Minimum removals required to make a given array Bitonic

Last Updated : 02 Mar, 2023

Given an array arr[] of size N, the task is to find the minimum number of array elements required to be removed from the array such that the given array is converted to a bitonic array.

Examples:

Input: arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 }
Output: 3
Explanation:
Removing arr[0], arr[1] and arr[5] modifies arr[] to { 1, 5, 6, 3, 1 }
Since the array elements follow an increasing order followed by a decreasing order, the required output is 3.

Input: arr[] = { 1, 3, 1 }
Output: 0
Explanation:
The given array is already a bitonic array. Therefore, the required output is 3.

Approach: The problem can be solved based on the concept of the longest increasing subsequence problem. Follow the steps below to solve the problem:

Initialize a variable, say left[], such that left[i] stores the length of the longest increasing subsequence up to the i^th index.
Initialize a variable, say right[], such that right[i] stores the length of the longest decreasing subsequence over the range [i, N].
Traverse left[] and right[] array using variable i and find the maximum value of (left[i] + right[i] – 1) and store it in a variable, say maxLen.
Finally, print the value of N – maxLen.

Below is the implementation of the above approach:

C++14

   // C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
    // left[i]: Stores the length
    // of LIS up to i-th index
    vector<int> left(N, 1);
 
    // right[i]: Stores the length
    // of decreasing subsequence
    // over the range [i, N]
    vector<int> right(N, 1);
 
    // Calculate the length of LIS
    // up to i-th index
    for (int i = 1; i < N; i++) {
 
        // Traverse the array
        // upto i-th index
        for (int j = 0; j < i; j++) {
 
            // If arr[j] is less than arr[i]
            if (arr[j] < arr[i]) {
 
                // Update left[i]
                left[i] = max(left[i],
                              left[j] + 1);
            }
        }
    }
 
    // Calculate the length of decreasing
    // subsequence over the range [i, N]
    for (int i = N - 2; i >= 0;
         i--) {
 
        // Traverse right[] array
        for (int j = N - 1; j > i;
             j--) {
 
            // If arr[i] is greater
            // than arr[j]
            if (arr[i] > arr[j]) {
 
                // Update right[i]
                right[i] = max(right[i],
                               right[j] + 1);
            }
        }
    }
 
    // Stores length of the
    // longest bitonic array
    int maxLen = 0;
 
    // Traverse left[] and right[] array
    for (int i = 1; i < N - 1; i++) {
 
        // Update maxLen
        maxLen = max(maxLen, left[i] + right[i] - 1);
    }
 
    cout << (N - maxLen) << "\n";
}
 
// Function to print minimum removals
// required to make given array bitonic
void makeBitonic(int arr[], int N)
{
    if (N == 1) {
        cout << "0" << endl;
        return;
    }
 
    if (N == 2) {
        if (arr[0] != arr[1])
            cout << "0" << endl;
        else
            cout << "1" << endl;
 
        return;
    }
 
    min_element_removal(arr, N);
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    makeBitonic(arr, N);
 
    return 0;
}
 
  

C

   // C program to implement
// the above approach
#include <stdio.h>
#define max(a,b) ((a) > (b) ? (a) : (b)) //defining max
 
// Function to count minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
   
  // left[i]: Stores the length
  // of LIS up to i-th index
  int left[N];
  for (int i = 0; i < N; i++)
    left[i] = 1;
 
  // right[i]: Stores the length
  // of decreasing subsequence
  // over the range [i, N]
  int right[N];
  for (int i = 0; i < N; i++)
    right[i] = 1;
 
  // Calculate the length of LIS
  // up to i-th index
  for (int i = 1; i < N; i++) {
 
    // Traverse the array
    // upto i-th index
    for (int j = 0; j < i; j++) {
 
      // If arr[j] is less than arr[i]
      if (arr[j] < arr[i]) {
 
        // Update left[i]
        left[i] = max(left[i], left[j] + 1);
      }
    }
  }
 
  // Calculate the length of decreasing
  // subsequence over the range [i, N]
  for (int i = N - 2; i >= 0;
       i--) {
 
    // Traverse right[] array
    for (int j = N - 1; j > i;
         j--) {
 
      // If arr[i] is greater
      // than arr[j]
      if (arr[i] > arr[j]) {
 
        // Update right[i]
        right[i] = max(right[i],  right[j] + 1);
      }
    }
  }
 
  // Stores length of the
  // longest bitonic array
  int maxLen = 0;
 
  // Traverse left[] and right[] array
  for (int i = 1; i < N - 1; i++) {
 
    // Update maxLen
    maxLen = max(maxLen, left[i] + right[i] - 1);
  }
 
  printf("%d\n", (N - maxLen));
}
 
// Function to print minimum removals
// required to make given array bitonic
void makeBitonic(int arr[], int N)
{
  if (N == 1) {
    printf("0\n");
    return;
  }
 
  if (N == 2) {
    if (arr[0] != arr[1])
      printf("0\n");
    else
      printf("1\n");
 
    return;
  }
 
  min_element_removal(arr, N);
}
 
// Driver Code
int main()
{
  int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
  int N = sizeof(arr) / sizeof(arr[0]);
 
  makeBitonic(arr, N);
 
  return 0;
}
 
// This code is contributed by phalashi.
 
