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# Check if an Array is made up of Subarrays of continuous repetitions of every distinct element

Given an array arr[], consisting of N integers, the task is to check whether the entire array is only made up of subarrays such that each subarray consists of consecutive repetitions of a single element and every distinct element in the array is part of such subarray.

Examples:

Input: N = 10, arr[] = {1, 1, 1, 1, 2, 2, 3, 3, 3, 3}
Output: Yes
Explanation:
The given array consists of 3 distinct elements {1, 2, 3} and subarrays {1, 1, 1, 1}, {2, 2}, {3, 3, 3, 3}.
Therefore, the given array satisfies the conditions.

Input: N = 10, arr[] = {1, 1, 1, 2, 2, 2, 2, 1, 3, 3}
Output: No
Explanation:
The given array consists of 3 distinct elements {1, 2, 3} and subarrays {1, 1, 1}, {2, 2, 2, 2}, {1}, {3, 3}.
Since the subarray {1} does not contain any repetition, the given array does not satisfy the conditions.

Approach:
Follow the steps below to solve the problem:

• Initialize a variable curr = 0 to store the size of every subarray of a single repeating element is encountered.
• If any such index is found where arr[i] ≠ arr[i – 1], check if curr is greater than 1 or not. If so, reset curr to 0 and continue. Otherwise, print “No” as a subarray exists of a single element without repetition.
• Otherwise, increase curr.
• After traversing the array, check if curr is greater than 1 or not. If curr is equal to 1, this ensures that the last element is different from the second last element. Therefore, print “No”.
• Otherwise, print “Yes”.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above problem``#include ``using` `namespace` `std;` `// Function to check if the``// array is made up of``// subarrays of repetitions``bool` `ContinuousElements(``int` `a[],``                        ``int` `n)``{` `    ``// Base Case``    ``if` `(n == 1)``        ``return` `false``;` `    ``// Stores the size of``    ``// current subarray``    ``int` `curr = 1;``    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// If a different element``        ``// is encountered``        ``if` `(a[i] != a[i - 1]) {` `            ``// If the previous subarray``            ``// was a single element``            ``if` `(curr == 1)``                ``return` `false``;` `            ``// Reset to new subarray``            ``else``                ``curr = 0;``        ``}` `        ``// Increase size of subarray``        ``curr++;``    ``}` `    ``// If last element differed from``    ``// the second last element``    ``if` `(curr == 1)``        ``return` `false``;` `    ``return` `true``;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 1, 1, 2, 2, 1, 3, 3 };``    ``int` `n = ``sizeof``(a)``            ``/ ``sizeof``(a);` `    ``if` `(ContinuousElements(a, n))``        ``cout << ``"Yes"` `<< endl;``    ``else``        ``cout << ``"No"` `<< endl;` `    ``return` `0;``}`

## Java

 `// Java Program to implement``// the above approach``class` `GFG{`` ` `// Function to check if the``// array is made up of``// subarrays of repetitions``static` `boolean` `ContinuousElements(``int` `a[],``                                  ``int` `n)``{`` ` `    ``// Base Case``    ``if` `(n == ``1``)``        ``return` `false``;`` ` `    ``// Stores the size of``    ``// current subarray``    ``int` `curr = ``1``;``    ``for` `(``int` `i = ``1``; i < n; i++)``    ``{`` ` `        ``// If a different element``        ``// is encountered``        ``if` `(a[i] != a[i - ``1``])``        ``{`` ` `            ``// If the previous subarray``            ``// was a single element``            ``if` `(curr == ``1``)``                ``return` `false``;`` ` `            ``// Reset to new subarray``            ``else``                ``curr = ``0``;``        ``}`` ` `        ``// Increase size of subarray``        ``curr++;``    ``}`` ` `    ``// If last element differed from``    ``// the second last element``    ``if` `(curr == ``1``)``        ``return` `false``;`` ` `    ``return` `true``;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``1``, ``1``, ``2``, ``2``, ``1``, ``3``, ``3` `};``    ``int` `n = a.length;`` ` `    ``if` `(ContinuousElements(a, n))``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}``}` `// This code is contributed by rock_cool`

