For each value in [1, N] find Minimum element present in all Subarray of that size

Given an array A[] of size N, the task is to find the minimum element present in all subarrays for all sizes from 1 to N where all the elements in the array are in range 1 to N

Examples:

Input: A[ ] = {1, 2, 3}
Output: [-1, 2, 1]
Explanation: All subarrays of size 1 {{1}, {2}, {3}} there is no common value 
For subarrays of size 2 {{1, 2}, {2, 3}} the minimum common element is 2 
For subarrays of size 3 {{1, 2, 3}} the minimum common element is 1 Hence, ans=[-1, 2, 1]

Input: A[ ] = {1, 2, 1, 3, 1}
Output: [-1, 1, 1, 1, 1]

Naive Approach: The basic idea to solve the problem is to find all the subarrays for all the sizes in the range [1, N]. Now for all subarrays of the same size find the minimum common element in those subarrays. Follow the steps mentioned below to solve the problem:

• Iterate a loop from i = 1 to N:
  • Create every possible subarray of size i.
  • Count the frequency of each element in all the subarrays.
  • Check if the occurrence of any element is equal to the total number of subarrays of that size
  • Store the first element satisfying  the above conditions
• Return the resultant array of the minimum common elements.

Below is the implementation of the above approach.

C++

// C++ code for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// the minimum element array
// from size 1 to N
vector<int> Calculate_Min_array(vector<int>& A,
                                int N)
{
    // Minimum element array(ans)
    // Count array for every subsegment(cnt)
    // Total occurrence array in all
    // subsegment of a given size(res)
    vector<int> ans(N + 1, -1), cnt(N + 1, 0),
        res(N + 1, 0);
    for (int i = 1; i <= N; i++) {

        // Counting all the elements
        // for every subsegment of size i
        for (int j = 0; j < N - i + 1;
             j++) {
            for (int k = j; k < j + i;
                 k++) {
                cnt[A[k]]++;
            }

            // If count of element is
            // greater than 0 then
            // increment its occurrence
            for (int k = 1; k <= N; k++) {
                if (cnt[k]) {

                    // If element is present
                    // increase its count
                    res[k]++;
                    cnt[k] = 0;
                }
            }
        }

        // When occurrence of an element
        // is equal to total subsegment
        // of size i then we will get the
        // desired val for that subsegment
        for (int j = 1; j <= N; j++) {
            if (res[j] == (N - i + 1)) {
                ans[i] = j;
                break;
            }
            res[j] = 0;
        }
    }

    // Final array
    return ans;
}

// Print Function
void print(vector<int> vec, int N)
{
    vector<int> ans
        = Calculate_Min_array(vec, N);

    // Output
    for (int i = 1; i <= N; i++)
        cout << ans[i] << " ";
    cout << "\n";
}

// Driver code
int main()
{
    // Initialization of array
    vector<int> A = { 1, 2, 3 };
    int N = 3;

    // Calling function
    print(A, N);
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
import java.util.*;

class GFG {

  // Function to calculate
  // the minimum element array
  // from size 1 to N
  public static int[] Calculate_Min_array(int A[], int N)
  {
    // Minimum element array(ans)
    // Count array for every subsegment(cnt)
    // Total occurrence array in all
    // subsegment of a given size(res)
    int ans[] = new int[N + 1];
    int cnt[] = new int[N + 1];
    int res[] = new int[N + 1];

    for (int i = 0; i < N + 1; i++) {
      ans[i] = -1;
    }
    for (int i = 1; i <= N; i++) {

      // Counting all the elements
      // for every subsegment of size i
      for (int j = 0; j < N - i + 1; j++) {
        for (int k = j; k < j + i; k++) {
          cnt[A[k]]++;
        }

        // If count of element is
        // greater than 0 then
        // increment its occurrence
        for (int k = 1; k <= N; k++) {
          if (cnt[k] != 0) {

            // If element is present
            // increase its count
            res[k]++;
            cnt[k] = 0;
          }
        }
      }

