Count of Missing Numbers in a sorted array
Given a sorted array arr[], the task is to calculate the number of missing numbers between the first and last element of the sorted array.
Examples:
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
Naive Approach:
The simplest approach to solve the problem is to iterate through the array and calculate the sum of all the adjacent differences of the elements of the array.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach:
To optimize the above approach, the idea is to observe that the total count of numbers in the range of [arr[0], arr[N – 1]] is given by arr[N-1] – arr[0] + 1. Since the size of the array is N, the count of missing integers in the array is given by arr[N-1] – arr[0] + 1 – N.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach #include <bits/stdc++.h> 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 codepublic 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 |
32
Time Complexity: O(1)
Auxiliary Space: O(1)
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