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Length of smallest subarray required to be removed to make remaining elements consecutive
• Last Updated : 04 May, 2021

Given an array arr[] consisting of N integers, the task is to find the length of the smallest subarray required to be removed to make the remaining array elements consecutive.

Examples:

Input: arr[] = {1, 2, 3, 7, 5, 4, 5}
Output: 2
Explanation:
Removing the subarray {7, 5} from the array arr[] modifies the array to {1, 2, 3, 4, 5}, which makes all array elements consecutive. Therefore, the length of the subarray removed is 2, which is minimum.

Input: arr[] = {4, 5, 6, 8, 9, 10}
Output: 3

Naive Approach: The simplest approach to solve the given problem is to remove generate all possible subarrays of the array arr[] and for each of them, check if their removal makes the remaining array elements consecutive or not. After checking for all the subarrays, print the length of the minimum subarray obtained that satisfies the condition.

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by storing the length of the longest prefix and suffix of consecutive elements and then find the minimum length of the subarray required to be removed such that the concatenation of the prefix and suffix forms a sequence of consecutive elements.
Follow the below steps to solve the problem:

• Initialize two variables, say L as 0 and R as (N – 1) to store the ending indices of the longest prefix and starting index of the longest suffix of consecutive elements respectively.
• Update the value of L to the first index where arr[i] + 1 is not equal to arr[i + 1] such that arr[0, …, L] is a consecutive prefix array.
• Update the value of R to the first index from the end where arr[i] is not equal to arr[i – 1] + 1 such that arr[R, …, N – 1] is a consecutive suffix array.
• Initialize a variable, say ans, to store the minimum of (N – L – 1) and R to store the required result.
• If the value of arr[R] ≤ arr[L] + 1, then store the right index, R1 as arr[0, …, L, R1, …, N – 1] is a consecutive array.
• If the value of (R1 – L – 1) is less than the value of ans, then update the value of ans to (R1 – L – 1).
• After completing the above steps, print the value of the ans as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the length of the``// smallest subarray to be removed to``// make remaining array elements consecutive``void` `shortestSubarray(``int``* A, ``int` `N)``{``    ``int` `i;` `    ``// Store the ending index of the``    ``// longest prefix consecutive array``    ``int` `left_index;` `    ``// Traverse the array to find the``    ``// longest prefix consecutive sequence``    ``for` `(i = 0; i < N - 1; i++) {``        ``if` `(A[i] + 1 != A[i + 1])``            ``break``;``    ``}` `    ``// A[0...left_index] is the``    ``// prefix consecutive sequence``    ``left_index = i;` `    ``// Store the starting index of the``    ``// longest suffix consecutive sequence``    ``int` `right_index;` `    ``// Traverse the array to find the``    ``// longest suffix consecutive sequence``    ``for` `(i = N - 1; i >= 1; i--) {``        ``if` `(A[i] != A[i - 1] + 1)``            ``break``;``    ``}` `    ``// A[right_index...N-1] is``    ``// the consecutive sequence``    ``right_index = i;` `    ``int` `updated_right;` `    ``// Store the smallest subarray``    ``// required to be removed``    ``int` `minLength = min(N - left_index - 1,``                        ``right_index);` `    ``// Check if subarray from the``    ``// middle can be removed``    ``if` `(A[right_index]``        ``<= A[left_index] + 1) {` `        ``// Update the right index s.t.``        ``// A[0, N-1] is consecutive``        ``updated_right = right_index``                        ``+ A[left_index]``                        ``- A[right_index] + 1;` `        ``// If updated_right < N, then``        ``// update the minimumLength``        ``if` `(updated_right < N)``            ``minLength = min(minLength,``                            ``updated_right``                                ``- left_index - 1);``    ``}` `    ``// Print the required result``    ``cout << minLength;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 7, 4, 3, 5 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``shortestSubarray(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{` `// Function to find the length of the``// smallest subarray to be removed to``// make remaining array elements consecutive``static` `void` `shortestSubarray(``int` `A[], ``int` `N)``{``    ``int` `i;` `    ``// Store the ending index of the``    ``// longest prefix consecutive array``    ``int` `left_index;` `    ``// Traverse the array to find the``    ``// longest prefix consecutive sequence``    ``for``(i = ``0``; i < N - ``1``; i++)``    ``{``        ``if` `(A[i] + ``1` `!= A[i + ``1``])``            ``break``;``    ``}` `    ``// A[0...left_index] is the``    ``// prefix consecutive sequence``    ``left_index = i;` `    ``// Store the starting index of the``    ``// longest suffix consecutive sequence``    ``int` `right_index;` `    ``// Traverse the array to find the``    ``// longest suffix consecutive sequence``    ``for``(i = N - ``1``; i >= ``1``; i--)``    ``{``        ``if` `(A[i] != A[i - ``1``] + ``1``)``            ``break``;``    ``}` `    ``// A[right_index...N-1] is``    ``// the consecutive sequence``    ``right_index = i;` `    ``int` `updated_right;` `    ``// Store the smallest subarray``    ``// required to be removed``    ``int` `minLength = Math.min(N - left_index - ``1``,``                             ``right_index);` `    ``// Check if subarray from the``    ``// middle can be removed``    ``if` `(A[right_index] <= A[left_index] + ``1``)``    ``{``        ` `        ``// Update the right index s.t.``        ``// A[0, N-1] is consecutive``        ``updated_right = right_index + A[left_index] -``                     ``A[right_index] + ``1``;` `        ``// If updated_right < N, then``        ``// update the minimumLength``        ``if` `(updated_right < N)``            ``minLength = Math.min(minLength,``                                 ``updated_right -``                                 ``left_index - ``1``);``    ``}` `    ``// Print the required result``    ``System.out.println(minLength);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``7``, ``4``, ``3``, ``5` `};``    ``int` `N = arr.length;``    ` `    ``shortestSubarray(arr, N);``}``}` `// This code is contributed by Kingash`

