Minimum value of maximum absolute difference of all adjacent pairs in an Array

Given an array arr, containing non-negative integers and (-1)s, of size N, the task is to replace those (-1)s with a common non-negative integer such that the maximum absolute difference of all adjacent pairs is minimum. Print this minimum possible value of the maximum absolute difference.

Examples:

Input: arr = {-1, -1, 11, -1, 3, -1}
Output: 4
Replace every -1 element with 7. Now the maximum absolute difference of all adjacent pairs is minimum which is equal to 4



Input: arr = {4, -1}
Output: 0

Approach:

  1. Consider only those non-missing elements that are adjacent to at least one missing element.
  2. Find the maximum element and the minimum element among them.
  3. We need to find a value that minimizes the maximum absolute difference between the common value and these values.
  4. The optimal value is equals to
    (minimum element + maximum element) / 2

Below is the implementation of the above approach:

C++

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// C++ program to find the minimum value
// of maximum absolute difference of
// all adjacent pairs in an Array
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the minimum possible
// value of the maximum absolute difference.
int maximumAbsolute(int arr[], int n)
{
    // To store minimum and maximum elements
    int mn = INT_MAX;
    int mx = INT_MIN;
  
    for (int i = 0; i < n; i++) {
        // If right side element is equals -1
        // and left side is not equals -1
        if (i > 0
            && arr[i] == -1
            && arr[i - 1] != -1) {
            mn = min(mn, arr[i - 1]);
            mx = max(mx, arr[i - 1]);
        }
  
        // If left side element is equals -1
        // and right side is not equals -1
        if (i < n - 1
            && arr[i] == -1
            && arr[i + 1] != -1) {
            mn = min(mn, arr[i + 1]);
            mx = max(mx, arr[i + 1]);
        }
    }
  
    // Calculating the common integer
    // which needs to be replaced with
    int common_integer = (mn + mx) / 2;
  
    // Replace all -1 elements
    // with the common integer
    for (int i = 0; i < n; i++) {
        if (arr[i] == -1)
            arr[i] = common_integer;
    }
  
    int max_diff = 0;
  
    // Calculating the maximum
    // absolute difference
    for (int i = 0; i < n - 1; i++) {
        int diff = abs(arr[i] - arr[i + 1]);
  
        if (diff > max_diff)
            max_diff = diff;
    }
  
    // Return the maximum absolute difference
    return max_diff;
}
  
// Driver Code
int main()
{
    int arr[] = { -1, -1, 11, -1, 3, -1 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << maximumAbsolute(arr, n);
  
    return 0;
}

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Java

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// Java program to find the minimum value
// of maximum absolute difference of
// all adjacent pairs in an Array
import java.util.*;
  
class GFG{
   
// Function to find the minimum possible
// value of the maximum absolute difference.
static int maximumAbsolute(int arr[], int n)
{
    // To store minimum and maximum elements
    int mn = Integer.MAX_VALUE;
    int mx = Integer.MIN_VALUE;
   
    for (int i = 0; i < n; i++) {
  
        // If right side element is equals -1
        // and left side is not equals -1
        if (i > 0
            && arr[i] == -1
            && arr[i - 1] != -1) {
            mn = Math.min(mn, arr[i - 1]);
            mx = Math.max(mx, arr[i - 1]);
        }
   
        // If left side element is equals -1
        // and right side is not equals -1
        if (i < n - 1
            && arr[i] == -1
            && arr[i + 1] != -1) {
            mn = Math.min(mn, arr[i + 1]);
            mx = Math.max(mx, arr[i + 1]);
        }
    }
   
    // Calculating the common integer
    // which needs to be replaced with
    int common_integer = (mn + mx) / 2;
   
    // Replace all -1 elements
    // with the common integer
    for (int i = 0; i < n; i++) {
        if (arr[i] == -1)
            arr[i] = common_integer;
    }
   
    int max_diff = 0;
   
