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Smallest element present in every subarray of all possible lengths
  • Last Updated : 02 Mar, 2021
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Given an array arr[] of length N, the task for every possible length of subarray is to find the smallest element present in every subarray of that length.

Examples:

Input: N = 10, arr[] = {2, 3, 5, 3, 2, 3, 1, 3, 2, 7}
Output: -1-1 3 2 2 2 1 1 1 1
Explanation:
For length = 1, no element is common in every subarray of length 1. Therefore, output is -1.
For length = 2, no element is common in every subarray of length 2. Therefore, output is -1.
For length = 3, the common element in every subarray is 3. Therefore, the output is 3.
For length = 4, both 2 and 3 are common in every subarray of length 4. 2 being the smaller, is the required output.
Similarly, for lengths 5 and 6, the smallest common element in every subarray of these lengths is 2.
For lengths 7, 8, 9 and 10, the smallest common element in every subarray of these lengths is 1.

Input: N = 3, arr[] = {2, 2, 2}
Output: 2 2 2

Naive Approach: The idea is to find the common elements in all the subarrays of size K for each possible value of K ( 1 ≤ K ≤ N) and print the smallest common element. Follow the steps below to solve the problem:



  1. Add the count of every unique number for every subarray of length K.
  2. Check if the count of numbers is equal to the number of subarrays i.e., N – K – 1.
  3. If found to be true, then that particular element has occurred in every subarray of size K.
  4. For multiple such elements, print the smallest amongst them.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to add count of numbers
// in the map for a subarray of length k
void uniqueElements(int arr[], int start, int K,
                    map<int, int>& mp)
{
    // Set to store unique elements
    set<int> st;
 
    // Add elements to the set
    for (int i = 0; i < K; i++)
        st.insert(arr[start + i]);
 
    // Iterator of the set
    set<int>::iterator itr = st.begin();
 
    // Adding count in map
    for (; itr != st.end(); itr++)
        mp[*itr]++;
}
 
// Funtion to check if there is any number
// which repeats itself in every subarray
// of length K
void checkAnswer(map<int, int>& mp, int N, int K)
{
    // Check all number starting from 1
    for (int i = 1; i <= N; i++) {
 
        // Check if i occured n-k+1 times
        if (mp[i] == (N - K + 1)) {
 
            // Print the smallest number
            cout << i << " ";
            return;
        }
    }
 
    // Print -1, if no such number found
    cout << -1 << " ";
}
 
// Function to count frequency of each
// number in each subarray of length K
void smallestPresentNumber(int arr[], int N, int K)
{
    map<int, int> mp;
 
    // Traverse all subarrays of length K
    for (int i = 0; i <= N - K; i++) {
        uniqueElements(arr, i, K, mp);
    }
 
    // Check and print the smallest number
    // present in every subarray and print it
    checkAnswer(mp, N, K);
}
 
// Function to generate the value of K
void generateK(int arr[], int N)
{
    for (int k = 1; k <= N; k++)
 
        // Functoin call
        smallestPresentNumber(arr, N, k);
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    generateK(arr, N);
 
    return (0);
}

Java




// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG
{
 
  // Function to add count of numbers
  // in the map for a subarray of length k
  static void uniqueElements(int arr[], int start, int K,
                             Map<Integer,Integer> mp)
  {
    // Set to store unique elements
    Set<Integer> st=new HashSet<>();
 
    // Add elements to the set
    for (int i = 0; i < K; i++)
      st.add(arr[start + i]);
 
    // Iterator of the set
    Iterator itr = st.iterator();
 
    // Adding count in map
    while(itr.hasNext())
    {
      Integer t = (Integer)itr.next();
      mp.put(t,mp.getOrDefault(t, 0) + 1);
    }
 
  }
 
  // Funtion to check if there is any number
  // which repeats itself in every subarray
  // of length K
  static void checkAnswer(Map<Integer,Integer> mp, int N, int K)
  {
 
    // Check all number starting from 1
    for (int i = 1; i <= N; i++)
    {
 
      // Check if i occured n-k+1 times
      if(mp.containsKey(i))   
        if (mp.get(i) == (N - K + 1))
        {
 
          // Print the smallest number
          System.out.print(i + " ");
          return;
        }
    }
 
    // Print -1, if no such number found
    System.out.print(-1 + " ");
  }
 
  // Function to count frequency of each
  // number in each subarray of length K
  static void smallestPresentNumber(int arr[], int N, int K)
  {
    Map<Integer, Integer> mp = new HashMap<>();
 
