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:
- Add the count of every unique number for every subarray of length K.
- Check if the count of numbers is equal to the number of subarrays i.e., N – K – 1.
- If found to be true, then that particular element has occurred in every subarray of size K.
- 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 kvoid 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]++;}// Function to check if there is any number// which repeats itself in every subarray// of length Kvoid 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 occurred 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 Kvoid 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 Kvoid generateK(int arr[], int N){ for (int k = 1; k <= N; k++) // Function call smallestPresentNumber(arr, N, k);}// Driver Codeint 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 approachimport 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); } } // Function 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 occurred 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++) // Function 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 kdef 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;# Function to check if there is any number# which repeats itself in every subarray# of length Kdef checkAnswer(mp, N, K) : # Check all number starting from 1 for i in range(1, N + 1) : if i in mp : # Check if i occurred 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 Kdef 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 Kdef generateK(arr, N) : for k in range(1, N + 1) : # Function call smallestPresentNumber(arr, N, k);# Driver Codeif __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 approachusing 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; } } } // Function 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 occurred 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++) // Function 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. |
Javascript
<script>// Javascript program for the above approach // Function to add count of numbers // in the map for a subarray of length k function uniqueElements(arr, start, K, mp) { // Set to store unique elements let st = new Set(); // add elements to the set for (let i = 0; i < K; i++) st.add(arr[start + i]); // adding count in map for(let itr of st) { if(mp.has(itr)) { mp.set(itr, mp.get(itr) + 1); } else{ mp.set(itr, 1); } } } // Function to check if there is any number // which repeats itself in every subarray // of length K function checkAnswer(mp, N, K) { // Check all number starting from 1 for (let i = 1; i <= N; i++) { // Check if i occurred n-k+1 times if(mp.has(i)) if (mp.get(i) == (N - K + 1)) { // Print the smallest number document.write(i + " "); return; } } // Print -1, if no such number found document.write(-1 + " "); } // Function to count frequency of each // number in each subarray of length K function smallestPresentNumber(arr, N, K) { let mp = new Map(); // Traverse all subarrays of length K for (let 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 function generateK(arr, N) { for (let k = 1; k <= N; k++) // Function call smallestPresentNumber(arr, N, k); } // Driver code // Given array let arr = [ 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 ]; // Size of array let N = arr.length; // Function call generateK(arr, N); // This code is contributed by gfgking</script> |
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:
- Initialize an array, say indices[], to store the index where a particular number occurs corresponding to that index number.
- 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.
- 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.
- Initialize an answer[] array of size N+1 with -1 where answer[i] represents the answer for subarrays of length K.
- 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.
- 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.
- 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 lengthsvoid 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 nvoid 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 Nvoid 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 occurrence and // gaps between occurrences int start = -1; int gap = -1; // To cover the distance between last // occurrence and the end of the array indices[itr].push_back(N); // To find the distance // between any two occurrences 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 lengthsvoid 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 Codeint 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 approachimport 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 occurrence and // gaps between occurrences int start = -1; int gap = -1; int t = (int)itr.next(); // To cover the distance between last // occurrence and the end of the array indices.get(t).add(N); // To find the distance // between any two occurrences 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 |
Python3
# Python program of the above approach# Function to print the common# elements for all subarray lengthsdef 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 ndef 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 Ndef 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 occurrence and # gaps between occurrences start = -1 gap = -1 # To cover the distance between last # occurrence and the end of the array indices[itr].append(N) # To find the distance # between any two occurrences 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 lengthsdef 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 arrayarr = [2, 3, 5, 3, 2, 3, 1, 3, 2, 7]# Size of arrayN = len(arr)# Function CallsmallestPresentNumber(arr, N)# This code is contributed by avanitrachhadiya2155 |
C#
// C# program for above approachusing System;using System.Collections.Generic;class GFG{// Function to print the common// elements for all subarray lengthsstatic 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 nstatic 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 Nstatic 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 occurrence and // gaps between occurrences int start = -1; int gap = -1; int t = itr; // To cover the distance between last // occurrence and the end of the array indices[t].Add(N); // To find the distance // between any two occurrences 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 lengthsstatic 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 codestatic 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 |
Javascript
<script>// Javascript program of the above approach// Function to print the common// elements for all subarray lengthsfunction printAnswer(answer, N) { for(let i = 1; i <= N; i++) { document.write(answer[i] + " "); }}// Function to find and store the// minimum element present in all// subarrays of all lengths from 1 to nfunction updateAnswerArray(answer, N){ let 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 let 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 Nfunction lengthOfSubarray(indices, st, N){ // Stores the minimum common // elements for all subarray lengths let answer = new Array(N + 1).fill(-1); // Find for every element, the minimum length // such that the number is present in every // subsequence of that particular length or more for(let itr of st) { // To store first occurrence and // gaps between occurrences let start = -1; let gap = -1; // To cover the distance between last // occurrence and the end of the array indices[itr].push(N); // To find the distance // between any two occurrences for(let i = 0; i < indices[itr].length; i++) { gap = Math.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 lengthsfunction smallestPresentNumber(arr, N){ // Initializing indices array let indices = new Array(N + 1).fill(0).map(() => []); // Store the numbers present // in the array let elements = new Set(); // Push the index in the indices[A[i]] and // also store the numbers in set to get // the numbers present in input array for(let i = 0; i < N; i++) { indices[arr[i]].push(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// Given arraylet arr = [ 2, 3, 5, 3, 2, 3, 1, 3, 2, 7 ];// Size of arraylet N = arr.length;// Function CallsmallestPresentNumber(arr, N);// This code is contributed by gfgking</script> |
Output:
-1 -1 3 2 2 2 1 1 1 1
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
