给定一个已排序的数组arr [],任务是计算已排序数组的第一个元素与最后一个元素之间缺少的数字数量。
例子:
Input: arr[] = { 1, 4, 5, 8 }
Output: 4
Explanation:
The missing integers in the array are {2, 3, 6, 7}.
Therefore, the count is 4.
Input: arr[] = {5, 10, 20, 40}
Output: 32
天真的方法:
解决问题的最简单方法是遍历数组并计算数组元素所有相邻差的总和。
时间复杂度: O(N)
辅助空间: O(1)
高效方法:
为了优化上述方法,我们的想法是观察在[arr [0],arr [N – 1]]范围内的总数为arr [N-1] – arr [0] + 1 。由于数组的大小为N ,因此数组中丢失整数的计数由arr [N-1] – arr [0] + 1 – N给出。
下面是上述方法的实现:
C++
// C++ Program for the above approach
#include
using namespace std;
// Function that find the count of
// missing numbers in array a[]
void countMissingNum(int a[], int N)
{
// Calculate the count of missing
// numbers in the array
int count = a[N - 1] - a[0] + 1 - N;
cout << count << endl;
}
// Driver Code
int main()
{
int arr[] = { 5, 10, 20, 40 };
int N = sizeof(arr) / sizeof(arr[0]);
countMissingNum(arr, N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function that find the count of
// missing numbers in array a[]
public static void countMissingNum(int[] a,
int N)
{
// Calculate the count of missing
// numbers in the array
int count = a[N - 1] - a[0] + 1 - N;
System.out.println(count);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 5, 10, 20, 40 };
int N = arr.length;
countMissingNum(arr, N);
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program for the above approach
# Function that find the count of
# missing numbers in array a[]
def countMissingNum(a, N):
# Calculate the count of missing
# numbers in the array
count = a[N - 1] - a[0] + 1 - N
print(count)
# Driver Code
arr = [ 5, 10, 20, 40 ]
N = len(arr)
countMissingNum(arr, N)
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function that find the count of
// missing numbers in array a[]
public static void countMissingNum(int[] a,
int N)
{
// Calculate the count of missing
// numbers in the array
int count = a[N - 1] - a[0] + 1 - N;
Console.Write(count);
}
// Driver code
public static void Main(string[] args)
{
int []arr = { 5, 10, 20, 40 };
int N = arr.Length;
countMissingNum(arr, N);
}
}
// This code is contributed by rutvik_56
输出:
32
时间复杂度: O(1)
辅助空间: O(1)