Find minimum subarray length to reduce frequency
Last Updated : 22 Aug, 2023
Given an array arr[] of length N and a positive integer k, the task is to find the minimum length of the subarray that needs to be removed from the given array such that the frequency of the remaining elements in the array is less than or equal to k.
Examples:
Input: n = 4, arr[] = {3, 1, 3, 6}, k = 1
Output: 1
Explanation: We can see that only 3 is having frequency 2 that is greater than k. So we can remove the 3 at the start of the array to that frequency of all elements become less than or equal to k. Thus the minimum length of the subarray to be removed is 1.
Input: n = 6, arr[] = {1, 2, 3, 3, 2, 1}, k = 1
Output: 3
Explanation: We Can remove the subarray {1, 2, 3} which is the minimum possible length subarray after being removed making the frequency of remaining elements less than or equal to k.
Approach: This can be solved with the following idea:
This problem can be solved using Binary Search and Sliding Window Technique.
Below are the steps involved in the implementation of the code:
- Create a Hash Table to store the frequency of the elements of the given array.
- Create a variable cnt to store the count of numbers to be removed for satisfying the condition that the frequency of remaining elements after removing the subarray becomes less than or equal to k.
- Store the frequency of the array elements in the Hash table.
- If the frequency of an element becomes greater than k then increment cnt.
- Now the subarray size lies in the range [0, n] where n is the array length. So we can apply binary search to it.
- Initialize the answer as n.
- Find the mid element on every iteration of binary search and create a window of size mid and check if it is possible that removing this window size from the array can make cnt equal to zero.
- If it is possible update the answer as a minimum of answer and mid and now reduce the search space to [left, mid-1] to find if a smaller length is possible.
- If it is not possible then reduce the search space to [mid+1, r] to find for larger length than mid.
- After the end of the binary search, we will get the required answer.
Below is the implementation of the above approach:
C++ #include <bits/stdc++.h> using namespace std; // Function to find minimum length // needed to be removed int minLength(int n, int k, vector<int>& arr) { // Map to keep the count of // Frequency unordered_map<int, int> f; int cnt = 0; for (int i = 0; i < n; i++) { f[arr[i]]++; if (f[arr[i]] > k) cnt++; } int l = 0, r = n; int ans = n; // Binary search while (l <= r) { int mid = l + (r - l) / 2; for (int i = 0; i < mid; i++) { if (f[arr[i]] > k) cnt--; f[arr[i]]--; } if (cnt == 0) { ans = min(ans, mid); r = mid - 1; for (int i = 0; i < mid; i++) { f[arr[i]]++; if (f[arr[i]] > k) cnt++; } continue; } bool flag = false; // If particular length is possible for (int i = mid; i < n; i++) { if (f[arr[i - mid]] >= k) cnt++; f[arr[i - mid]]++; if (f[arr[i]] > k) cnt--; f[arr[i]]--; if (cnt == 0) { for (int j = i - mid + 1; j <= i; j++) { f[arr[j]]++; if (f[arr[j]] > k) cnt++; } flag = true; break; } } if (flag) { ans = min(ans, mid); r = mid - 1; } else { l = mid + 1; for (int i = n - mid; i < n; i++) { f[arr[i]]++; if (f[arr[i]] > k) cnt++; } } } // Return the ans return ans; } // Driver code int main() { int n = 4; int k = 1; vector<int> arr = { 3, 1, 2, 3 }; // Function call cout << minLength(n, k, arr); return 0; } import java.util.HashMap; import java.util.Map; import java.util.Vector; class GFG { // Function to find minimum length // needed to be removed static int minLength(int n, int k, Vector<Integer> arr) { // Map to keep the count of Frequency Map<Integer, Integer> f = new HashMap<>(); int cnt = 0; for (int i = 0; i < n; i++) { f.put(arr.get(i), f.getOrDefault(arr.get(i), 0) + 1); if (f.get(arr.get(i)) > k) cnt++; } int l = 0, r = n; int ans = n; // Binary search while (l <= r) { int mid = l + (r - l) / 2; for (int i = 0; i < mid; i++) { if (f.get(arr.get(i)) > k) cnt--; f.put(arr.get(i), f.get(arr.get(i)) - 1); } if (cnt == 0) { ans = Math.min(ans, mid); r = mid - 1; for (int i = 0; i < mid; i++) { f.put(arr.get(i), f.get(arr.get(i)) + 1); if (f.get(arr.get(i)) > k) cnt++; } continue; } boolean flag = false; // If particular length is possible for (int i = mid; i < n; i++) { if (f.get(arr.get(i - mid)) >= k) cnt++; f.put(arr.get(i - mid), f.get(arr.get(i - mid)) + 1); if (f.get(arr.get(i)) > k) cnt--; f.put(arr.get(i), f.get(arr.get(i)) - 1); if (cnt == 0) { for (int j = i - mid + 1; j <= i; j++) { f.put(arr.get(j), f.get(arr.get(j)) + 1); if (f.get(arr.get(j)) > k) cnt++; } flag = true; break; } } if (flag) { ans = Math.min(ans, mid); r = mid - 1; } else { l = mid + 1; for (int i = n - mid; i < n; i++) { f.put(arr.get(i), f.get(arr.get(i)) + 1); if (f.get(arr.get(i)) > k) cnt++; } } } // Return the ans return ans; } // Nikunj Sonigara public static void main(String[] args) { int n = 4; int k = 1; Vector<Integer> arr = new Vector<>(4); arr.add(3); arr.add(1); arr.add(2); arr.add(3); // Function call System.out.println(minLength(n, k, arr)); } } # Python code for the above approach from typing import List from collections import defaultdict # Function to find minimum length needed to be removed def min_length(n: int, k: int, arr: List[int]) -> int: # Dictionary to keep the count of frequency f = defaultdict(int) cnt = 0 for i in range(n): f[arr[i]] += 1 if f[arr[i]] > k: cnt += 1 l = 0 r = n ans = n # Binary search while l <= r: mid = l + (r - l) // 2 for i in range(mid): if f[arr[i]] > k: cnt -= 1 f[arr[i]] -= 1 if cnt == 0: ans = min(ans, mid) r = mid - 1 for i in range(mid): f[arr[i]] += 1 if f[arr[i]] > k: cnt += 1 continue flag = False # If a particular length is possible for i in range(mid, n): if f[arr[i - mid]] >= k: cnt += 1 f[arr[i - mid]] += 1 if f[arr[i]] > k: cnt -= 1 f[arr[i]] -= 1 if cnt == 0: for j in range(i - mid + 1, i + 1): f[arr[j]] += 1 if f[arr[j]] > k: cnt += 1 flag = True break if flag: ans = min(ans, mid) r = mid - 1 else: l = mid + 1 for i in range(n - mid, n): f[arr[i]] += 1 if f[arr[i]] > k: cnt += 1 # Return the answer return ans # Driver code if __name__ == "__main__": n = 4 k = 1 arr = [3, 1, 2, 3] # Function call print(min_length(n, k, arr))
// c# code for the above approach using System; using System.Collections.Generic; class GFG { // Function to find minimum length // needed to be removed static int MinLength(int n, int k, List<int> arr) { // Dictionary to keep the count of Frequency Dictionary<int, int> f = new Dictionary<int, int>(); int cnt = 0; for (int i = 0; i < n; i++) { if (f.ContainsKey(arr[i])) f[arr[i]]++; else f[arr[i]] = 1; if (f[arr[i]] > k) cnt++; } int l = 0, r = n; int ans = n; // Binary search while (l <= r) { int mid = l + (r - l) / 2; for (int i = 0; i < mid; i++) { if (f[arr[i]] > k) cnt--; f[arr[i]]--; } if (cnt == 0) { ans = Math.Min(ans, mid); r = mid - 1; for (int i = 0; i < mid; i++) { f[arr[i]]++; if (f[arr[i]] > k) cnt++; } continue; } bool flag = false; // If a particular length is possible for (int i = mid; i < n; i++) { if (f[arr[i - mid]] >= k) cnt++; f[arr[i - mid]]++; if (f[arr[i]] > k) cnt--; f[arr[i]]--; if (cnt == 0) { for (int j = i - mid + 1; j <= i; j++) { f[arr[j]]++; if (f[arr[j]] > k) cnt++; } flag = true; break; } } if (flag) { ans = Math.Min(ans, mid); r = mid - 1; } else { l = mid + 1; for (int i = n - mid; i < n; i++) { f[arr[i]]++; if (f[arr[i]] > k) cnt++; } } } // Return the ans return ans; } // Nikunj Sonigara public static void Main(string[] args) { int n = 4; int k = 1; List<int> arr = new List<int> { 3, 1, 2, 3 }; // Function call Console.WriteLine(MinLength(n, k, arr)); } } //JavaScript code function minLength(n, k, arr) { const f = new Map(); let cnt = 0; for (let i = 0; i < n; i++) { if (!f.has(arr[i])) { f.set(arr[i], 1); } else { f.set(arr[i], f.get(arr[i]) + 1); if (f.get(arr[i]) > k) { cnt++; } } } let l = 0, r = n; let ans = n; while (l <= r) { const mid = l + Math.floor((r - l) / 2); for (let i = 0; i < mid; i++) { if (f.get(arr[i]) > k) { cnt--; } f.set(arr[i], f.get(arr[i]) - 1); } if (cnt === 0) { ans = Math.min(ans, mid); r = mid - 1; for (let i = 0; i < mid; i++) { f.set(arr[i], f.get(arr[i]) + 1); if (f.get(arr[i]) > k) { cnt++; } } continue; } let flag = false; for (let i = mid; i < n; i++) { if (f.get(arr[i - mid]) >= k) { cnt++; } f.set(arr[i - mid], f.get(arr[i - mid]) + 1); if (f.get(arr[i]) > k) { cnt--; } f.set(arr[i], f.get(arr[i]) - 1); if (cnt === 0) { for (let j = i - mid + 1; j <= i; j++) { f.set(arr[j], f.get(arr[j]) + 1); if (f.get(arr[j]) > k) { cnt++; } } flag = true; break; } } if (flag) { ans = Math.min(ans, mid); r = mid - 1; } else { l = mid + 1; for (let i = n - mid; i < n; i++) { f.set(arr[i], f.get(arr[i]) + 1); if (f.get(arr[i]) > k) { cnt++; } } } } return ans; } // Driver code const n = 4; const k = 1; const arr = [3, 1, 2, 3]; // Function call console.log(minLength(n, k, arr)); //Contributed by Aditi Tyagi Time Complexity: O(N*logN)
