给定一个数组arr[]和一个数字K ,任务是找到绝对差为K 的两个元素之间的最大距离。如果无法找到任何最大距离,则打印“-1” 。
例子:
Input: arr[] = {3, 5, 1, 4, 2, 2}
Output: 5
Explanation:
The max distance between the two elements (3, 2) at index 0 and 5 is 5.
Input: arr[] = {11, 2, 3, 8, 5, 2}
Output: 3
Explanation:
The max distance between the two elements (3, 2) at index 2 and 5 is 3.
朴素的方法:这个想法是一个一个地选择每个元素(比如arr[i] ),搜索前一个元素(arr[i] – K)和下一个元素(arr[i] + K) 。如果存在两个元素中的任何一个,则找到当前元素与其下一个或前一个元素之间的最大距离。
时间复杂度: O(N 2 )
辅助空间: O(1)
Efficient Approach:优化上面的方法,思路是使用 哈希图。以下是步骤:
- 遍历数组并将第一次出现的索引存储在 HashMap 中。
- 现在,对于数组arr[]中的每个元素,检查arr[i] + K和arr[i] – K是否存在于哈希图中。
- 如果存在,则使用arr[i]更新与两个元素的距离并检查下一个元素。
- 打印上述步骤后得到的最大距离。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function that find maximum distance
// between two elements whose absolute
// difference is K
int maxDistance(int arr[], int K, int N)
{
// To store the first index
map Map;
// Traverse elements and find
// maximum distance between 2
// elements whose absolute
// difference is K
int maxDist = 0;
for(int i = 0; i < N; i++)
{
// If this is first occurrence
// of element then insert
// its index in map
if (Map.find(arr[i]) == Map.end())
Map[arr[i]] = i;
// If next element present in
// map then update max distance
if (Map.find(arr[i] + K) != Map.end())
{
maxDist = max(maxDist,
i - Map[arr[i] + K]);
}
// If previous element present
// in map then update max distance
if (Map.find(arr[i] - K) != Map.end())
{
maxDist = max(maxDist,
i - Map[arr[i] - K]);
}
}
// Return the answer
if (maxDist == 0)
return -1;
else
return maxDist;
}
// Driver code
int main()
{
// Given array arr[]
int arr[] = { 11, 2, 3, 8, 5, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
// Given difference K
int K = 2;
// Function call
cout << maxDistance(arr, K, N);
return 0;
}
// This code is contributed by divyeshrabadiya07
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function that find maximum distance
// between two elements whose absolute
// difference is K
static int maxDistance(int[] arr, int K)
{
// To store the first index
Map map
= new HashMap<>();
// Traverse elements and find
// maximum distance between 2
// elements whose absolute
// difference is K
int maxDist = 0;
for (int i = 0; i < arr.length; i++) {
// If this is first occurrence
// of element then insert
// its index in map
if (!map.containsKey(arr[i]))
map.put(arr[i], i);
// If next element present in
// map then update max distance
if (map.containsKey(arr[i] + K)) {
maxDist
= Math.max(
maxDist,
i - map.get(arr[i] + K));
}
// If previous element present
// in map then update max distance
if (map.containsKey(arr[i] - K)) {
maxDist
= Math.max(maxDist,
i - map.get(arr[i] - K));
}
}
// Return the answer
if (maxDist == 0)
return -1;
else
return maxDist;
}
// Driver Code
public static void main(String args[])
{
// Given array arr[]
int[] arr = { 11, 2, 3, 8, 5, 2 };
// Given difference K
int K = 2;
// Function call
System.out.println(
maxDistance(arr, K));
}
}
Python3
# Python3 program for the above approach
# Function that find maximum distance
# between two elements whose absolute
# difference is K
def maxDistance(arr, K):
# To store the first index
map = {}
# Traverse elements and find
# maximum distance between 2
# elements whose absolute
# difference is K
maxDist = 0
for i in range(len(arr)):
# If this is first occurrence
# of element then insert
# its index in map
if not arr[i] in map:
map[arr[i]] = i
# If next element present in
# map then update max distance
if arr[i] + K in map:
maxDist = max(maxDist,
i - map[arr[i] + K])
# If previous element present
# in map then update max distance
if arr[i] - K in map:
maxDist = max(maxDist,
i - map[arr[i] - K])
# Return the answer
if maxDist == 0:
return -1
else:
return maxDist
# Driver code
# Given array arr[]
arr = [ 11, 2, 3, 8, 5, 2 ]
# Given difference K
K = 2
# Function call
print(maxDistance(arr,K))
# This code is contributed by Stuti Pathak
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that find maximum distance
// between two elements whose absolute
// difference is K
static int maxDistance(int[] arr, int K)
{
// To store the first index
Dictionary map = new Dictionary();
// Traverse elements and find
// maximum distance between 2
// elements whose absolute
// difference is K
int maxDist = 0;
for (int i = 0; i < arr.Length; i++)
{
// If this is first occurrence
// of element then insert
// its index in map
if (!map.ContainsKey(arr[i]))
map.Add(arr[i], i);
// If next element present in
// map then update max distance
if (map.ContainsKey(arr[i] + K))
{
maxDist = Math.Max(maxDist,
i - map[arr[i] + K]);
}
// If previous element present
// in map then update max distance
if (map.ContainsKey(arr[i] - K))
{
maxDist = Math.Max(maxDist,
i - map[arr[i] - K]);
}
}
// Return the answer
if (maxDist == 0)
return -1;
else
return maxDist;
}
// Driver Code
public static void Main(String []args)
{
// Given array []arr
int[] arr = { 11, 2, 3, 8, 5, 2 };
// Given difference K
int K = 2;
// Function call
Console.WriteLine(
maxDistance(arr, K));
}
}
// This code is contributed by Princi Singh
Javascript
输出:
2
时间复杂度: O(N)
辅助空间: O(N)
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