    // Calculating the maximum
    // absolute difference
    for (int i = 0; i < n - 1; i++) {
        int diff = Math.abs(arr[i] - arr[i + 1]);
   
        if (diff > max_diff)
            max_diff = diff;
    }
   
    // Return the maximum absolute difference
    return max_diff;
}
   
// Driver Code
public static void main(String[] args)
{
    int arr[] = { -1, -1, 11, -1, 3, -1 };
    int n = arr.length;
   
    // Function call
    System.out.print(maximumAbsolute(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program to find the minimum value
# of maximum absolute difference of
# all adjacent pairs in an Array
  
# Function to find the minimum possible
# value of the maximum absolute difference.
def maximumAbsolute(arr, n):
  
    # To store minimum and maximum elements
    mn = 10**9
    mx = -10**9
  
    for i in range(n):
      
        # If right side element is equals -1
        # and left side is not equals -1
        if (i > 0
            and arr[i] == -1
            and arr[i - 1] != -1):
            mn = min(mn, arr[i - 1])
            mx = max(mx, arr[i - 1])
  
        # If left side element is equals -1
        # and right side is not equals -1
        if (i < n - 1
            and arr[i] == -1
            and arr[i + 1] != -1):
            mn = min(mn, arr[i + 1])
            mx = max(mx, arr[i + 1])
  
    # Calculating the common integer
    # which needs to be replaced with
    common_integer = (mn + mx) // 2
  
    # Replace all -1 elements
    # with the common integer
    for i in range(n):
        if (arr[i] == -1):
            arr[i] = common_integer
  
    max_diff = 0
  
    # Calculating the maximum
    # absolute difference
    for i in range(n-1):
        diff = abs(arr[i] - arr[i + 1])
  
        if (diff > max_diff):
            max_diff = diff
  
    # Return the maximum absolute difference
    return max_diff
  
# Driver Code
if __name__ == '__main__':
    arr=[-1, -1, 11, -1, 3, -1]
    n = len(arr)
  
    # Function call
    print(maximumAbsolute(arr, n))
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to find the minimum value
// of maximum absolute difference of
// all adjacent pairs in an Array
using System;
  
class GFG{
   
    // Function to find the minimum possible
    // value of the maximum absolute difference.
    static int maximumAbsolute(int []arr, int n)
    {
        // To store minimum and maximum elements
        int mn = int.MaxValue;
        int mx = int.MinValue;
       
        for (int i = 0; i < n; i++) {
      
            // If right side element is equals -1
            // and left side is not equals -1
            if (i > 0
                && arr[i] == -1
                && arr[i - 1] != -1) {
                mn = Math.Min(mn, arr[i - 1]);
                mx = Math.Max(mx, arr[i - 1]);
            }
       
            // If left side element is equals -1
            // and right side is not equals -1
            if (i < n - 1
                && arr[i] == -1
                && arr[i + 1] != -1) {
                mn = Math.Min(mn, arr[i + 1]);
                mx = Math.Max(mx, arr[i + 1]);
            }
        }
       
        // Calculating the common integer
        // which needs to be replaced with
        int common_integer = (mn + mx) / 2;
       
        // Replace all -1 elements
        // with the common integer
        for (int i = 0; i < n; i++) {
            if (arr[i] == -1)
                arr[i] = common_integer;
        }
       
        int max_diff = 0;
       
        // Calculating the maximum
        // absolute difference
        for (int i = 0; i < n - 1; i++) {
            int diff = Math.Abs(arr[i] - arr[i + 1]);
       
            if (diff > max_diff)
                max_diff = diff;
        }
       
        // Return the maximum absolute difference
        return max_diff;
    }
       
    // Driver Code
    public static void Main(string[] args)
    {
        int []arr = { -1, -1, 11, -1, 3, -1 };
        int n = arr.Length;
       
        // Function call
        Console.Write(maximumAbsolute(arr, n));
    }
}
  
// This code is contributed by Yash_R

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Output:

4

Time complexity: O(N)

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