Auxiliary Space: O(N)
Similar Reads
Binary Search on Answer Tutorial with Problems Binary Search on Answer is the algorithm in which we are finding our answer with the help of some particular conditions. We have given a search space in which we take an element [mid] and check its validity as our answer, if it satisfies our given condition in the problem then we store its value and
15+ min read
Time Crunch Challenge Geeks for Geeks is organizing a hackathon consisting of N sections, each containing K questions. The duration of the hackathon is H hours. Each participant can determine their speed of solving questions, denoted as S (S = questions-per-hour). During each hour, a participant can choose a section and
10 min read
Find minimum subarray length to reduce frequency Given an array arr[] of length N and a positive integer k, the task is to find the minimum length of the subarray that needs to be removed from the given array such that the frequency of the remaining elements in the array is less than or equal to k. Examples: Input: n = 4, arr[] = {3, 1, 3, 6}, k =
10 min read
Maximize minimum sweetness in cake cutting Given that cake consists of N chunks, whose individual sweetness is represented by the sweetness[] array, the task is to cut the cake into K + 1 pieces to maximize the minimum sweetness. Examples: Input: N = 6, K = 2, sweetness[] = {6, 3, 2, 8, 7, 5}Output: 9Explanation: Divide the cake into [6, 3],
7 min read
Maximize minimum element of an Array using operations Given an array A[] of N integers and two integers X and Y (X ≤ Y), the task is to find the maximum possible value of the minimum element in an array A[] of N integers by adding X to one element and subtracting Y from another element any number of times, where X ≤ Y. Examples: Input: N= 3, A[] = {1,
8 min read
Maximize the minimum element and return it Given an array A[] of size N along with W, K. We can increase W continuous elements by 1 where we are allowed to perform this operation K times, the task is to maximize the minimum element and return the minimum element after operations. Examples: Input: N = 6, K = 2, W = 3, A[] = {2, 2, 2, 2, 1, 1}
8 min read
Find the maximum value of the K-th smallest usage value in Array Given an array arr[] of size N and with integers M, K. You are allowed to perform an operation where you can increase the value of the least element of the array by 1. You are to do this M times. The task is to find the largest possible value for the Kth smallest value among arr[] after M operations
7 min read
Aggressive Cows Given an array stalls[] which denotes the position of a stall and an integer k which denotes the number of aggressive cows. The task is to assign stalls to k cows such that the minimum distance between any two of them is the maximum possible.Examples: Input: stalls[] = [1, 2, 4, 8, 9], k = 3Output:
15+ min read
Minimum time to complete at least K tasks when everyone rest after each task Given an array arr[] of size n representing the time taken by a person to complete a task. Also, an array restTime[] which denotes the amount of time one person takes to rest after finishing a task. Each person is independent of others i.e. they can work simultaneously on different tasks at the same
8 min read
Maximum count of integers to be chosen from given two stacks having sum at most K Given two stacks stack1[] and stack2[] of size N and M respectively and an integer K, The task is to count the maximum number of integers from two stacks having sum less than or equal to K. Examples: Input: stack1[ ] = { 60, 90, 120 }stack2[ ] = { 100, 10, 10, 250 }, K = 130Output: 3Explanation: Tak
9 min read