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 <bits/stdc++.h>using namespace std;// Function to check if the// array is made up of// subarrays of repetitionsbool 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 codeint main(){ int a[] = { 1, 1, 2, 2, 1, 3, 3 }; int n = sizeof(a) / sizeof(a[0]); if (ContinuousElements(a, n)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java Program to implement// the above approachclass GFG{ // Function to check if the// array is made up of// subarrays of repetitionsstatic 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 codepublic 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 repetitionsdef 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 codeif __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 approachusing System;class GFG{// Function to check if the// array is made up of// subarrays of repetitionsstatic 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 codepublic 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
<script>// Javascript program to implement// the above problem// Function to check if the// array is made up of// subarrays of repetitionsfunction ContinuousElements(a, n){ // Base Case if (n == 1) return false; // Stores the size of // current subarray let curr = 1; for(let 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 let a = [ 1, 1, 2, 2, 1, 3, 3 ];let n = a.length;if (ContinuousElements(a, n)) document.write("Yes");else document.write("No"); // This code is contributed by divyesh072019</script> |
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 <iostream>#include <unordered_map>#include <vector>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[0]); 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<Integer, Integer> 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<Integer, Integer> 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 Truearr = [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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