    // Traverse all subarrays of length K
    for (int i = 0; i <= N - K; i++)
    {
      uniqueElements(arr, i, K, mp);
    }
 
    // Check and print the smallest number
    // present in every subarray and print it
    checkAnswer(mp, N, K);
  }
 
  // Function to generate the value of K
  static void generateK(int arr[], int N)
  {
    for (int k = 1; k <= N; k++)
 
      // Functoin call
      smallestPresentNumber(arr, N, k);
  }
 
  // Driver code
  public static void main (String[] args)
  {
    // Given array
    int arr[] = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
 
    // Size of array
    int N = arr.length;
 
    // Function call
    generateK(arr, N);
 
  }
}
 
// This code is contributed by offbeat.

Python3




# Python3 program for the above approach
 
# Function to add count of numbers
# in the map for a subarray of length k
def uniqueElements(arr, start, K, mp) :
 
    # Set to store unique elements
    st = set();
 
    # Add elements to the set
    for i in range(K) :
        st.add(arr[start + i]);
 
    # Adding count in map
    for itr in st:
        if itr in mp :
            mp[itr] += 1;
        else:
            mp[itr] = 1;
 
# Funtion to check if there is any number
# which repeats itself in every subarray
# of length K
def checkAnswer(mp, N, K) :
 
    # Check all number starting from 1
    for i in range(1, N + 1) :
         
        if i in mp :
           
            # Check if i occured n-k+1 times
            if (mp[i] == (N - K + 1)) :
     
                # Print the smallest number
                print(i, end = " ");
                return;
 
    # Print -1, if no such number found
    print(-1, end = " ");
 
# Function to count frequency of each
# number in each subarray of length K
def smallestPresentNumber(arr, N,  K) :
    mp = {};
 
    # Traverse all subarrays of length K
    for i in range(N - K + 1) :
        uniqueElements(arr, i, K, mp);
 
    # Check and print the smallest number
    # present in every subarray and print it
    checkAnswer(mp, N, K);
 
# Function to generate the value of K
def generateK(arr, N) :
 
    for k in range(1, N + 1) :
 
        # Functoin call
        smallestPresentNumber(arr, N, k);
 
# Driver Code
if __name__ == "__main__" :
 
    # Given array
    arr = [ 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 ];
 
    # Size of array
    N = len(arr);
 
    # Function call
    generateK(arr, N);
     
    # This code is contributed by AnkThon

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to add count of numbers
  // in the map for a subarray of length k
  static void uniqueElements(int[] arr, int start,
                             int K, Dictionary<int, int> mp)
  {
 
    // Set to store unique elements
    HashSet<int> st = new HashSet<int>();
 
    // Add elements to the set
    for (int i = 0; i < K; i++)
      st.Add(arr[start + i]);
 
    // Adding count in map
    foreach(int itr in st)
    {
      if(mp.ContainsKey(itr))
      {
        mp[itr]++;
      }
      else{
        mp[itr] = 1;
      }
    }
  }
 
  // Funtion to check if there is any number
  // which repeats itself in every subarray
  // of length K
  static void checkAnswer(Dictionary<int, int> mp, int N, int K)
  {
 
    // Check all number starting from 1
    for (int i = 1; i <= N; i++)
    {
 
      // Check if i occured n-k+1 times
      if(mp.ContainsKey(i))   
        if (mp[i] == (N - K + 1))
        {
 
          // Print the smallest number
          Console.Write(i + " ");
          return;
        }
    }
 
    // Print -1, if no such number found
    Console.Write(-1 + " ");
  }
 
  // Function to count frequency of each
  // number in each subarray of length K
  static void smallestPresentNumber(int[] arr, int N, int K)
  {
    Dictionary<int, int> mp = new Dictionary<int, int>();
 
    // Traverse all subarrays of length K
    for (int i = 0; i <= N - K; i++)
    {
      uniqueElements(arr, i, K, mp);
    }
 
    // Check and print the smallest number
    // present in every subarray and print it
    checkAnswer(mp, N, K);
  }
 
  // Function to generate the value of K
  static void generateK(int[] arr, int N)
  {
    for (int k = 1; k <= N; k++)
 
      // Functoin call
      smallestPresentNumber(arr, N, k);
  }
 
  // Driver code
  static void Main()
  {
 
    // Given array
    int[] arr = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
 
    // Size of array
    int N = arr.Length;
 
    // Function call
    generateK(arr, N);
  }
}
 
// This code is contributed by divyesh072019.
Output: 
-1 -1 3 2 2 2 1 1 1 1

 

Time Complexity: O(N3)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized if all the indices where the particular number is present in the array are stored using an array and find the minimum length so that it is present in every subarray of length 1 ≤ K ≤ N.