      // When occurrence of an element
      // is equal to total subsegment
      // of size i then we will get the
      // desired val for that subsegment
      for (int j = 1; j <= N; j++) {
        if (res[j] == (N - i + 1)) {
          ans[i] = j;
          break;
        }
        res[j] = 0;
      }
    }

    // Final array
    return ans;
  }

  // Print Function
  public static void print(int vec[], int N)
  {
    int ans[] = Calculate_Min_array(vec, N);

    // Output
    for (int i = 1; i <= N; i++)
      System.out.print(ans[i] + " ");
    System.out.println();
  }
  public static void main(String[] args)
  {
    int A[] = { 1, 2, 3 };
    int N = 3;

    // Calling function
    print(A, N);
  }
}

// This code is contributed by Rohit Pradhan

Python3

# Python code for the above approach

# Function to calculate
# the minimum element array
# from size 1 to N
def Calculate_Min_array(A, N):

    # Minimum element array(ans)
    # Count array for every subsegment(cnt)
    # Total occurrence array in all
    # subsegment of a given size(res)
    ans = [-1 for i in range(N + 1)]
    cnt = [0 for i in range(N + 1)]
    res = [0 for i in range(N + 1)]
    for i in range(1, N + 1):

        # Counting all the elements
        # for every subsegment of size i
        for j in range(N - i + 1):
            for k in range(j, j + i):
                cnt[A[k]] = cnt[A[k]] + 1

            # If count of element is
            # greater than 0 then
            # increment its occurrence
            for k in range(1, N + 1):
                if (cnt[k]):

                    # If element is present
                    # increase its count
                    res[k] += 1
                    cnt[k] = 0

        # When occurrence of an element
        # is equal to total subsegment
        # of size i then we will get the
        # desired val for that subsegment
        for j in range(1,N+1):
            if (res[j] == (N - i + 1)):
                ans[i] = j
                break
            res[j] = 0

    # Final array
    return ans

# Print Function
def Print(vec,N):

    ans = Calculate_Min_array(vec, N)

    # Output
    for i in range(1,N+1):
        print(ans[i] ,end = " ")
    print("")

# Driver code

# Initialization of array
A = [ 1, 2, 3 ]
N = 3

# Calling function
Print(A, N)

# This code is contributed by shinjanpatra

C#

// C# code for the above approach
using System;
class GFG {

    // Function to calculate
    // the minimum element array
    // from size 1 to N
    static int[] Calculate_Min_array(int[] A, int N)
    {
        // Minimum element array(ans)
        // Count array for every subsegment(cnt)
        // Total occurrence array in all
        // subsegment of a given size(res)
        int[] ans = new int[N + 1];
        int[] cnt = new int[N + 1];
        int[] res = new int[N + 1];

        for (int i = 0; i < N + 1; i++) {
            ans[i] = -1;
        }
        for (int i = 1; i <= N; i++) {

            // Counting all the elements
            // for every subsegment of size i
            for (int j = 0; j < N - i + 1; j++) {
                for (int k = j; k < j + i; k++) {
                    cnt[A[k]]++;
                }

                // If count of element is
                // greater than 0 then
                // increment its occurrence
                for (int k = 1; k <= N; k++) {
                    if (cnt[k] != 0) {

                        // If element is present
                        // increase its count
                        res[k]++;
                        cnt[k] = 0;
                    }
                }
            }

            // When occurrence of an element
            // is equal to total subsegment
            // of size i then we will get the
            // desired val for that subsegment
            for (int j = 1; j <= N; j++) {
                if (res[j] == (N - i + 1)) {
                    ans[i] = j;
                    break;
                }
                res[j] = 0;
            }
        }

        // Final array
        return ans;
    }

    // Print Function
    static void print(int[] vec, int N)
    {
        int[] ans = Calculate_Min_array(vec, N);

        // Output
        for (int i = 1; i <= N; i++)
            Console.Write(ans[i] + " ");
        Console.WriteLine();
    }
    public static int Main()
    {
        int[] A = new int[] { 1, 2, 3 };
        int N = 3;