  

Java

   // Java program to implement
// the above approach
class GFG {
     
    // Function to count minimum array elements
    // required to be removed to make an array bitonic
    static void min_element_removal(int arr[], int N)
    {
        // left[i]: Stores the length
        // of LIS up to i-th index
        int left[] = new int[N]; 
         
        for(int i = 0; i < N; i++)
            left[i] = 1;
     
        // right[i]: Stores the length
        // of decreasing subsequence
        // over the range [i, N]
        int right[] = new int[N]; 
         
        for(int i = 0; i < N; i++)
            right[i] = 1;
             
        // Calculate the length of LIS
        // up to i-th index
        for (int i = 1; i < N; i++) {
     
            // Traverse the array
            // upto i-th index
            for (int j = 0; j < i; j++) {
     
                // If arr[j] is less than arr[i]
                if (arr[j] < arr[i]) {
     
                    // Update left[i]
                    left[i] = Math.max(left[i],
                                  left[j] + 1);
                }
            }
        }
     
        // Calculate the length of decreasing
        // subsequence over the range [i, N]
        for (int i = N - 2; i >= 0;
             i--) {
     
            // Traverse right[] array
            for (int j = N - 1; j > i;
                 j--) {
     
                // If arr[i] is greater
                // than arr[j]
                if (arr[i] > arr[j]) {
     
                    // Update right[i]
                    right[i] = Math.max(right[i],
                                   right[j] + 1);
                }
            }
        }
     
        // Stores length of the
        // longest bitonic array
        int maxLen = 0;
     
        // Traverse left[] and right[] array
        for (int i = 1; i < N - 1; i++) {
     
            // Update maxLen
            maxLen = Math.max(maxLen, left[i] + right[i] - 1);
        }
     
        System.out.println(N - maxLen);
    }
     
    // Function to print minimum removals
    // required to make given array bitonic
    static void makeBitonic(int arr[], int N)
    {
        if (N == 1) {
            System.out.println("0");
            return;
        }
     
        if (N == 2) {
            if (arr[0] != arr[1])
                System.out.println("0");
            else
                System.out.println("1");
     
            return;
        }
     
        min_element_removal(arr, N);
    }
     
    // Driver Code
    public static void main (String[] args) {
         
        int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
         
        int N = arr.length;
     
        makeBitonic(arr, N);
    }
}
 
// This code is contributed by AnkitRai01
 
  

Python3

   # Python3 program to implement
# the above approach
 
# Function to count minimum array elements
# required to be removed to make an array bitonic
def min_element_removal(arr, N):
     
    # left[i]: Stores the length
    # of LIS up to i-th index
    left = [1] * N
 
    # right[i]: Stores the length
    # of decreasing subsequence
    # over the range [i, N]
    right = [1] * (N)
 
    #Calculate the length of LIS
    #up to i-th index
    for i in range(1, N):
 
        #Traverse the array
        #upto i-th index
        for j in range(i):
 
            #If arr[j] is less than arr[i]
            if (arr[j] < arr[i]):
 
                #Update left[i]
                left[i] = max(left[i], left[j] + 1)
 
    # Calculate the length of decreasing
    # subsequence over the range [i, N]
    for i in range(N - 2, -1, -1):
 
        # Traverse right[] array
        for j in range(N - 1, i, -1):
 
            # If arr[i] is greater
            # than arr[j]
            if (arr[i] > arr[j]):
 
                # Update right[i]
                right[i] = max(right[i], right[j] + 1)
 
    # Stores length of the
    # longest bitonic array
    maxLen = 0
 
    # Traverse left[] and right[] array
    for i in range(1, N - 1):
 
        # Update maxLen
        maxLen = max(maxLen, left[i] + right[i] - 1)
 
    print((N - maxLen))
 
# Function to print minimum removals
# required to make given array bitonic
def makeBitonic(arr, N):
    if (N == 1):
        print("0")
        return
 
    if (N == 2):
        if (arr[0] != arr[1]):
            print("0")
        else:
            print("1")
 
        return
 
    min_element_removal(arr, N)
 