Follow the steps below to solve the problem:

  1. Initialize an array, say indices[], to store the index where a particular number occurs corresponding to that index number.
  2. Now, for every number which is present in the given array, find the minimum length so that it is present in every subarray of that length.
  3. Minimum length can be found by finding the maximum interval at which that particular number repeats itself in the given array. Similarly, find the same for other numbers of the array.
  4. Initialize an answer[] array of size N+1 with -1 where answer[i] represents the answer for subarrays of length K.
  5. Now, the indices[] array gives the number which was present in every subarray of length, say K, then update answer[K] with the same number if answer[K] was -1.
  6. After traversing, update answer[] array such that if a number is present in all the subarrays of length K, then that particular number will also be present in all the subarrays of length greater than K.
  7. After updating answer[] array, print all the elements present in that array as the answer for every subarray of length K.

Below is the implementation of the above approach: 

C++




// C++ program of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the common
// elements for all subarray lengths
void printAnswer(int answer[], int N)
{
    for (int i = 1; i <= N; i++) {
        cout << answer[i] << " ";
    }
}
 
// Function to find and store the
// minimum element present in all
// subarrays of all lengths from 1 to n
void updateAnswerArray(int answer[], int N)
{
    int i = 0;
 
    // Skip lengths for which
    // answer[i] is -1
    while (answer[i] == -1)
        i++;
 
    // Initialize minimum as the first
    // element where answer[i] is not -1
    int minimum = answer[i];
 
    // Updating the answer array
    while (i <= N) {
 
        // If answer[i] is -1, then minimum
        // can be substituted in that place
        if (answer[i] == -1)
            answer[i] = minimum;
 
        // Find minimum answer
        else
            answer[i] = min(minimum, answer[i]);
        minimum = min(minimum, answer[i]);
        i++;
    }
}
 
// Function to find the minimum number
// corresponding to every subarray of
// length K, for every K from 1 to N
void lengthOfSubarray(vector<int> indices[],
                      set<int> st, int N)
{
    // Stores the minimum common
    // elements for all subarray lengths
    int answer[N + 1];
 
    // Initialize with -1.
    memset(answer, -1, sizeof(answer));
 
    // Find for every element, the minimum length
    // such that the number is present in every
    // subsequence of that particular length or more
    for (auto itr : st) {
 
        // To store first occurence and
        // gaps between occurences
        int start = -1;
        int gap = -1;
 
        // To cover the distance between last
        // occurence and the end of the array
        indices[itr].push_back(N);
 
        // To find the distance
        // between any two occurences
        for (int i = 0; i < indices[itr].size(); i++) {
            gap = max(gap, indices[itr][i] - start);
            start = indices[itr][i];
        }
        if (answer[gap] == -1)
            answer[gap] = itr;
    }
 
    // Update and store the answer
    updateAnswerArray(answer, N);
 
    // Print the required answer
    printAnswer(answer, N);
}
 
// Function to find the smallest
// element common in all subarrays for
// every possible subarray lengths
void smallestPresentNumber(int arr[], int N)
{
    // Initializing indices array
    vector<int> indices[N + 1];
 
    // Store the numbers present
    // in the array
    set<int> elements;
 
    // Push the index in the indices[A[i]] and
    // also store the numbers in set to get
    // the numbers present in input array
    for (int i = 0; i < N; i++) {
        indices[arr[i]].push_back(i);
        elements.insert(arr[i]);
    }
 
    // Function call to calculate length of
    // subarray for which a number is present
    // in every subarray of that length
    lengthOfSubarray(indices, elements, N);
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    smallestPresentNumber(arr, N);
 
    return (0);
}

Java




// Java program for above approach
import java.util.*;
import java.lang.*;
 
class GFG
{
 
  // Function to print the common
  // elements for all subarray lengths
  static void printAnswer(int answer[], int N)
  {
    for (int i = 1; i <= N; i++)
    {
      System.out.print(answer[i]+" ");
    }
  }
 
  // Function to find and store the
  // minimum element present in all
  // subarrays of all lengths from 1 to n
  static void updateAnswerArray(int answer[], int N)
  {
    int i = 0;
 