        // Calling function
        print(A, N);
        return 0;
    }
}

// This code is contributed by Taranpreet

Javascript

<script>
    // JavaScript code for the above approach

    // Function to calculate
    // the minimum element array
    // from size 1 to N
    const Calculate_Min_array = (A, N) => {

        // Minimum element array(ans)
        // Count array for every subsegment(cnt)
        // Total occurrence array in all
        // subsegment of a given size(res)
        let ans = new Array(N + 1).fill(-1), cnt = new Array(N + 1).fill(0),
            res = new Array(N + 1).fill(0);
        for (let i = 1; i <= N; i++) {

            // Counting all the elements
            // for every subsegment of size i
            for (let j = 0; j < N - i + 1;
                j++) {
                for (let k = j; k < j + i;
                    k++) {
                    cnt[A[k]]++;
                }

                // If count of element is
                // greater than 0 then
                // increment its occurrence
                for (let k = 1; k <= N; k++) {
                    if (cnt[k]) {

                        // If element is present
                        // increase its count
                        res[k]++;
                        cnt[k] = 0;
                    }
                }
            }

            // When occurrence of an element
            // is equal to total subsegment
            // of size i then we will get the
            // desired val for that subsegment
            for (let j = 1; j <= N; j++) {
                if (res[j] == (N - i + 1)) {
                    ans[i] = j;
                    break;
                }
                res[j] = 0;
            }
        }

        // Final array
        return ans;
    }

    // Print Function
    const print = (vec, N) => {
        let ans
            = Calculate_Min_array(vec, N);

        // Output
        for (let i = 1; i <= N; i++)
            document.write(`${ans[i]} `);
        document.write("<br/>");
    }

    // Driver code

    // Initialization of array
    let A = [1, 2, 3];
    let N = 3;

    // Calling function
    print(A, N);

// This code is contributed by rakeshsahni

</script>

Output

-1 2 1 

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

Efficient Approach: The idea to solve the problem efficiently is as follows:

If two consecutive occurrence of a value A[i] is maximum x, then it is part of all the subarrays of size x but not of all subarrays with size less than x. 

Follow the steps below to implement the above idea:

• Create an array (say pos[i]) for each ith element to store the positions of the ith element the array.
• Then calculate the value of the maximum adjacent difference for every element.
• Iterate from i = 1 to N:
  • Start another loop from j = maximum adjacent difference two i to N:
    • If the answer for jth size subarray is not found then i is the minimum common element for all j sized subarray and continue for higher values of j.
    • Otherwise, break from the loop because all the higher values must also be filled
• Return the resultant array.

Below is the implementation of the above approach:

C++

// C++ code for the above approach:

#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// the minimum element array
// from size 1 to N
vector<int> Calculate_Min_array(vector<int>& A,
                                int N)
{

    vector<int> mx(N + 1, -1), ans(N + 1, -1);
    vector<int> pos[N + 1];

    // Inserting the first position
    // of elements
    for (int i = 1; i <= N; i++) {
        pos[i].push_back(0);
    }

    // Inserting the diff position
    // of elements
    for (int i = 0; i < N; i++) {
        int x = A[i];
        pos[x].push_back(i + 1);
    }

    // Inserting the last position
    // of elements
    for (int i = 1; i <= N; i++) {
        pos[i].push_back(N + 1);
    }

    // Calculating max adjacent diff
    // of elements
    for (int i = 1; i <= N; i++) {
        for (int j = 0;
             j < pos[i].size() - 1; j++) {
            mx[i] = max(mx[i],
                        pos[i][j + 1]
                            - pos[i][j]);
        }
    }

    // Calculating ans for every subarray size
    for (int i = 1; i <= N; i++) {
        for (int j = mx[i]; j <= N; j++) {

            // If ans[j] is already present
            // move to next element
            if (ans[j] != -1)
                break;

            // Otherwise store the ans[j]=i
            ans[j] = i;
        }
    }

    // Final array
    return ans;
}

// Print Function
void print(vector<int> A, int N)
{
    // Calculation Minimum element array
    // For Every subsegment length from
    // 1 to N
    vector<int> ans
        = Calculate_Min_array(A, N);

    // Output
    for (int i = 1; i <= N; i++)
        cout << ans[i] << " ";
    cout << "\n";
}

