给定由N个正整数和一个正整数K组成的数组arr [] ,任务是找到最大可能的整数X ,以使任何数组元素和X之间的绝对差最大为K。如果不存在这样的X值,则打印“ -1” 。
例子:
Input: arr[] = {6, 4, 8, 5}, K = 2
Output: 6
Explanation: Considering X to be 6, the absolute difference between every array element and X(= 6) is at most K (= 2), as illustrated below:
- Absolute difference between arr[0](= 6) and X(= 6) = |6 – 6| = 0.
- Absolute difference between arr[1](= 4) and X(= 6) = |4 – 6| = 2.
- Absolute difference between arr[2](= 8) and X(= 6) = |8 – 6| = 2.
- Absolute difference between arr[3](= 5) and X(= 6) = |5 – 6| = 1.
Input: arr[] = {1, 2, 5}, K = 2
Output: 3
方法:可以根据以下观察结果解决给定问题:
- 考虑到数组元素为arr [i] , | arr [i] – X |的值最多为K。
- 如果arr [i]> X ,则X≤(arr [i] – K) 。否则, X≤(arr [i] + K) 。
- 根据以上两个方程, X的最大值必须为arr [i]和K的最小值之和。
请按照以下步骤解决问题:
- 在变量S中找到给定数组arr []的最小元素。
- 将X的最大值视为(S + K) 。
- 遍历给定数组arr [] ,如果arr [i]与X的绝对差值为K ,则将X更新为-1并退出循环。否则,请继续迭代。
- 完成上述步骤后,打印X的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find maximum value
// of X such that |A[i] - X| ≤ K
int maximumNumber(int arr[], int N,
int K)
{
// Stores the smallest array element
int minimum = *min_element(arr,
arr + N);
// Store the possible value of X
int ans = minimum + K;
// Traverse the array A[]
for (int i = 0; i < N; i++) {
// If required criteria is not satisfied
if (abs(arr[i] - ans) > K) {
// Update ans
ans = -1;
break;
}
}
// Print the result
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 5 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
maximumNumber(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find maximum value
// of X such that |A[i] - X| ≤ K
static void maximumNumber(int arr[], int N,
int K)
{
// Stores the smallest array element
int minimum = Arrays.stream(arr).min().getAsInt();
// Store the possible value of X
int ans = minimum + K;
// Traverse the array A[]
for(int i = 0; i < N; i++)
{
// If required criteria is not satisfied
if (Math.abs(arr[i] - ans) > K)
{
// Update ans
ans = -1;
break;
}
}
// Print the result
System.out.print(ans);
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 1, 2, 5 };
int K = 2;
int N = arr.length;
maximumNumber(arr, N, K);
}
}
// This code is contributed by sanjoy_62
Python3
# Python3 program for the above approach
# Function to find maximum value
# of X such that |A[i] - X| ≤ K
def maximumNumber(arr, N, K):
# Stores the smallest array element
minimum = min(arr)
# Store the possible value of X
ans = minimum + K
# Traverse the array A[]
for i in range(N):
# If required criteria is not satisfied
if (abs(arr[i] - ans) > K):
# Update ans
ans = -1
break
# Print the result
print(ans)
# Driver Code
if __name__ == '__main__':
arr = [1, 2, 5]
K = 2
N = len(arr)
maximumNumber(arr, N, K)
# This code is contributed by SURENDRA_GANGWAR
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find maximum value
// of X such that |A[i] - X| ≤ K
static void maximumNumber(int []arr, int N,
int K)
{
// Stores the smallest array element
int mn = 100000000;
for(int i = 0; i < N; i++)
{
if (arr[i] < mn)
mn = arr[i];
}
// Store the possible value of X
int ans = mn + K;
// Traverse the array A[]
for(int i = 0; i < N; i++)
{
// If required criteria is not satisfied
if (Math.Abs(arr[i] - ans) > K)
{
// Update ans
ans = -1;
break;
}
}
// Print the result
Console.Write(ans);
}
// Driver Code
public static void Main()
{
int []arr = { 1, 2, 5 };
int K = 2;
int N = arr.Length;
maximumNumber(arr, N, K);
}
}
// This code is contributed by ipg2016107
Javascript
输出:
3
时间复杂度: O(N)
辅助空间: O(1)