Split array into maximum subarrays such that every distinct element lies in a single subarray
Last Updated :
01 Dec, 2022
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:
- Initialize an array, say hash[] to store the index of the last occurrence of every array element.
- 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.
- 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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(X) where X = 1000001
Related Topic: Subarrays, Subsequences, and Subsets in Array
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