    // Skip lengths for which
    // answer[i] is -1
    while (answer[i] == -1)
      i++;
 
    // Initialize minimum as the first
    // element where answer[i] is not -1
    int minimum = answer[i];
 
    // Updating the answer array
    while (i <= N) {
 
      // If answer[i] is -1, then minimum
      // can be substituted in that place
      if (answer[i] == -1)
        answer[i] = minimum;
 
      // Find minimum answer
      else
        answer[i] = Math.min(minimum, answer[i]);
      minimum = Math.min(minimum, answer[i]);
      i++;
    }
  }
 
  // Function to find the minimum number
  // corresponding to every subarray of
  // length K, for every K from 1 to N
  static void lengthOfSubarray(ArrayList<ArrayList<Integer>> indices,
                               Set<Integer> st, int N)
  {
    // Stores the minimum common
    // elements for all subarray lengths
    int[] answer = new int[N + 1];
 
    // Initialize with -1.
    Arrays.fill(answer, -1);
 
    // Find for every element, the minimum length
    // such that the number is present in every
    // subsequence of that particular length or more
    Iterator itr = st.iterator();
    while (itr.hasNext())
    {
 
      // To store first occurence and
      // gaps between occurences
      int start = -1;
      int gap = -1;
      int t = (int)itr.next();
 
      // To cover the distance between last
      // occurence and the end of the array
      indices.get(t).add(N);
 
      // To find the distance
      // between any two occurences
      for (int i = 0; i < indices.get(t).size(); i++)
      {
        gap = Math.max(gap, indices.get(t).get(i) - start);
        start = indices.get(t).get(i);
      }
      if (answer[gap] == -1)
        answer[gap] = t;
    }
 
    // Update and store the answer
    updateAnswerArray(answer, N);
 
    // Print the required answer
    printAnswer(answer, N);
  }
 
  // Function to find the smallest
  // element common in all subarrays for
  // every possible subarray lengths
  static void smallestPresentNumber(int arr[], int N)
  {
    // Initializing indices array
    ArrayList<ArrayList<Integer>> indices = new ArrayList<>();
 
    for(int i = 0; i <= N; i++)
      indices.add(new ArrayList<Integer>());
 
    // Store the numbers present
    // in the array
    Set<Integer> elements = new HashSet<>();
 
    // Push the index in the indices[A[i]] and
    // also store the numbers in set to get
    // the numbers present in input array
    for (int i = 0; i < N; i++)
    {
      indices.get(arr[i]).add(i);
      elements.add(arr[i]);
    }
 
    // Function call to calculate length of
    // subarray for which a number is present
    // in every subarray of that length
    lengthOfSubarray(indices, elements, N);
  }
 
  // Driver function
  public static void main (String[] args)
  {
 
    // Given array
    int arr[] = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
 
    // Size of array
    int N = arr.length;
 
    // Function Call
    smallestPresentNumber(arr, N);
  }
}
 
// This code is contributed by offbeat

C#




// C# program for above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to print the common
// elements for all subarray lengths
static void printAnswer(int[] answer, int N)
{
    for(int i = 1; i <= N; i++)
    {
        Console.Write(answer[i] + " ");
    }
}
 
// Function to find and store the
// minimum element present in all
// subarrays of all lengths from 1 to n
static void updateAnswerArray(int[] answer, int N)
{
    int i = 0;
     
    // Skip lengths for which
    // answer[i] is -1
    while (answer[i] == -1)
        i++;
     
    // Initialize minimum as the first
    // element where answer[i] is not -1
    int minimum = answer[i];
     
    // Updating the answer array
    while (i <= N)
    {
         
        // If answer[i] is -1, then minimum
        // can be substituted in that place
        if (answer[i] == -1)
            answer[i] = minimum;
         
        // Find minimum answer
        else
            answer[i] = Math.Min(minimum, answer[i]);
            minimum = Math.Min(minimum, answer[i]);
            i++;
    }
}
 
// Function to find the minimum number
// corresponding to every subarray of
// length K, for every K from 1 to N
static void lengthOfSubarray(List<List<int>> indices,
                               HashSet<int> st, int N)
{
    // Stores the minimum common
    // elements for all subarray lengths
    int[] answer = new int[N + 1];
     
    // Initialize with -1.
    Array.Fill(answer, -1);
     
    // Find for every element, the minimum length
    // such that the number is present in every
    // subsequence of that particular length or more
    foreach(int itr in st)
    {
         
