Minimum value of maximum absolute difference of all adjacent pairs in an Array
Last Updated :
25 Apr, 2023
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:
- Consider only those non-missing elements that are adjacent to at least one missing element.
- Find the maximum element and the minimum element among them.
- We need to find a value that minimizes the maximum absolute difference between the common value and these values.
- The optimal value is equals to
(minimum element + maximum element) / 2
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maximumAbsolute( int arr[], int n)
{
int mn = INT_MAX;
int mx = INT_MIN;
for ( int i = 0; i < n; i++) {
if (i > 0
&& arr[i] == -1
&& arr[i - 1] != -1) {
mn = min(mn, arr[i - 1]);
mx = max(mx, arr[i - 1]);
}
if (i < n - 1
&& arr[i] == -1
&& arr[i + 1] != -1) {
mn = min(mn, arr[i + 1]);
mx = max(mx, arr[i + 1]);
}
}
int common_integer = (mn + mx) / 2;
for ( int i = 0; i < n; i++) {
if (arr[i] == -1)
arr[i] = common_integer;
}
int max_diff = 0;
for ( int i = 0; i < n - 1; i++) {
int diff = abs (arr[i] - arr[i + 1]);
if (diff > max_diff)
max_diff = diff;
}
return max_diff;
}
int main()
{
int arr[] = { -1, -1, 11, -1, 3, -1 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maximumAbsolute(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int maximumAbsolute( int arr[], int n)
{
int mn = Integer.MAX_VALUE;
int mx = Integer.MIN_VALUE;
for ( int i = 0 ; i < n; i++) {
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 (i < n - 1
&& arr[i] == - 1
&& arr[i + 1 ] != - 1 ) {
mn = Math.min(mn, arr[i + 1 ]);
mx = Math.max(mx, arr[i + 1 ]);
}
}
int common_integer = (mn + mx) / 2 ;
for ( int i = 0 ; i < n; i++) {
if (arr[i] == - 1 )
arr[i] = common_integer;
}
int max_diff = 0 ;
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 max_diff;
}
public static void main(String[] args)
{
int arr[] = { - 1 , - 1 , 11 , - 1 , 3 , - 1 };
int n = arr.length;
System.out.print(maximumAbsolute(arr, n));
}
}
|
Python3
def maximumAbsolute(arr, n):
mn = 10 * * 9
mx = - 10 * * 9
for i in range (n):
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 (i < n - 1
and arr[i] = = - 1
and arr[i + 1 ] ! = - 1 ):
mn = min (mn, arr[i + 1 ])
mx = max (mx, arr[i + 1 ])
common_integer = (mn + mx) / / 2
for i in range (n):
if (arr[i] = = - 1 ):
arr[i] = common_integer
max_diff = 0
for i in range (n - 1 ):
diff = abs (arr[i] - arr[i + 1 ])
if (diff > max_diff):
max_diff = diff
return max_diff
if __name__ = = '__main__' :
arr = [ - 1 , - 1 , 11 , - 1 , 3 , - 1 ]
n = len (arr)
print (maximumAbsolute(arr, n))
|
C#
using System;
class GFG{
static int maximumAbsolute( int []arr, int n)
{
int mn = int .MaxValue;
int mx = int .MinValue;
for ( int i = 0; i < n; i++) {
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 (i < n - 1
&& arr[i] == -1
&& arr[i + 1] != -1) {
mn = Math.Min(mn, arr[i + 1]);
mx = Math.Max(mx, arr[i + 1]);
}
}
int common_integer = (mn + mx) / 2;
for ( int i = 0; i < n; i++) {
if (arr[i] == -1)
arr[i] = common_integer;
}
int max_diff = 0;
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 max_diff;
}
public static void Main( string [] args)
{
int []arr = { -1, -1, 11, -1, 3, -1 };
int n = arr.Length;
Console.Write(maximumAbsolute(arr, n));
}
}
|
Javascript
<script>
function maximumAbsolute(arr, n)
{
var mn = Number.MAX_VALUE;
var mx = Number.MIN_VALUE;
for (i = 0; i < n; i++)
{
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 (i < n - 1 && arr[i] == -1 &&
arr[i + 1] != -1)
{
mn = Math.min(mn, arr[i + 1]);
mx = Math.max(mx, arr[i + 1]);
}
}
var common_integer = (mn + mx) / 2;
for (i = 0; i < n; i++)
{
if (arr[i] == -1)
arr[i] = common_integer;
}
var max_diff = 0;
for (i = 0; i < n - 1; i++)
{
var diff = Math.abs(arr[i] - arr[i + 1]);
if (diff > max_diff)
max_diff = diff;
}
return max_diff;
}
var arr = [ -1, -1, 11, -1, 3, -1 ];
var n = arr.length;
document.write(maximumAbsolute(arr, n));
</script>
|
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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