## Python3

 `# Python3 program to implement``# the above problem` `# Function to check if the``# array is made up of``# subarrays of repetitions``def` `ContinuousElements(a, n):` `  ``# Base Case``  ``if` `(n ``=``=` `1``):``    ``return` `False` `  ``# Stores the size of``  ``# current subarray``  ``curr ``=` `1``  ``for` `i ``in` `range` `(``1``, n):` `    ``# If a different element``    ``# is encountered``    ``if` `(a[i] !``=` `a[i ``-` `1``]):` `      ``# If the previous subarray``      ``# was a single element``      ``if` `(curr ``=``=` `1``):``        ``return` `False` `      ``# Reset to new subarray``      ``else``:``        ``curr ``=` `0` `        ``# Increase size of subarray``        ``curr ``+``=` `1` `        ``# If last element differed from``        ``# the second last element``        ``if` `(curr ``=``=` `1``):``          ``return` `False` `        ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` `    ``a ``=` `[``1``, ``1``, ``2``, ``2``, ``1``, ``3``, ``3``]``    ``n ``=` `len``(a)` `    ``if` `(ContinuousElements(a, n)):``          ``print` `(``"Yes"``)``    ``else``:``          ``print` `(``"No"``)` `# This code is contributed by Chitranayal`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to check if the``// array is made up of``// subarrays of repetitions``static` `Boolean ContinuousElements(``int` `[]a,``                                  ``int` `n)``{` `    ``// Base Case``    ``if` `(n == 1)``        ``return` `false``;` `    ``// Stores the size of``    ``// current subarray``    ``int` `curr = 1;``    ``for``(``int` `i = 1; i < n; i++)``    ``{` `        ``// If a different element``        ``// is encountered``        ``if` `(a[i] != a[i - 1])``        ``{` `            ``// If the previous subarray``            ``// was a single element``            ``if` `(curr == 1)``                ``return` `false``;` `            ``// Reset to new subarray``            ``else``                ``curr = 0;``        ``}` `        ``// Increase size of subarray``        ``curr++;``    ``}` `    ``// If last element differed from``    ``// the second last element``    ``if` `(curr == 1)``        ``return` `false``;` `    ``return` `true``;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = { 1, 1, 2, 2, 1, 3, 3 };``    ``int` `n = a.Length;` `    ``if` `(ContinuousElements(a, n))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``

Output:

`No`

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

New Approach:- Another approach to solving this problem is to use a hash table to keep track of the frequency of each distinct element in the array. Then, we can iterate through the hash table and check if the frequency of any element is not equal to the length of any subarray made up of that element. If such an element exists, then the array is not made up of subarrays of continuous repetitions of every distinct element.

Here’s the implementation of this approach:-

## C++

 `#include ``#include ``#include ` `using` `namespace` `std;` `bool` `checkSubarrays(``int` `arr[], ``int` `n) {``    ``unordered_map<``int``, ``int``> freq;` `    ``// Count frequency of each distinct element``    ``for` `(``int` `i = 0; i < n; i++) {``        ``freq[arr[i]]++;``    ``}` `    ``// Check if frequency of each distinct element``    ``// is equal to the length of any subarray made up``    ``// of that element``    ``for` `(``auto` `it = freq.begin(); it != freq.end(); it++) {``        ``int` `elem = it->first;``        ``int` `count = it->second;` `        ``int` `len = 0;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(arr[i] == elem) {``                ``len++;``            ``} ``else` `{``                ``if` `(len != count) {``                    ``return` `false``;``                ``}``                ``len = 0;``            ``}``        ``}``        ``if` `(len != count) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `int` `main() {``    ``int` `arr[] = {1, 1, 2, 2, 1, 3, 3};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``if` `(checkSubarrays(arr, n)) {``        ``cout << ``"Yes\n"``;``    ``} ``else` `{``        ``cout << ``"No\n"``;``    ``}` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``static` `boolean` `checkSubarrays(``int``[] arr, ``int` `n)``    ``{``        ``Map freq = ``new` `HashMap<>();``        ``// Count frequency of each distinct element``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``freq.put(arr[i],``                     ``freq.getOrDefault(arr[i], ``0``) + ``1``);``        ``}` `        ``// Check if frequency of each distinct element``        ``// is equal to the length of any subarray made up``        ``// of that element``        ``for` `(Map.Entry entry :``             ``freq.entrySet()) {``            ``int` `elem = entry.getKey();``            ``int` `count = entry.getValue();` `            ``int` `len = ``0``;``            ``for` `(``int` `i = ``0``; i < n; i++) {``                ``if` `(arr[i] == elem) {``                    ``len++;``                ``}``                ``else` `{``                    ``if` `(len != count) {``                        ``return` `false``;``                    ``}``                    ``len = ``0``;``                ``}``            ``}``            ``if` `(len != count) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``1``, ``1``, ``2``, ``2``, ``1``, ``3``, ``3` `};``        ``int` `n = arr.length;` `        ``if` `(checkSubarrays(arr, n)) {``            ``System.out.println(``"Yes"``);``        ``}``        ``else` `{``            ``System.out.println(``"No"``);``        ``}``    ``}``}`

