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📜  排序数组中缺失数字的计数

📅  最后修改于: 2021-09-07 02:52:10             🧑  作者: Mango

给定一个排序数组arr[],任务是计算排序数组的第一个和最后一个元素之间缺失数字的数量。

例子:

天真的方法:
解决问题最简单的方法是遍历数组,计算数组元素的所有相邻差值之和。

时间复杂度: O(N)
辅助空间: O(1)

有效的方法:
为了优化上述方法,我们的想法是观察[arr[0], arr[N – 1]]范围内的数字总数由arr[N-1] – arr[0] + 1 给出.由于数组的大小为N ,数组中缺失整数的计数由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)