# Driver Code
if __name__ == '__main__':
    arr=[2, 1, 1, 5, 6, 2, 3, 1]
    N = len(arr)
 
    makeBitonic(arr, N)
 
    # This code is contributed by mohit kumar 29
 
  

C#

   // C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to count minimum array elements
// required to be removed to make an array bitonic
static void min_element_removal(int []arr, int N)
{
     
    // left[i]: Stores the length
    // of LIS up to i-th index
    int []left = new int[N]; 
     
    for(int i = 0; i < N; i++)
        left[i] = 1;
 
    // right[i]: Stores the length
    // of decreasing subsequence
    // over the range [i, N]
    int []right = new int[N]; 
     
    for(int i = 0; i < N; i++)
        right[i] = 1;
         
    // Calculate the length of LIS
    // up to i-th index
    for(int i = 1; i < N; i++)
    {
         
        // Traverse the array
        // upto i-th index
        for(int j = 0; j < i; j++) 
        {
             
            // If arr[j] is less than arr[i]
            if (arr[j] < arr[i]) 
            {
                 
                // Update left[i]
                left[i] = Math.Max(left[i],
                                   left[j] + 1);
            }
        }
    }
 
    // Calculate the length of decreasing
    // subsequence over the range [i, N]
    for(int i = N - 2; i >= 0; i--) 
    {
         
        // Traverse right[] array
        for(int j = N - 1; j > i; j--) 
        {
             
            // If arr[i] is greater
            // than arr[j]
            if (arr[i] > arr[j]) 
            {
                 
                // Update right[i]
                right[i] = Math.Max(right[i],
                                    right[j] + 1);
            }
        }
    }
 
    // Stores length of the
    // longest bitonic array
    int maxLen = 0;
 
    // Traverse left[] and right[] array
    for(int i = 1; i < N - 1; i++)
    {
         
        // Update maxLen
        maxLen = Math.Max(maxLen, left[i] + 
                                 right[i] - 1);
    }
    Console.WriteLine(N - maxLen);
}
 
// Function to print minimum removals
// required to make given array bitonic
static void makeBitonic(int []arr, int N)
{
    if (N == 1) 
    {
        Console.WriteLine("0");
        return;
    }
 
    if (N == 2)
    {
        if (arr[0] != arr[1])
            Console.WriteLine("0");
        else
            Console.WriteLine("1");
             
        return;
    }
    min_element_removal(arr, N);
}
 
// Driver Code
public static void Main(String[] args) 
{
    int []arr = { 2, 1, 1, 5, 6, 2, 3, 1 };
    int N = arr.Length;
 
    makeBitonic(arr, N);
}
}
 
// This code is contributed by AnkThon
 
  

Javascript

   <script>
 
// Javascript program to implement
// the above approach
 
// Function to count minimum array elements
// required to be removed to make an array bitonic
function min_element_removal(arr, N)
{
    // left[i]: Stores the length
    // of LIS up to i-th index
    var left = Array(N).fill(1);
 
    // right[i]: Stores the length
    // of decreasing subsequence
    // over the range [i, N]
    var right = Array(N).fill(1);
 
    // Calculate the length of LIS
    // up to i-th index
    for (var i = 1; i < N; i++) {
 
        // Traverse the array
        // upto i-th index
        for (var j = 0; j < i; j++) {
 
            // If arr[j] is less than arr[i]
            if (arr[j] < arr[i]) {
 
                // Update left[i]
                left[i] = Math.max(left[i],
                              left[j] + 1);
            }
        }
    }
 
    // Calculate the length of decreasing
    // subsequence over the range [i, N]
    for (var i = N - 2; i >= 0;
         i--) {
 
        // Traverse right[] array
        for (var j = N - 1; j > i;
             j--) {
 
            // If arr[i] is greater
            // than arr[j]
            if (arr[i] > arr[j]) {
 
                // Update right[i]
                right[i] = Math.max(right[i],
                               right[j] + 1);
            }
        }
    }
 
    // Stores length of the
    // longest bitonic array
    var maxLen = 0;
 
    // Traverse left[] and right[] array
    for (var i = 1; i < N - 1; i++) {
 
        // Update maxLen
        maxLen = Math.max(maxLen, left[i] + right[i] - 1);
    }
 
    document.write((N - maxLen) + "<br>");
}
 
// Function to print minimum removals
// required to make given array bitonic
function makeBitonic(arr, N)
{
    if (N == 1) {
        document.write( "0" + "<br>");
        return;
    }
 
    if (N == 2) {
        if (arr[0] != arr[1])
            document.write( "0" + "<br>");
        else
            document.write( "1" + "<br>");
 
        return;
    }
 
    min_element_removal(arr, N);
}
 
// Driver Code
var arr = [2, 1, 1, 5, 6, 2, 3, 1];
var N = arr.length;
makeBitonic(arr, N);
 
// This code is contributed by rutvik_56.
</script>
 
  

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

Optimized Approach:

This approach has a time complexity of O(n), where n is the number of elements in the array, since it takes a single pass through the array to find the local maximum and local minimum, and it performs a constant amount of work for each iteration of the loop.

