数组中不同元素之间的最大绝对差
原文：https://www.geeksforgeeks.org/maximum-absolute-difference-between-distinct-elements-in-an-array/
给定N个整数的数组arr[]，任务是找到数组不同元素之间的最大绝对差。
示例：
输入：
arr[] = {12, 10, 9, 45, 2, 10, 10, 45, 10}输出：10
说明：
给定数组的不同元素是 12、9、2。
因此，它们之间的最大绝对差为
12 – 2 = 10。输入：
arr[] = {2, -1, 10, 3, -2, -1, 10}输出：5
说明：
给定数组的不同元素是 2、3，-2。
因此，它们之间的最大绝对差为
3 – (-2)= 5。
朴素的方法：朴素的方法是将不同元素存储在数组temp[]中，并打印出数组temp[]的最大和最小元素差。
时间复杂度：O(N ^ 2)。
辅助空间：O(n)。
有效方法：可以使用散列来优化上述幼稚方法。 步骤如下：
-
将数组
arr[]的每个元素的频率存储在HashMap中。 -
现在，使用上面创建的
HashMap找到频率为1的数组的最大值和最小值。 -
打印上一步中获得的最大值和最小值之间的差。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum
// absolute difference between
// distinct elements in arr[]
int MaxAbsDiff(int arr[], int n)
{
// Map to store each element
// with their occurrence in array
unordered_map<int, int> map;
// maxElement and minElement to
// store maximum and minimum
// distinct element in arr[]
int maxElement = INT_MIN;
int minElement = INT_MAX;
// Traverse arr[] and update each
// element frequency in Map
for(int i = 0; i < n; i++)
{
map[arr[i]]++;
}
// Traverse Map and check if
// value of any key appears 1
// then update maxElement and
// minElement by that key
for(auto itr = map.begin();
itr != map.end(); itr++)
{
if (itr -> second == 1)
{
maxElement = max(maxElement,
itr -> first);
minElement = min(minElement,
itr -> first);
}
}
// Return absolute difference of
// maxElement and minElement
return abs(maxElement - minElement);
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 12, 10, 9, 45, 2,
10, 10, 45, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << MaxAbsDiff(arr, n) << "\n";
return 0;
}
// This code is contributed by akhilsaini
Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the maximum
// absolute difference between
// distinct elements in arr[]
static int MaxAbsDiff(int[] arr, int n)
{
// HashMap to store each element
// with their occurrence in array
Map<Integer, Integer> map
= new HashMap<>();
// maxElement and minElement to
// store maximum and minimum
// distinct element in arr[]
int maxElement = Integer.MIN_VALUE;
int minElement = Integer.MAX_VALUE;
// Traverse arr[] and update each
// element frequency in HashMap
for (int i = 0; i < n; i++) {
map.put(arr[i],
map.getOrDefault(arr[i], 0)
+ 1);
}
// Traverse HashMap and check if
// value of any key appears 1
// then update maxElement and
// minElement by that key
for (Map.Entry<Integer, Integer> k :
map.entrySet()) {
if (k.getValue() == 1) {
maxElement
= Math.max(maxElement,
k.getKey());
minElement
= Math.min(minElement,
k.getKey());
}
}
// Return absolute difference of
// maxElement and minElement
return Math.abs(maxElement
- minElement);
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 12, 10, 9, 45, 2,
10, 10, 45, 10 };
int n = arr.length;
// Function Call
System.out.println(MaxAbsDiff(arr, n));
}
}
Python3
# Python3 program for
# the above approach
import sys
from collections import defaultdict
# Function to find the maximum
# absolute difference between
# distinct elements in arr[]
def MaxAbsDiff(arr, n):
# HashMap to store each element
# with their occurrence in array
map = defaultdict (int)
# maxElement and minElement to
# store maximum and minimum
# distinct element in arr[]
maxElement = -sys.maxsize - 1
minElement = sys.maxsize
# Traverse arr[] and update each
# element frequency in HashMap
for i in range (n):
map[arr[i]] += 1
# Traverse HashMap and check if
# value of any key appears 1
# then update maxElement and
# minElement by that key
for k in map:
if (map[k] == 1):
maxElement = max(maxElement, k)
minElement = min(minElement, k)
# Return absolute difference of
# maxElement and minElement
return abs(maxElement - minElement)
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [12, 10, 9, 45, 2,
10, 10, 45, 10]
n = len( arr)
# Function Call
print(MaxAbsDiff(arr, n))
# This code is contributed by Chitranayal
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum
// absolute difference between
// distinct elements in []arr
static int MaxAbsDiff(int[] arr, int n)
{
// Dictionary to store each element
// with their occurrence in array
Dictionary<int,
int> map = new Dictionary<int,
int>();
// maxElement and minElement to
// store maximum and minimum
// distinct element in []arr
int maxElement = int.MinValue;
int minElement = int.MaxValue;
// Traverse []arr and update each
// element frequency in Dictionary
for(int i = 0; i < n; i++)
{
if(map.ContainsKey(arr[i]))
map[arr[i]] = map[arr[i]] + 1;
else
map.Add(arr[i], 1);
}
// Traverse Dictionary and check if
// value of any key appears 1
// then update maxElement and
// minElement by that key
foreach (KeyValuePair<int, int> k in map)
{
if (k.Value == 1)
{
maxElement = Math.Max(maxElement,
k.Key);
minElement = Math.Min(minElement,
k.Key);
}
}
// Return absolute difference of
// maxElement and minElement
return Math.Abs(maxElement - minElement);
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int[] arr = { 12, 10, 9, 45, 2,
10, 10, 45, 10 };
int n = arr.Length;
// Function call
Console.WriteLine(MaxAbsDiff(arr, n));
}
}
// This code is contributed by Princi Singh
输出：
10
时间复杂度：O(n)；
辅助空间：O(n)。
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