# 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 ` `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 = { 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);`   `    ``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``;`   `  ``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`

Output

`3`

Time Complexity: O(N)
Auxiliary Space: O(X) where X, is the maximum element in the given array.

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