Minimum common element in subarrays of all possible lengths

• Last Updated : 20 May, 2021

Given an array arr[] consisting of N integers from the range [1, N]( repetition allowed ), the task is to find the minimum common element for each possible subarray length. If no such element exists for any particular length of the subarray, then print -1.

Examples:

Input: arr[] = {1, 3, 4, 5, 6, 7}
Output: -1 -1 -1 4 3 1
Explanation:
K = 1: No common element exists. Therefore, print -1.
K = 2: No common element exists. Therefore, print -1.
K = 3: No common element exists. Therefore, print -1.
K = 4: Since 4 is common in all subarrays of size 4, print 4.
K = 5: Since 3 and 4 is common in all subarrays of size 5, print 3 as it is the minimum.
K = 6: Print 1 as it is the minimum element in the array.

Input: arr[]: {1, 2, 2, 2, 1}
Output: -1 2 2 1 1

Approach: Follow the steps below to solve the problem:

• Traverse the array and store the last occurrence of every element in a Map.
• Initialize an array temp[] and store in it for each value, the maximum distance between any pair of consecutive repetitions of it in the array.
• Once the above step is completed, update temp[] by comparing temp[i] with the distance of the last occurrence of i from the end of the array.
• Now, store the minimum comment element for all subarrays of length 1 to N one by one and print them.

Below is the implementation of the above approach:

C++

 // C++ Program to implement the// above approach #include using namespace std; // Function to find maximum distance// between every two elementvoid max_distance(int a[], int temp[], int n){    // Stores index of last occurence    // of each array element    map mp;     // Initialize temp[] with -1    for (int i = 1; i <= n; i++) {        temp[i] = -1;    }     // Traverse the array    for (int i = 0; i < n; i++) {         // If array element has        // not occurred previously        if (mp.find(a[i]) == mp.end())             // Update index in temp            temp[a[i]] = i + 1;         // Otherwise        else             // Compare temp[a[i]] with distance            // from its previous occurence and            // store the maximum            temp[a[i]] = max(temp[a[i]],                             i - mp[a[i]]);         mp[a[i]] = i;    }     for (int i = 1; i <= n; i++) {         // Compare temp[i] with distance        // of its last occurence from the end        // of the array and store the maximum        if (temp[i] != -1)            temp[i] = max(temp[i], n - mp[i]);    }} // Function to find the minimum common// element in subarrays of all possible lengthsvoid min_comm_ele(int a[], int ans[],                  int temp[], int n){    // Function call to find a the maximum    // distance between every pair of repetition    max_distance(a, temp, n);     // Initialize ans[] to -1    for (int i = 1; i <= n; i++) {        ans[i] = -1;    }     for (int i = 1; i <= n; i++) {         // Check if subarray of length        // temp[i] contains i as one        // of the common elements        if (ans[temp[i]] == -1)            ans[temp[i]] = i;    }     for (int i = 1; i <= n; i++) {         // Find the minimum of all        // common elements        if (i > 1 && ans[i - 1] != -1) {             if (ans[i] == -1)                ans[i] = ans[i - 1];            else                ans[i] = min(ans[i],                             ans[i - 1]);        }         cout << ans[i] << " ";    }} // Driver Codeint main(){    int N = 6;    int a[] = { 1, 3, 4, 5, 6, 7 };    int temp, ans;    min_comm_ele(a, ans, temp, N);     return 0;}

Java

 // Java program to implement the// above approachimport java.util.*; class GFG{    // Function to find maximum distance// between every two elementstatic void max_distance(int a[], int temp[], int n){         // Stores index of last occurence    // of each array element    Map mp = new HashMap();     // Initialize temp[] with -1    for(int i = 1; i <= n; i++)    {        temp[i] = -1;    }     // Traverse the array    for(int i = 0; i < n; i++)    {                 // If array element has        // not occurred previously        if (mp.get(a[i]) == null)                     // Update index in temp            temp[a[i]] = i + 1;         // Otherwise        else             // Compare temp[a[i]] with distance            // from its previous occurence and            // store the maximum            temp[a[i]] = Math.max(temp[a[i]],                   i - mp.getOrDefault(a[i], 0));         mp.put(a[i], i);    }     for(int i = 1; i <= n; i++)    {                 // Compare temp[i] with distance        // of its last occurence from the end        // of the array and store the maximum        if (temp[i] != -1)            temp[i] = Math.max(temp[i],                               n - mp.getOrDefault(i, 0));    }} // Function to find the minimum common// element in subarrays of all possible lengthsstatic void min_comm_ele(int a[], int ans[],                         int temp[], int n){         // Function call to find a the maximum    // distance between every pair of repetition    max_distance(a, temp, n);     // Initialize ans[] to -1    for(int i = 1; i <= n; i++)    {        ans[i] = -1;    }     for(int i = 1; i <= n; i++)    {                 // Check if subarray of length        // temp[i] contains i as one        // of the common elements        if (temp[i] >= 0 && ans[temp[i]] == -1)            ans[temp[i]] = i;    }     for(int i = 1; i <= n; i++)    {                 // Find the minimum of all        // common elements        if (i > 1 && ans[i - 1] != -1)        {            if (ans[i] == -1)                ans[i] = ans[i - 1];            else                ans[i] = Math.min(ans[i],                                  ans[i - 1]);        }        System.out.print(ans[i] + " ");    }} // Driver Codepublic static void main(String args[]){    int N = 6;    int a[] = { 1, 3, 4, 5, 6, 7 };         int []temp = new int;    Arrays.fill(temp, 0);         int []ans = new int;    Arrays.fill(ans, 0);         min_comm_ele(a, ans, temp, N);}} // This code is contributed by SURENDRA_GANGWAR

