Split array into maximum subarrays such that every distinct element lies in a single subarray

Given an array, arr[] of size N, the task is to split the array into the maximum number of subarrays such that the first and the last occurrence of all distinct array element lies in a single subarray.

Examples:

Input: arr[] = {1, 1, 2, 2}
Output: 2
Explanation:
Split the array into subarrays {1, 1} and {2, 2}.
Therefore, the required output is 2.

Input: arr[] = {1, 2, 4, 1, 4, 7, 7, 8}
Output: 3
Explanation:
Split the array into subarrays {1, 2, 4, 1, 4}, {7, 7} and {8}.
Therefore, the required output is 3.

Approach: The idea is to use Hashing to store the index of the last occurrence of every array element. Follow the steps below to solve the problem:

1. Initialize an array, say hash[] to store the index of the last occurrence of every array element.
2. Traverse the array and check if the maximum index of the last occurrence of all the previous elements of the current subarray is less than or equal to the current index, then increment the count by 1.
3. Finally, print the value of count.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to maximize the
// count of subarrays
int maxCtSubarrays(int arr[], int N)
{
    // Store the last index of
    // every array element
    int hash[1000001] = { 0 };

    for (int i = 0; i < N; i++) {
        hash[arr[i]] = i;
    }

    // Store the maximum index of the
    // last occurrence of all elements
    int maxIndex = -1;

    // Store the count of subarrays
    int res = 0;

    for (int i = 0; i < N; i++) {
        maxIndex = max(maxIndex,
                       hash[arr[i]]);

        // If maximum of last indices
        // previous elements is equal
        // to the current index
        if (maxIndex == i) {
            res++;
        }
    }

    // Return the count
    // of subarrays
    return res;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 1,
                  4, 7, 7, 8 };
    int N = sizeof(arr)
            / sizeof(arr[0]);

    cout << maxCtSubarrays(arr, N);
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG {

// Function to maximize the
// count of subarrays
static int maxCtSubarrays(int arr[],
                          int N)
{
  // Store the last index of
  // every array element
  int hash[] = new int[1000001];

  for (int i = 0; i < N; i++)
  {
    hash[arr[i]] = i;
  }

  // Store the maximum index of the
  // last occurrence of all elements
  int maxIndex = -1;

  // Store the count of subarrays
  int res = 0;

  for (int i = 0; i < N; i++)
  {
    maxIndex = Math.max(maxIndex,
                        hash[arr[i]]);

    // If maximum of last indices
    // previous elements is equal
    // to the current index
    if (maxIndex == i)
    {
      res++;
    }
  }

  // Return the count
  // of subarrays
  return res;
}

// Driver Code
public static void main(String[] args)
{
  int arr[] = {1, 2, 4, 1,
               4, 7, 7, 8};
  int N = arr.length;
  System.out.print(maxCtSubarrays(arr, N));
}
}

// This code is contributed by Chitranayal

Python3

# Python3 program to implement
# the above approach

# Function to maximize the
# count of subarrays
def maxCtSubarrays(arr, N):

    # Store the last index of
    # every array element
    hash = [0] * (1000001)

    for i in range(N):
        hash[arr[i]] = i

    # Store the maximum index of the
    # last occurrence of all elements
    maxIndex = -1

    # Store the count of subarrays
    res = 0

    for i in range(N):
        maxIndex = max(maxIndex,
                       hash[arr[i]])

        # If maximum of last indices
        # previous elements is equal
        # to the current index
        if (maxIndex == i):
            res += 1

    # Return the count
    # of subarrays
    return res

# Driver Code
if __name__ == '__main__':

    arr = [ 1, 2, 4, 1,
            4, 7, 7, 8 ]
    N = len(arr)

    print(maxCtSubarrays(arr, N))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG {

// Function to maximize the
// count of subarrays
static int maxCtSubarrays(int []arr,
                          int N)
{
  // Store the last index of
  // every array element
  int []hash = new int[1000001];

  for (int i = 0; i < N; i++)
  {
    hash[arr[i]] = i;
  }

  // Store the maximum index of the
  // last occurrence of all elements
  int maxIndex = -1;

  // Store the count of subarrays
  int res = 0;

  for (int i = 0; i < N; i++)
  {
    maxIndex = Math.Max(maxIndex,
                        hash[arr[i]]);

    // If maximum of last indices
    // previous elements is equal
    // to the current index
    if (maxIndex == i)
    {
      res++;
    }
  }

  // Return the count
  // of subarrays
  return res;
}

// Driver Code
public static void Main(String[] args)
{
  int []arr = {1, 2, 4, 1,
               4, 7, 7, 8};
  int N = arr.Length;
  Console.Write(maxCtSubarrays(arr, N));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to maximize the
// count of subarrays
function maxCtSubarrays(arr, N)
{
  // Store the last index of
  // every array element
  let hash = new Array(1000001).fill(0);

  for (let i = 0; i < N; i++)
  {
    hash[arr[i]] = i;
  }

  // Store the maximum index of the
  // last occurrence of all elements
  let maxIndex = -1;

  // Store the count of subarrays
  let res = 0;

  for (let i = 0; i < N; i++)
  {
    maxIndex = Math.max(maxIndex,
                        hash[arr[i]]);

    // If maximum of last indices
    // previous elements is equal
    // to the current index
    if (maxIndex == i)
    {
      res++;
    }
  }

  // Return the count
  // of subarrays
  return res;
}

// Driver Code

        let arr = [1, 2, 4, 1,
               4, 7, 7, 8];
          let N = arr.length;
          document.write(maxCtSubarrays(arr, N));

// This code is contributed by avijitmondal1998.
</script>

Output

3

Time Complexity: O(N)
Auxiliary Space: O(X) where X = 1000001

Related Topic: Subarrays, Subsequences, and Subsets in Array