C++

   #include <bits/stdc++.h>
using namespace std;
 
// Function to count minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
    int l = 0, r = N-1, count = 0;
    while(l < r)
    {
        // find the first local maximum from the left side
        while(l < N-1 && arr[l] <= arr[l+1])
            l++;
 
        // find the first local minimum from the right side
        while(r > 0 && arr[r] <= arr[r-1])
            r--;
 
        // if both indices have not crossed each other,
        // remove the elements between them
        if(l < r)
        {
            l++;
            r--;
            count++;
        }
    }
 
    // print the minimum number of removals
    cout << count << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 5,7,1,3,6,2,1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    min_element_removal(arr, N);
 
    return 0;
}
 
  

Java

   import java.util.*;
 
class Main {
    // Function to count minimum array elements
    // required to be removed to make an array bitonic
    public static void min_element_removal(int arr[], int N)
    {
        int l = 0, r = N - 1, count = 0;
        while (l < r) {
            // find the first local maximum from the left
            // side
            while (l < N - 1 && arr[l] <= arr[l + 1])
                l++;
 
            // find the first local minimum from the right
            // side
            while (r > 0 && arr[r] <= arr[r - 1])
                r--;
 
            // if both indices have not crossed each other,
            // remove the elements between them
            if (l < r) {
                l++;
                r--;
                count++;
            }
        }
 
        // print the minimum number of removals
        System.out.println(count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 5, 7, 1, 3, 6, 2, 1 };
        int N = arr.length;
 
        min_element_removal(arr, N);
    }
}
 
  

Python3

   # Function to count minimum array elements
# required to be removed to make an array bitonic
def min_element_removal(arr, N):
    l, r, count = 0, N-1, 0
    while l < r:
       
        # find the first local maximum from the left side
        while l < N-1 and arr[l] <= arr[l+1]:
            l += 1
 
        # find the first local minimum from the right side
        while r > 0 and arr[r] <= arr[r-1]:
            r -= 1
 
        # if both indices have not crossed each other,
        # remove the elements between them
        if l < r:
            l += 1
            r -= 1
            count += 1
 
    # print the minimum number of removals
    print(count)
 
 
# Driver Code
arr = [5, 7, 1, 3, 6, 2, 1]
N = len(arr)
 
min_element_removal(arr, N)
 
  

Javascript

   // Function to count minimum array elements
// required to be removed to make an array bitonic
function min_element_removal(arr, N) {
  let l = 0, r = N-1, count = 0;
  while (l < r) {
       
    // find the first local maximum from the left side
    while (l < N-1 && arr[l] <= arr[l+1]) {
      l++;
    }
 
    // find the first local minimum from the right side
    while (r > 0 && arr[r] <= arr[r-1]) {
      r--;
    }
 
    // if both indices have not crossed each other,
    // remove the elements between them
    if (l < r) {
      l++;
      r--;
      count++;
    }
  }
 
  // print the minimum number of removals
  console.log(count);
}
 
// Driver Code
let arr = [5, 7, 1, 3, 6, 2, 1];
let N = arr.length;
 
min_element_removal(arr, N);
 
  

C#

   using System;
 
class Program
{
    static void Main(string[] args)
    {
        int[] arr = { 5, 7, 1, 3, 6, 2, 1 };
        int N = arr.Length;
 
        int l = 0, r = N - 1, count = 0;
        while (l < r)
        {
            // find the first local maximum from the left side
            while (l < N - 1 && arr[l] <= arr[l + 1])
                l++;
 
            // find the first local minimum from the right side
            while (r > 0 && arr[r] <= arr[r - 1])
                r--;
 
            // if both indices have not crossed each other,
            // remove the elements between them
            if (l < r)
            {
                l++;
                r--;
                count++;
            }
        }
 
        // print the minimum number of removals
        Console.WriteLine(count);
    }
}
 
  

Output

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

Minimum time required to cover a Binary Array

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Minimum removals required to make a given array Bitonic

C++14

C

Java

Python3

C#

Javascript

C++

Java

Python3

Javascript

C#

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