Python3

 # Python3 Program to implement# the above approach # Function to find maximum# distance between every# two elementdef max_distance(a, temp, n):     # Stores index of last    # occurence of each    # array element    mp = {}     # Initialize temp[]    # with -1    for i in range(1, n + 1):        temp[i] = -1     # Traverse the array    for i in range(n):         # If array element has        # not occurred previously        if (a[i] not in mp):             # Update index in temp            temp[a[i]] = i + 1         # Otherwise        else:             # Compare temp[a[i]] with            # distance from its previous            # occurence and store the maximum            temp[a[i]] = max(temp[a[i]],                             i - mp[a[i]])         mp[a[i]] = i     for i in range(1, n + 1):         # Compare temp[i] with        # distance of its last        # occurence from the end        # of the array and store        # the maximum        if (temp[i] != -1):            temp[i] = max(temp[i],                          n - mp[i]) # Function to find the minimum# common element in subarrays# of all possible lengthsdef min_comm_ele(a, ans,                 temp, n):     # Function call to find    # a the maximum distance    # between every pair of    # repetition    max_distance(a, temp, n)     # Initialize ans[] to -1    for i in range(1, n + 1):        ans[i] = -1     for i in range(1, n + 1):         # Check if subarray of length        # temp[i] contains i as one        # of the common elements        if (ans[temp[i]] == -1):            ans[temp[i]] = i     for i in range(1, n + 1):         # Find the minimum of all        # common elements        if (i > 1 and            ans[i - 1] != -1):             if (ans[i] == -1):                ans[i] = ans[i - 1]            else:                ans[i] = min(ans[i],                             ans[i - 1])         print(ans[i], end = " ") # Driver Codeif __name__ == "__main__":     N = 6    a = [1, 3, 4, 5, 6, 7]    temp =  * 100    ans =  * 100    min_comm_ele(a, ans,                 temp, N) # This code is contributed by Chitranayal

C#

 // C# program to implement the// above approachusing System;using System.Collections.Generic;class GFG {         // Function to find maximum distance    // between every two element    static void max_distance(int[] a, int[] temp, int n)    {                  // Stores index of last occurence        // of each array element        Dictionary mp = new Dictionary();               // Initialize temp[] with -1        for(int i = 1; i <= n; i++)        {            temp[i] = -1;        }              // Traverse the array        for(int i = 0; i < n; i++)        {                          // If array element has            // not occurred previously            if (!mp.ContainsKey(a[i]))                              // Update index in temp                temp[a[i]] = i + 1;                  // Otherwise            else                      // Compare temp[a[i]] with distance                // from its previous occurence and                // store the maximum                temp[a[i]] = Math.Max(temp[a[i]], i - mp[a[i]]);                         if(mp.ContainsKey(a[i]))            {                mp[a[i]] = i;            }            else{                mp.Add(a[i], i);            }        }              for(int i = 1; i <= n; i++)        {                          // Compare temp[i] with distance            // of its last occurence from the end            // of the array and store the maximum            if (temp[i] != -1)            {                if(mp.ContainsKey(i))                {                    temp[i] = Math.Max(temp[i], n - mp[i]);                }                else{                    temp[i] = Math.Max(temp[i], n);                }            }        }    }          // Function to find the minimum common    // element in subarrays of all possible lengths    static void min_comm_ele(int[] a, int[] ans,                             int[] temp, int n)    {                  // Function call to find a the maximum        // distance between every pair of repetition        max_distance(a, temp, n);              // Initialize ans[] to -1        for(int i = 1; i <= n; i++)        {            ans[i] = -1;        }              for(int i = 1; i <= n; i++)        {                          // Check if subarray of length            // temp[i] contains i as one            // of the common elements            if (temp[i] >= 0 && ans[temp[i]] == -1)                ans[temp[i]] = i;        }              for(int i = 1; i <= n; i++)        {                          // Find the minimum of all            // common elements            if (i > 1 && ans[i - 1] != -1)            {                if (ans[i] == -1)                    ans[i] = ans[i - 1];                else                    ans[i] = Math.Min(ans[i],                                      ans[i - 1]);            }            Console.Write(ans[i] + " ");        }    }   // Driver code  static void Main()  {               int N = 6;        int[] a = { 1, 3, 4, 5, 6, 7 };                  int[] temp = new int;        Array.Fill(temp, 0);                  int[] ans = new int;        Array.Fill(ans, 0);                  min_comm_ele(a, ans, temp, N);  }} // This code is contributed by divyeshrabadiya07

Javascript


Output:
-1 -1 -1 4 3 1

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

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