## Javascript

 `function` `checkSubarrays(arr, n) {``  ``const freq = ``new` `Map();` `  ``// Count frequency of each distinct element``  ``for` `(let i = 0; i < n; i++) {``    ``freq.set(arr[i], (freq.get(arr[i]) || 0) + 1);``  ``}` `  ``// Check if frequency of each distinct element``  ``// is equal to the length of any subarray made up``  ``// of that element``  ``for` `(const [elem, count] of freq.entries()) {``    ``let len = 0;``    ``for` `(let i = 0; i < n; i++) {``      ``if` `(arr[i] === elem) {``        ``len++;``      ``} ``else` `{``        ``if` `(len !== count) {``          ``return` `false``;``        ``}``        ``len = 0;``      ``}``    ``}``    ``if` `(len !== count) {``      ``return` `false``;``    ``}``  ``}``  ``return` `true``;``}` `const arr = [1, 1, 2, 2, 1, 3, 3];``const n = arr.length;` `if` `(checkSubarrays(arr, n)) {``  ``console.log(``"Yes"``);``} ``else` `{``  ``console.log(``"No"``);``}`

## C#

 `using` `System;``using` `System.Collections.Generic;` `public` `class` `Program {``    ``public` `static` `bool` `CheckSubarrays(``int``[] arr, ``int` `n)``    ``{``        ``Dictionary<``int``, ``int``> freq``            ``= ``new` `Dictionary<``int``, ``int``>();``        ``// Count frequency of each distinct element``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(!freq.ContainsKey(arr[i])) {``                ``freq[arr[i]] = 1;``            ``}``            ``else` `{``                ``freq[arr[i]]++;``            ``}``        ``}` `        ``// Check if frequency of each distinct element``        ``// is equal to the length of any subarray made up``        ``// of that element``        ``foreach``(``var` `item ``in` `freq)``        ``{``            ``int` `elem = item.Key;``            ``int` `count = item.Value;` `            ``int` `len = 0;``            ``for` `(``int` `i = 0; i < n; i++) {``                ``if` `(arr[i] == elem) {``                    ``len++;``                ``}``                ``else` `{``                    ``if` `(len != count) {``                        ``return` `false``;``                    ``}``                    ``len = 0;``                ``}``            ``}``            ``if` `(len != count) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 1, 1, 2, 2, 1, 3, 3 };``        ``int` `n = arr.Length;` `        ``if` `(CheckSubarrays(arr, n)) {``            ``Console.WriteLine(``"Yes"``);``        ``}``        ``else` `{``            ``Console.WriteLine(``"No"``);``        ``}``    ``}``}`

## Python3

 `def` `checkSubarrays(arr, n):``    ``freq ``=` `{}` `    ``# Count frequency of each distinct element``    ``for` `i ``in` `range``(n):``        ``if` `arr[i] ``in` `freq:``            ``freq[arr[i]] ``+``=` `1``        ``else``:``            ``freq[arr[i]] ``=` `1` `    ``# Check if frequency of each distinct element``    ``# is equal to the length of any subarray made up``    ``# of that element``    ``for` `elem, count ``in` `freq.items():``        ``length ``=` `0``        ``for` `i ``in` `range``(n):``            ``if` `arr[i] ``=``=` `elem:``                ``length ``+``=` `1``            ``else``:``                ``if` `length !``=` `count:``                    ``return` `False``                ``length ``=` `0``        ``if` `length !``=` `count:``            ``return` `False``    ``return` `True`  `arr ``=` `[``1``, ``1``, ``2``, ``2``, ``1``, ``3``, ``3``]``n ``=` `len``(arr)` `if` `checkSubarrays(arr, n):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)`

Output:-

`No`

Time Complexity: O(n^2), where n is the length of the input array. This is because we are iterating over each distinct element in the array and then checking the length of all subarrays made up of that element. In the worst case, each element could be distinct, and there could be n such elements, leading to a time complexity of O(n^2).

Auxiliary Space: O(n), where n is the length of the input array. This is because we are using an unordered map to store the frequency of each distinct element, which can have at most n entries. Additionally, we are using a variable len to keep track of the length of the current subarray, which could be at most n. Therefore, the total space complexity is O(n + n) = O(n).

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