## Python3

 `# Python3 program for the above approach` `# Function to find the length of the``# smallest subarray to be removed to``# make remaining array elements consecutive``def` `shortestSubarray(A, N):``    ` `    ``i ``=` `0` `    ``# Store the ending index of the``    ``# longest prefix consecutive array``    ``left_index ``=` `0` `    ``# Traverse the array to find the``    ``# longest prefix consecutive sequence``    ``for` `i ``in` `range``(N ``-` `1``):``        ``if` `(A[i] ``+` `1` `!``=` `A[i ``+` `1``]):``            ``break` `    ``# A[0...left_index] is the``    ``# prefix consecutive sequence``    ``left_index ``=` `i` `    ``# Store the starting index of the``    ``# longest suffix consecutive sequence``    ``right_index ``=` `0` `    ``# Traverse the array to find the``    ``# longest suffix consecutive sequence``    ``i ``=` `N ``-` `1``    ` `    ``while` `(i >``=` `1``):``        ``if` `(A[i] !``=` `A[i ``-` `1``] ``+` `1``):``            ``break``        ` `        ``i ``-``=` `1` `    ``# A[right_index...N-1] is``    ``# the consecutive sequence``    ``right_index ``=` `i` `    ``updated_right ``=` `0` `    ``# Store the smallest subarray``    ``# required to be removed``    ``minLength ``=` `min``(N ``-` `left_index ``-` `1``, right_index)` `    ``# Check if subarray from the``    ``# middle can be removed``    ``if` `(A[right_index] <``=` `A[left_index] ``+` `1``):``        ` `        ``# Update the right index s.t.``        ``# A[0, N-1] is consecutive``        ``updated_right ``=` `(right_index ``+` `A[left_index] ``-``                                       ``A[right_index] ``+` `1``)` `        ``# If updated_right < N, then``        ``# update the minimumLength``        ``if` `(updated_right < N):``            ``minLength ``=` `min``(minLength, updated_right ``-``                                       ``left_index ``-` `1``)` `    ``# Print the required result``    ``print``(minLength)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``7``, ``4``, ``3``, ``5` `]``    ``N ``=` `len``(arr)``    ` `    ``shortestSubarray(arr, N)` `# This code is contributed by SURENDRA_GANGWAR`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to find the length of the``// smallest subarray to be removed to``// make remaining array elements consecutive``static` `void` `shortestSubarray(``int``[] A, ``int` `N)``{``    ``int` `i;` `    ``// Store the ending index of the``    ``// longest prefix consecutive array``    ``int` `left_index;` `    ``// Traverse the array to find the``    ``// longest prefix consecutive sequence``    ``for``(i = 0; i < N - 1; i++)``    ``{``        ``if` `(A[i] + 1 != A[i + 1])``            ``break``;``    ``}` `    ``// A[0...left_index] is the``    ``// prefix consecutive sequence``    ``left_index = i;` `    ``// Store the starting index of the``    ``// longest suffix consecutive sequence``    ``int` `right_index;` `    ``// Traverse the array to find the``    ``// longest suffix consecutive sequence``    ``for``(i = N - 1; i >= 1; i--)``    ``{``        ``if` `(A[i] != A[i - 1] + 1)``            ``break``;``    ``}` `    ``// A[right_index...N-1] is``    ``// the consecutive sequence``    ``right_index = i;` `    ``int` `updated_right;` `    ``// Store the smallest subarray``    ``// required to be removed``    ``int` `minLength = Math.Min(N - left_index - 1,``                             ``right_index);` `    ``// Check if subarray from the``    ``// middle can be removed``    ``if` `(A[right_index] <= A[left_index] + 1)``    ``{``        ` `        ``// Update the right index s.t.``        ``// A[0, N-1] is consecutive``        ``updated_right = right_index + A[left_index] -``                     ``A[right_index] + 1;` `        ``// If updated_right < N, then``        ``// update the minimumLength``        ``if` `(updated_right < N)``            ``minLength = Math.Min(minLength,``                                 ``updated_right -``                                 ``left_index - 1);``    ``}` `    ``// Print the required result``    ``Console.WriteLine(minLength);``}` `// Driver code``static` `public` `void` `Main()``{``    ``int``[] arr = { 1, 2, 3, 7, 4, 3, 5 };``    ``int` `N = arr.Length;``    ` `    ``shortestSubarray(arr, N);``}``}` `// This code is contributed by offbeat`

## Javascript

 ``
Output:
`4`

Time Complexity: O(N)
Auxiliary Space: O(1)

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