// Driver code
int main()
{
    int N = 3;

    // Initialization of array
    vector<int> A = { 1, 2, 3 };
    print(A, N);
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;
class GFG
{

  // Function to calculate
  // the minimum element array
  // from size 1 to N
  static int[] Calculate_Min_array(int[] A, int N)
  {
    int mx[] = new int[N + 1];
    int ans[] = new int[N + 1];
    int[][] pos = new int[N + 1][N + 1];

    for (int i = 0; i <= N; i++) {
      mx[i] = -1;
      ans[i] = -1;
    }

    HashMap<Integer, Integer> map = new HashMap<>();

    // Inserting the first position
    // of elements
    for (int i = 1; i <= N; i++) {
      pos[i][0] = 0;
      map.put(i, 1);
    }

    // Inserting the diff position
    // of elements
    for (int i = 0; i < N; i++) {
      int x = A[i];
      int ind = map.get(x);
      pos[x][ind] = i + 1;
      map.put(x, ++ind);
    }

    // Inserting the last position
    // of elements
    for (int i = 1; i <= N; i++) {
      int ind = map.get(i);
      pos[i][ind] = N + 1;
      map.put(i, ++ind);
    }

    // Calculating max adjacent diff
    // of elements
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j < map.get(i) - 1; j++) {
        mx[i]  = Math.max(mx[i], pos[i][j + 1]
                          - pos[i][j]);
      }
    }

    // Calculating ans for every subarray size
    for (int i = 1; i <= N; i++) {
      for (int j = mx[i]; j <= N; j++) {

        // If ans[j] is already present
        // move to next element
        if (ans[j] != -1)
          break;

        // Otherwise store the ans[j]=i
        ans[j] = i;
      }
    }

    // Final array
    return ans;
  }

  // Print Function
  static void print(int[] A, int N)
  {

    // Calculation Minimum element array
    // For Every subsegment length from
    // 1 to N
    int[] ans = Calculate_Min_array(A, N);

    // Output
    for (int i = 1; i <= N; i++)
      System.out.print(ans[i] + " ");
  }

  // Driver code
  public static void main(String[] args)
  {

    // Initialization of array
    int N = 3;
    int A[] = { 1, 2, 3 };

    // Driver code
    print(A, N);
  }
}

// This code is contributed by phasing17

Python3

# Python code for the above approach:

# Function to calculate
# the minimum element array
# from size 1 to N
def Calculate_Min_array(A, N):
    mx, ans, pos = [-1 for i in range(N + 1)] , [-1 for i in range(N + 1)] , [[] for i in range(N + 1)]

    # Inserting the first position
    # of elements
    for i in range(1, N + 1):
        pos[i].append(0)

    # Inserting the diff position
    # of elements
    for i in range(N):
        x = A[i]
        pos[x].append(i + 1)

    # Inserting the last position
    # of elements
    for i in range(1, N + 1):
        pos[i].append(N + 1)

    # Calculating max adjacent diff
    # of elements
    for i in range(1, N + 1):
        for j in range(len(pos[i]) - 1):
            mx[i] = max(mx[i], pos[i][j + 1] - pos[i][j])

    # Calculating ans for every subarray size
    for i in range(1, N + 1):
        for j in range(mx[i], N + 1):

            # If ans[j] is already present
            # move to next element
            if (ans[j] != -1):
                break

            # Otherwise store the ans[j]=i
            ans[j] = i

    # Final array
    return ans

# Print Function
def Print(A, N):

    # Calculation Minimum element array
    # For Every subsegment length from
    # 1 to N
    ans = Calculate_Min_array(A, N)

    # Output
    for i in range(1, N + 1):
        print(ans[i], end = " ")
    print("")

# Driver code
N = 3

# Initialization of array
A = [ 1, 2, 3 ]
Print(A, N)

# This code is contributed by shinjanpatra

C#

// C# code for the above approach:
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to calculate
  // the minimum element array
  // from size 1 to N
  static int[] Calculate_Min_array(int[] A, int N)
  {
    int []mx = new int[N + 1];
    int []ans = new int[N + 1];
    int[,] pos = new int[N + 1,N + 1];

    for (int i = 0; i <= N; i++) {
      mx[i] = -1;
      ans[i] = -1;
    }

    Dictionary<int, int> map = new Dictionary<int, int>();

