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

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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:
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:
 

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++




// 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;
}


Java




// 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


Python3




# 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


C#




// 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


Javascript




<script>
 
// Javascript 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.
function maximumAbsolute(arr, n)
{
     
    // To store minimum and maximum elements
    var mn = Number.MAX_VALUE;
    var mx = Number.MIN_VALUE;
 
    for(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
    var common_integer = (mn + mx) / 2;
 
    // Replace all -1 elements
    // with the common integer
    for(i = 0; i < n; i++)
    {
        if (arr[i] == -1)
            arr[i] = common_integer;
    }
 
    var max_diff = 0;
 
    // Calculating the maximum
    // absolute difference
    for(i = 0; i < n - 1; i++)
    {
        var 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
var arr = [ -1, -1, 11, -1, 3, -1 ];
var n = arr.length;
 
// Function call
document.write(maximumAbsolute(arr, n));
 
// This code is contributed by umadevi9616
 
</script>


Output: 

4

 

Time complexity: O(N), The time complexity of the given program is O(n), where n is the size of the input array. This is because the program iterates through the input array twice in two separate for-loops, which have a time complexity of O(n) each.

Auxiliary Space: O(1), The space complexity of the given program is O(1), which means it uses a constant amount of memory regardless of the size of the input array. This is because the program does not create any new data structures that depend on the size of the input array. It only uses a fixed number of integer variables to store the minimum, maximum, and common values, as well as the maximum absolute difference.
 


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Last Updated : 25 Apr, 2023
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