util.Arrays;

public class GFG {

    // Function to check given absolute difference
    // is possible or not
    static boolean isPos(int mid, int[] A, int N) {
        // declaring array containing X
        int[] sze = new int[]{A[0] + mid};

        // Iterating over N elements starting
        // from first index
        for (int i = 1; i < N; i++) {

            // Declaring answer variable for tracking
            int ans = Integer.MAX_VALUE;

            // Iterating on chosen elements
            for (int j = 0; j < sze.length; j++) {
                ans = Math.min(ans, Math.abs(A[i] - sze[j]));
            }

            // if answer is greater than mid
            if (ans > mid) {

                // Return false if elements required to
                // have maximum absolute difference mid
                // is greater than 3
                if (sze.length == 3)
                    return false;

                // push A[i] + mid to our array
                sze = Arrays.copyOf(sze, sze.length + 1);
                sze[sze.length - 1] = A[i] + mid;
            }
        }

        // Return true if it is possible
        return true;
    }

    // Function to find Minimize the
    // maximum value of absolute difference
    static int minimizeMaximumAbsDiff(int[] A, int N) {
        // sort the array
        Arrays.sort(A);

        // Range of binary search
        int low = 0, high = A[N - 1];

        // Running loop till high
        // is not equal to low
        while (high - low > 1) {

            // mid is average of low and high
            int mid = (low + high) / 2;

            // Checking test function
            if (isPos(mid, A, N)) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }

        // Checking whether low can be answer
        if (isPos(low, A, N))
            return low;

        // If not then it is high
        else
            return high;
    }

    // Driver Code
    public static void main(String[] args) {

        // Input 1
        int N = 6;
        int[] A = {1, 7, 7, 9, 9, 9};

        // Function Call
        System.out.println(minimizeMaximumAbsDiff(A, N));

        // Input 2
        int N1 = 6;
        int[] A1 = {5, 4, 2, 1, 30, 60};

        // Function Call
        System.out.println(minimizeMaximumAbsDiff(A1, N1));
    }
}
// This code is contributed by rohit singh

Python

# Function to check if given absolute difference
# is possible or not


def ispos(mid, A, N):
    # declaring array containing X
    sze = [A[0] + mid]

    # Iterating over N elements starting
    # from the first index
    for i in range(1, N):
        # Declaring answer variable for tracking
        ans = float('inf')

        # Iterating on chosen elements
        for j in range(len(sze)):
            ans = min(ans, abs(A[i] - sze[j]))

        # if answer is greater than mid
        if ans > mid:
            # Return false if elements required to
            # have the maximum absolute difference mid
            # is greater than 3
            if len(sze) == 3:
                return False

            # push A[i] + mid to our array
            sze.append(A[i] + mid)

    # Return true if it is possible
    return True


# Function to minimize the maximum value of absolute difference
def minimizeMaximumAbsDiff(A, N):
    # sort the array
    A.sort()

    # Range of binary search
    low, high = 0, A[N - 1]

    # Running loop until high
    # is not equal to low
    while high - low > 1:
        # mid is average of low and high
        mid = (low + high) // 2

        # Checking test function
        if ispos(mid, A, N):
            high = mid
        else:
            low = mid + 1

    # Checking whether low can be the answer
    if ispos(low, A, N):
        return low

    # If not then it is high
    else:
        return high


# Driver Code
# Input 1
N = 6
A = [1, 7, 7, 9, 9, 9]

# Function Call
print(minimizeMaximumAbsDiff(A, N))

# Input 2
N1 = 6
A1 = [5, 4, 2, 1, 30, 60]

# Function Call
print(minimizeMaximumAbsDiff(A1, N1))

using System;

public class GFG {
    // Function to check if given absolute difference
    // is possible or not
    static bool IsPos(int mid, int[] A, int N)
    {
        // declaring array containing X
        int[] sze = new int[] { A[0] + mid };

        // Iterating over N elements starting
        // from the first index
        for (int i = 1; i < N; i++) {
            // Declaring answer variable for tracking
            int ans = int.MaxValue;

            // Iterating on chosen elements
            for (int j = 0; j < sze.Length; j++) {
                ans = Math.Min(ans,
                               Math.Abs(A[i] - sze[j]));
            }

            // if answer is greater than mid
            if (ans > mid) {
                // Return false if elements required to
                // have maximum absolute difference mid
                // is greater than 3
                if (sze.Length == 3)
                    return false;

                // push A[i] + mid to our array
                Array.Resize(ref sze, sze.Length + 1);
                sze[sze.Length - 1] = A[i] + mid;
            }
        }

        // Return true if it is possible
        return true;
    }

    // Function to find Minimize the
    // maximum value of absolute difference
    static int MinimizeMaximumAbsDiff(int[] A, int N)
    {
        // sort the array
        Array.Sort(A);

        // Range of binary search
        int low = 0, high = A[N - 1];

        // Running loop until high
        // is not equal to low
        while (high - low > 1) {
            // mid is the average of low and high
            int mid = (low + high) / 2;

            // Checking test function
            if (IsPos(mid, A, N)) {
                high = mid;
            }
            else {
                low = mid + 1;
            }
        }

        // Checking whether low can be the answer
        if (IsPos(low, A, N))
            return low;

        // If not, then it is high
        else
            return high;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        // Input 1
        int N = 6;
        int[] A = { 1, 7, 7, 9, 9, 9 };

        // Function Call
        Console.WriteLine(MinimizeMaximumAbsDiff(A, N));

        // Input 2
        int N1 = 6;
        int[] A1 = { 5, 4, 2, 1, 30, 60 };

        // Function Call
        Console.WriteLine(MinimizeMaximumAbsDiff(A1, N1));
    }
}

JavaScript

// Function to check if the given 
// absolute difference is possible or not
function ispos(mid, A, N) {
    let sze = [A[0] + mid];

    for (let i = 1; i < N; i++) {
        let ans = Number.MAX_SAFE_INTEGER;

        for (let j = 0; j < sze.length; j++) {
            ans = Math.min(ans, Math.abs(A[i] - sze[j]));
        }

        if (ans > mid) {
            if (sze.length === 3) {
                return false;
            }
            sze.push(A[i] + mid);
        }
    }

    return true;
}

// Function to find Minimize the
// maximum value of absolute difference
function minimizeMaximumAbsDiff(A, N) {
    A.sort((a, b) => a - b);

    let low = 0,
        high = A[N - 1];

    while (high - low > 1) {
        let mid = Math.floor((low + high) / 2);

        if (ispos(mid, A, N)) {
            high = mid;
        } 
        else {
            low = mid + 1;
        }
    }

    if (ispos(low, A, N)) {
        return low;
    } 
    else {
        return high;
    }
}

// Driver Code

// Input 1
let N = 6, A = [1, 7, 7, 9, 9, 9];
console.log(minimizeMaximumAbsDiff(A, N));

// Input 2
let N1 = 6, A1 = [5, 4, 2, 1, 30, 60];
console.log(minimizeMaximumAbsDiff(A1, N1));

Output

0
2

Time Complexity: O(N*logN)
Auxiliary Space: O(1)

Minimum and Maximum sum of absolute differences of pairs

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Minimize the maximum value of the absolute difference

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