    // Inserting the first position
    // of elements
    for (int i = 1; i <= N; i++) {
      pos[i,0] = 0;
      map.Add(i, 1);
    }

    // Inserting the diff position
    // of elements
    for (int i = 0; i < N; i++) {
      int x = A[i];
      int ind = map[x];
      pos[x,ind] = i + 1;
      map[x] =  map[x] + ind++;
    }

    // Inserting the last position
    // of elements
    for (int i = 1; i <= N; i++) {
      int ind = map[i];
      pos[i,ind] = N + 1;
      map[i] =  map[i] + ind++;

    }

    // Calculating max adjacent diff
    // of elements
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j < map[i] - 1; j++) {
        mx[i]  = Math.Max(mx[i], pos[i,j + 1]
                          - pos[i,j]);
      }
    }

    // Calculating ans for every subarray size
    for (int i = 1; i <= N; i++) {
      for (int j = mx[i]; j <= N; j++) {

        // If ans[j] is already present
        // move to next element
        if (ans[j] != -1)
          break;

        // Otherwise store the ans[j]=i
        ans[j] = i;
      }
    }

    // Final array
    return ans;
  }

  // Print Function
  static void print(int[] A, int N)
  {

    // Calculation Minimum element array
    // For Every subsegment length from
    // 1 to N
    int[] ans = Calculate_Min_array(A, N);

    // Output
    for (int i = 1; i <= N; i++)
      Console.Write(ans[i] + " ");
  }

  // Driver code
  public static void Main(String[] args)
  {

    // Initialization of array
    int N = 3;
    int []A = { 1, 2, 3 };

    // Driver code
    print(A, N);
  }
}


// This code contributed by shikhasingrajput

Javascript

<script>

// JavaScript code for the above approach:

// Function to calculate
// the minimum element array
// from size 1 to N
function Calculate_Min_array(A,N)
{

    let mx = new Array(N + 1).fill(-1);
    let ans = new Array(N + 1).fill(-1);
    let pos = new Array(N + 1);
    for(let i=0;i<N + 1;i++){
        pos[i] = new Array();
    }

    // Inserting the first position
    // of elements
    for (let i = 1; i <= N; i++) {

        pos[i].push(0);
    }

    // Inserting the diff position
    // of elements
    for (let i = 0; i < N; i++) {
        let x = A[i];
        pos[x].push(i + 1);
    }

    // Inserting the last position
    // of elements
    for (let i = 1; i <= N; i++) {
        pos[i].push(N + 1);
    }

    // Calculating max adjacent diff
    // of elements
    for (let i = 1; i <= N; i++) {
        for (let j = 0;
            j < pos[i].length - 1; j++) {
            mx[i] = Math.max(mx[i],
                        pos[i][j + 1]
                            - pos[i][j]);
        }
    }

    // Calculating ans for every subarray size
    for (let i = 1; i <= N; i++) {
        for (let j = mx[i]; j <= N; j++) {

            // If ans[j] is already present
            // move to next element
            if (ans[j] != -1)
                break;

            // Otherwise store the ans[j]=i
            ans[j] = i;
        }
    }

    // Final array
    return ans;
}

// Print Function
function print(A,N)
{
    // Calculation Minimum element array
    // For Every subsegment length from
    // 1 to N
    let ans = Calculate_Min_array(A, N);

    // Output
    for (let i = 1; i <= N; i++){
        document.write(ans[i]," ");
    }
    document.write("</br>");
}

// Driver code
let N = 3;

// Initialization of array
let A = [ 1, 2, 3 ];
print(A, N)

// This code is contributed by shinjanpatra

</script>

Output

-1 2 1 

Time complexity: O(N) Because though nested loops are used but at most N points are filled and one point is visited at most twice
Auxiliary Space: O(N) Though array of vectors is used but total points stored is N