2

Check for each subarray whether it consists of all natural numbers up to its len...

 1 year ago
source link: https://www.geeksforgeeks.org/check-for-each-subarray-whether-it-consists-of-all-natural-numbers-up-to-its-length-or-not/
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
neoserver,ios ssh client

Check for each subarray whether it consists of all natural numbers up to its length or not

Given an array, arr[] representing a permutation of first N natural numbers in the range [1, N], the task for each ith index is to check if an i-length subarray exists or not such that it contains all the numbers in the range [1, i]. 
Note: 1 – based indexing used.

Examples:

Input: arr[] = {4, 5, 1, 3, 2}
Output: True False True False True
Explanation: 
For i = 1, the subarray {arr[2]} contains all the numbers from [1, i] and of size i(=1). Therefore, the required output is true. 
For i = 2, no subarray of size 2 exists which contains all the numbers in the range[1, i]. Therefore, the required output is false. 
For i = 3, the subarray {arr[2], arr[3], arr[4]} contains all the numbers from [1, i] and of size i(=3). Therefore, the required output is true. 
For i = 4, no subarray of size 4 exists which contains all the numbers in the range[1, i]. Therefore, the required output is false. 
For i = 5, the subarray {arr[0], arr[1], arr[2], arr[3], arr[4]} contains all the numbers from [1, i] and of size i(=5). Therefore, the required output is true. 

Input: arr = {1, 4, 3, 2}
Output: True False False True

Naive Approach: The idea is to traverse the array and for each index, check if there is a subarray of size i which contains all the numbers in the range [1, i] or not. If found to be true, then print True. Otherwise, print False. 

Time Complexity: O(N2) 
Auxiliary Space: O(N)

Efficient Approach: The problem can be solved using Hashing to store the position of each element of the given array efficiently. Follow the steps below to solve the problem:

  • Create a map, say, Map, to store the position of each element of the given array.
  • Traverse the array and store the position of each element of the array onto Map.
  • Create a set, say st to store the indexes of each element from the range [1, N].
  • Initialize two variables, say Min and Max, to store the smallest and the largest elements present in st.
  • Iterate over the range [1, N] and insert the value of Map[i] into st and check if Max – Min + 1 = i or not. If found to be true, then print True.
  • Otherwise, print False.

Below is the implementation of the above approach:

  • Python3
  • Javascript
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a subarray of size i exists
// that contain all the numbers in the range [1, i]
void checksubarrayExist1_N(int arr[], int N)
{
// Store the position
// of each element of arr[]
unordered_map<int, int> pos;
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert the position
// of arr[i]
pos[arr[i]] = i;
}
// Store position of each element
// from the range [1, N]
set<int> st;
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Insert the index of i into st
st.insert(pos[i]);
// Find the smallest element of st
int Min = *(st.begin());
// Find the largest element of st
int Max = *(st.rbegin());
// If distance between the largest
// and smallest element of arr[]
// till i-th index is equal to i
if (Max - Min + 1 == i) {
cout << "True ";
}
else {
cout << "False ";
}
}
}
// Driver Code
int main()
{
int arr[] = { 1, 4, 3, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
checksubarrayExist1_N(arr, N);
}
Output: 
True False False True

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)


report

Recommend

About Joyk


Aggregate valuable and interesting links.
Joyk means Joy of geeK