        // To store first occurence and
        // gaps between occurences
        int start = -1;
        int gap = -1;
        int t = itr;
         
        // To cover the distance between last
        // occurence and the end of the array
        indices[t].Add(N);
         
        // To find the distance
        // between any two occurences
        for(int i = 0; i < indices[t].Count; i++)
        {
            gap = Math.Max(gap, indices[t][i] - start);
            start = indices[t][i];
        }
        if (answer[gap] == -1)
            answer[gap] = t;
    }
     
    // Update and store the answer
    updateAnswerArray(answer, N);
     
    // Print the required answer
    printAnswer(answer, N);
}
 
// Function to find the smallest
// element common in all subarrays for
// every possible subarray lengths
static void smallestPresentNumber(int[] arr, int N)
{
     
    // Initializing indices array
    List<List<int>> indices = new List<List<int>>();
     
    for(int i = 0; i <= N; i++)
        indices.Add(new List<int>());
     
    // Store the numbers present
    // in the array
    HashSet<int> elements = new HashSet<int>();
     
    // Push the index in the indices[A[i]] and
    // also store the numbers in set to get
    // the numbers present in input array
    for(int i = 0; i < N; i++)
    {
        indices[arr[i]].Add(i);
        elements.Add(arr[i]);
    }
     
    // Function call to calculate length of
    // subarray for which a number is present
    // in every subarray of that length
    lengthOfSubarray(indices, elements, N);
}
 
// Driver code
static void Main()
{
     
    // Given array
    int[] arr = { 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 };
     
    // Size of array
    int N = arr.Length;
     
    // Function Call
    smallestPresentNumber(arr, N);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python program of the above approach
 
# Function to print the common
# elements for all subarray lengths
def printAnswer(answer, N):
    for i in range(N + 1):
        print(answer[i], end = " ")
 
# Function to find and store the
# minimum element present in all
# subarrays of all lengths from 1 to n
def updateAnswerArray(answer, N):
    i = 0
     
    # Skip lengths for which
    # answer[i] is -1
    while(answer[i] == -1):
        i += 1
     
    # Initialize minimum as the first
    # element where answer[i] is not -1
    minimum = answer[i]
     
    # Updating the answer array
    while(i <= N):
         
        # If answer[i] is -1, then minimum
        # can be substituted in that place
        if(answer[i] == -1):
            answer[i] = minimum
         
        # Find minimum answer
        else:
            answer[i] = min(minimum, answer[i])
        minimum = min(minimum, answer[i])
        i += 1
 
# Function to find the minimum number
# corresponding to every subarray of
# length K, for every K from 1 to N
def lengthOfSubarray(indices, st, N):
     
    # Stores the minimum common
    # elements for all subarray lengths
    #Initialize with -1.
    answer = [-1 for i in range(N + 1)]
     
    # Find for every element, the minimum length
    # such that the number is present in every
    # subsequence of that particular length or more
    for itr in st:
         
        # To store first occurence and
        # gaps between occurences
        start = -1
        gap = -1
         
        # To cover the distance between last
        # occurence and the end of the array
        indices[itr].append(N)
         
        # To find the distance
        # between any two occurences
        for i in range(len(indices[itr])):
            gap = max(gap, indices[itr][i] - start)
            start = indices[itr][i]
         
        if(answer[gap] == -1):
            answer[gap] = itr
     
    # Update and store the answer
    updateAnswerArray(answer, N)
     
    # Print the required answer
    printAnswer(answer, N)
     
# Function to find the smallest
# element common in all subarrays for
# every possible subarray lengths
def smallestPresentNumber(arr, N):
     
    # Initializing indices array
    indices = [[] for i in range(N + 1)]
     
    # Store the numbers present
    # in the array
    elements = []
     
    # Push the index in the indices[A[i]] and
    # also store the numbers in set to get
    # the numbers present in input array
    for i in range(N):
        indices[arr[i]].append(i)
        elements.append(arr[i])
     
    elements = list(set(elements))
     
    # Function call to calculate length of
    # subarray for which a number is present
    # in every subarray of that length
    lengthOfSubarray(indices, elements, N)
 
# Driver Code
 
# Given array
arr = [2, 3, 5, 3, 2, 3, 1, 3, 2, 7]
 
# Size of array
N = len(arr)
 
# Function Call
smallestPresentNumber(arr, N)
 
# This code is contributed by avanitrachhadiya2155
Output: 
-1 -1 3 2 2 2 1 1 1 1

 

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

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