Minimize maximum difference between adjacent elements possible by removing a single array element
Given an sorted array arr[] consisting of N elements, the task is to find the minimum of all maximum differences between adjacent elements of all arrays obtained by removal of any single array element.
Examples:
Input: arr[ ] = { 1, 3, 7, 8}
Output: 5
Explanation:
All possible arrays after removing a single element are as follows:
{3, 7, 8}: Difference between adjacent elements are { 4, 1}. Maximum = 4.
{ 1, 7, 8}: Difference between adjacent elements are { 6, 1}. Maximum = 6.
{ 1, 3, 8}: Difference between adjacent elements are { 2, 5}. Maximum = 5.
Finally, minimum of (4, 6, 5) is 4, which is the required output.
Input: arr[ ] = { 1, 2, 3, 4, 5}
Output: 1
Explanation:
All possible arrays after removing a single element are as follows:
{ 2, 3, 4, 5}: Difference between adjacent elements are { 1, 1, 1}. Maximum = 1.
{ 1, 3, 4, 5}: Difference between adjacent elements are { 2, 1, 1}. Maximum = 2.
{ 1, 2, 4, 5}: Difference between adjacent elements are { 1, 2, 1}. Maximum = 2.
{ 1, 2, 3, 5}: Difference between adjacent elements are { 1, 1, 2}. Maximum = 2.
Finally, minimum of (1, 2, 2, 2) is 1, which is the required output.
Approach: Follow the steps to solve the problem
- Declare a variable MinValue = INT_MAX to store the final answer.
- Traverse the array, for i in range [0, N – 1]
- Declare a vector new_arr which is a copy of arr[] except element arr[i]
- Store the maximum adjacent difference of new_arr in a variable diff
- Update MinValue = min(MinValue, diff)
- Return MinValue as the final answer.
Below is the implementation of above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int maxAdjacentDifference(vector< int > A)
{
int diff = 0;
for ( int i = 1; i < ( int )A.size(); i++) {
diff = max(diff, A[i] - A[i - 1]);
}
return diff;
}
int MinimumValue( int arr[], int N)
{
int MinValue = INT_MAX;
for ( int i = 0; i < N; i++) {
vector< int > new_arr;
for ( int j = 0; j < N; j++) {
if (i == j)
continue ;
new_arr.push_back(arr[j]);
}
MinValue
= min(MinValue,
maxAdjacentDifference(new_arr));
}
return MinValue;
}
int main()
{
int arr[] = { 1, 3, 7, 8 };
int N = sizeof (arr) / sizeof ( int );
cout << MinimumValue(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int maxAdjacentDifference(ArrayList<Integer> A)
{
int diff = 0 ;
for ( int i = 1 ; i < ( int )A.size(); i++)
{
diff = Math.max(diff, A.get(i) - A.get(i - 1 ));
}
return diff;
}
static int MinimumValue( int arr[], int N)
{
int MinValue = Integer.MAX_VALUE;
for ( int i = 0 ; i < N; i++) {
ArrayList<Integer> new_arr= new ArrayList<>();
for ( int j = 0 ; j < N; j++) {
if (i == j)
continue ;
new_arr.add(arr[j]);
}
MinValue
= Math.min(MinValue,
maxAdjacentDifference(new_arr));
}
return MinValue;
}
public static void main (String[] args) {
int arr[] = { 1 , 3 , 7 , 8 };
int N = arr.length;
System.out.print(MinimumValue(arr, N));
}
}
|
Python3
import sys
def maxAdjacentDifference(A):
diff = 0
for i in range ( 1 , len (A), 1 ):
diff = max (diff, A[i] - A[i - 1 ])
return diff
def MinimumValue(arr, N):
MinValue = sys.maxsize
for i in range (N):
new_arr = []
for j in range (N):
if (i = = j):
continue
new_arr.append(arr[j])
MinValue = min (MinValue,
maxAdjacentDifference(new_arr))
return MinValue
if __name__ = = '__main__' :
arr = [ 1 , 3 , 7 , 8 ]
N = len (arr)
print (MinimumValue(arr, N))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int maxAdjacentDifference(List< int > A)
{
int diff = 0;
for ( int i = 1; i < A.Count; i++)
{
diff = Math.Max(diff, A[i] - A[i - 1]);
}
return diff;
}
static int MinimumValue( int [] arr, int N)
{
int MinValue = Int32.MaxValue;
for ( int i = 0; i < N; i++)
{
List< int > new_arr = new List< int >();
for ( int j = 0; j < N; j++)
{
if (i == j)
continue ;
new_arr.Add(arr[j]);
}
MinValue = Math.Min(MinValue,
maxAdjacentDifference(new_arr));
}
return MinValue;
}
public static void Main( string [] args)
{
int [] arr = { 1, 3, 7, 8 };
int N = arr.Length;
Console.WriteLine(MinimumValue(arr, N));
}
}
|
Javascript
<script>
function maxAdjacentDifference(A)
{
let diff = 0;
for (let i = 1; i < A.length; i++)
{
diff = Math.max(diff, A[i] - A[i-1]);
}
return diff;
}
function MinimumValue(arr, N)
{
let MinValue = Number.MAX_VALUE;
for (let i = 0; i < N; i++) {
let new_arr=[];
for (let j = 0; j < N; j++) {
if (i == j)
continue ;
new_arr.push(arr[j]);
}
MinValue
= Math.min(MinValue,
maxAdjacentDifference(new_arr));
}
return MinValue;
}
let arr = [ 1, 3, 7, 8 ];
let N = arr.length;
document.write(MinimumValue(arr, N));
</script>
|
Time Complexity : O(N2)
Auxiliary Space: O(N)
Last Updated :
05 May, 2021
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