给定一些正整数的数组arr [] ,并且缺少由-1表示的特定整数的出现,任务是要找到该缺失数,以使相邻元素之间的最大绝对差最小。
例子:
Input: arr[] = {-1, 10, -1, 12, -1}
Output: 11
Explanation:
Differences of Adjacent elements –
=> a[0] – a[1] = 11 – 10 = 1
=> a[2] – a[1] = 11 – 10 = 1
=> a[3] – a[2] = 12 – 11 = 1
=> a[3] – a[4] = 12 – 11 = 1
Maximum absolute difference of adjacent elements – 2
Input: arr[] = {1,-1, 7, 5, 2, -1, 5}
Output: 4
Explanation:
Differences of Adjacent elements –
=> a[1] – a[0] = 4 – 1 = 3
=> a[2] – a[1] = 7 – 4 = 3
=> a[2] – a[3] = 7 – 5 = 2
=> a[3] – a[4] = 5 – 2 = 3
=> a[5] – a[4] = 4 – 2 = 2
=> a[6] – a[5] = 5 – 4 = 1
Maximum absolute difference of adjacent elements – 3
方法:想法是找到缺失数的最大和最小相邻元素,缺失数可以是这两个值的平均值,以使最大绝对差最小。
=> max = Maximum adjacent element
=> min = Minimum adjacent element
Missing number = (max + min)
-------------
2
下面是上述方法的实现:
C++
// C++ implementation of the missing
// number such that maximum absolute
// difference between adjacent element
// is minimum
#include
using namespace std;
// Function to find the missing
// number such that maximum
// absolute difference is minimum
int missingnumber(int n, int arr[])
{
int mn = INT_MAX, mx = INT_MIN;
// Loop to find the maximum and
// minimum adjacent element to
// missing number
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]);
}
}
long long int res = (mx + mn) / 2;
return res;
}
// Driver Code
int main()
{
int n = 5;
int arr[5] = { -1, 10, -1,
12, -1 };
int ans = 0;
// Function Call
int res = missingnumber(n, arr);
cout << res;
return 0;
}
Java
// Java implementation of the missing
// number such that maximum absolute
// difference between adjacent element
// is minimum
import java.util.*;
class GFG{
// Function to find the missing
// number such that maximum
// absolute difference is minimum
static int missingnumber(int n, int arr[])
{
int mn = Integer.MAX_VALUE,
mx = Integer.MIN_VALUE;
// Loop to find the maximum and
// minimum adjacent element to
// missing number
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 res = (mx + mn) / 2;
return res;
}
// Driver Code
public static void main(String[] args)
{
int n = 5;
int arr[] = { -1, 10, -1,
12, -1 };
// Function Call
int res = missingnumber(n, arr);
System.out.print(res);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the missing
# number such that maximum absolute
# difference between adjacent element
# is minimum
import sys
# Function to find the missing
# number such that maximum
# absolute difference is minimum
def missingnumber(n, arr) -> int:
mn = sys.maxsize;
mx = -sys.maxsize - 1;
# Loop to find the maximum and
# minimum adjacent element to
# missing number
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]);
res = (mx + mn) / 2;
return res;
# Driver Code
if __name__ == '__main__':
n = 5;
arr = [ -1, 10, -1, 12, -1 ];
# Function call
res = missingnumber(n, arr);
print(res);
# This code is contributed by amal kumar choubey
C#
// C# implementation of the missing
// number such that maximum absolute
// difference between adjacent element
// is minimum
using System;
class GFG{
// Function to find the missing
// number such that maximum
// absolute difference is minimum
static int missingnumber(int n, int []arr)
{
int mn = Int32.MaxValue,
mx = Int32.MinValue;
// Loop to find the maximum and
// minimum adjacent element to
// missing number
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 res = (mx + mn) / 2;
return res;
}
// Driver Code
public static void Main()
{
int n = 5;
int []arr = new int[]{ -1, 10, -1, 12, -1 };
// Function Call
int res = missingnumber(n, arr);
Console.WriteLine(res);
}
}
// This code is contributed by Nidhi_biet
11
时间复杂度: O(N)
辅助空间: O(1)
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