Length of smallest Subarray with at least one element repeated K times
Last Updated :
02 Nov, 2023
Given an array arr[] of length N and an integer K. The task is to find the minimum length of subarray such that at least one element of the subarray is repeated exactly K times in that subarray. If no such subarray exists, print -1.
Examples:
Input: arr[] = {1, 2, 1, 2, 1}, K = 2
Output: 3
Explanation: Subarray [1,2,1], have K = 2 occurrences of 1
Input: arr[] = {2, 2, 2, 3, 4}, K = 3
Output: 3
Brute Force/Naive Approach: Brute force approach is to iterate over all possible subarrays of the array, and for each subarray, count the occurrence of each element. If an element occurs exactly K times, then the length of the subarray is a candidate for the minimum length subarray.
- Initialize a variable mn to INT_MAX, which will store the minimum length subarray.
- Loop through each starting index i in the array, from 0 to N-1.
- Initialize a variable count to 0, which will count the occurrence of the element arr[i].
- Loop through each ending index j in the array, from i to N-1.
- If the current element arr[j] is the same as arr[i], increment the count.
- If the count is equal to K, update mn to the minimum value between mn and j-i+1 (the length of the current subarray).
- Continue looping until all subarrays with starting index i have been checked.
- After looping through all starting indices, if mn is still equal to INT_MAX, there is no subarray with an element occurring exactly K times. Return -1.
- Otherwise, return mn as the length of the minimum length subarray with an element occurring exactly K times.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum length of
// subarray having an element exactly K times
int minLengthKRepetitions(int* arr, int& N, int& K)
{
int mn = INT_MAX;
for (int i = 0; i < N; i++) {
int count = 0;
for (int j = i; j < N; j++) {
if (arr[j] == arr[i]) {
count++;
}
if (count == K) {
mn = min(mn, j - i + 1);
break;
}
}
}
return (mn == INT_MAX) ? -1 : mn;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << minLengthKRepetitions(arr, N, K);
return 0;
}
import java.util.*;
public class Main {
// Function to return the minimum length of
// subarray having an element exactly K times
public static int minLengthKRepetitions(int[] arr, int N, int K) {
int mn = Integer.MAX_VALUE;
for (int i = 0; i < N; i++) {
int count = 0;
for (int j = i; j < N; j++) {
if (arr[j] == arr[i]) {
count++;
}
if (count == K) {
mn = Math.min(mn, j - i + 1);
break;
}
}
}
return (mn == Integer.MAX_VALUE) ? -1 : mn;
}
// Driver code
public static void main(String[] args) {
int[] arr = { 1, 2, 2, 2, 1 };
int N = arr.length;
int K = 3;
System.out.println(minLengthKRepetitions(arr, N, K));
}
}
// This code is cntributed by shivregkec
import sys
# Function to return the minimum length of
# subarray having an element exactly K times
def minLengthKRepetitions(arr, N, K):
mn = sys.maxsize
for i in range(N):
count = 0
for j in range(i, N):
if arr[j] == arr[i]:
count += 1
if count == K:
mn = min(mn, j - i + 1)
break
return -1 if mn == sys.maxsize else mn
# Driver code
arr = [1, 2, 2, 2, 1]
N = len(arr)
K = 3
print(minLengthKRepetitions(arr, N, K))
# This code is contributed by shivhack999
using System;
public class MainClass {
// Function to return the minimum length of
// subarray having an element exactly K times
public static int
MinLengthKRepetitions(int[] arr, ref int N, ref int K)
{
int mn = int.MaxValue;
for (int i = 0; i < N; i++) {
int count = 0;
for (int j = i; j < N; j++) {
if (arr[j] == arr[i]) {
count++;
}
if (count == K) {
mn = Math.Min(mn, j - i + 1);
break;
}
}
}
return (mn == int.MaxValue) ? -1 : mn;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 2, 2, 1 };
int N = arr.Length;
int K = 3;
Console.WriteLine(
MinLengthKRepetitions(arr, ref N, ref K));
}
}
// This code is contributed by user_dtewbxkn77n
// Function to return the minimum length of
// subarray having an element exactly K times
function minLengthKRepetitions(arr, K) {
let mn = Number.MAX_VALUE;
const N = arr.length;
for (let i = 0; i < N; i++) {
let count = 0;
for (let j = i; j < N; j++) {
if (arr[j] === arr[i]) {
count++;
}
if (count === K) {
mn = Math.min(mn, j - i + 1);
break;
}
}
}
return mn === Number.MAX_VALUE ? -1 : mn;
}
// Driver code
const arr = [1, 2, 2, 2, 1];
const K = 3;
console.log(minLengthKRepetitions(arr, K));
Output:
3
Time Complexity: O(N*N)
Auxiliary Space: O(1)
Efficient Approach: In this problem, observe that the smallest length will be achieved when we have exactly one element in the subarray which has K frequency which means subarray will look something like [X . . . X] where X is an element of array arr. Now, follow the steps below to solve this problem:
- Create an array of pairs, such that the number (i.e. arr[i]) is the first element and its index (i.e. i) is second.
- Sort this array.
- Now, create a variable mn to store the answer and initialize it with INT_MAX.
- Now, traverse the array from i = 0 to i = (N - K) and in each iteration:
- If the element at i and (i+K-1) is equal then make mn equal to the minimum out of mn and the difference between the indexes of the following.
- Return mn as the final answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum length of
// subarray having an element exactly K times
int minLengthKRepetitions(int* arr, int& N,
int& K)
{
pair<int, int> indices[N];
int mn = INT_MAX, i;
for (i = 0; i < N; i++) {
indices[i].first = arr[i];
indices[i].second = i;
}
sort(indices, indices + N);
for (i = 0; i <= N - K; i++) {
if (indices[i].first == indices[i + K - 1].first)
mn = min(mn, indices[i + K - 1].second
- indices[i].second + 1);
}
return (mn == INT_MAX) ? -1 : mn;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << minLengthKRepetitions(arr, N, K);
return 0;
}
// Java code to implement the above approach
import java.util.*;
class GFG {
// Function to return the minimum length of
// subarray having an element exactly K times
public static int minLengthKRepetitions(int[] arr, int N, int K)
{
int[][] indices = new int[N][2] ;
int mn = Integer.MAX_VALUE, i;
for (i = 0; i < N; i++) {
indices[i][0] = arr[i];
indices[i][1] = i;
}
//Arrays.sort(indices);
for (i = 0; i <= N - K; i++) {
if (indices[i][0] == indices[i + K - 1][0])
mn = Math.min(mn, indices[i + K - 1][1]
- indices[i][1] + 1);
}
return (mn == Integer.MAX_VALUE) ? -1 : mn;
}
// Driver code
public static void main (String[] args)
{
int[] arr = { 1, 2, 2, 2, 1 };
int N = arr.length;
int K = 3;
System.out.println(minLengthKRepetitions(arr, N, K));
}
}
// This code is contributed by Shubham Singh
# Python program for the above approach
# Function to return the minimum length of
# subarray having an element exactly K times
def minLengthKRepetitions(arr, N, K):
indices = [[0,0] for i in range(N)]
mn = 2**32
for i in range(N):
indices[i][0] = arr[i]
indices[i][1] = i
indices.sort()
for i in range(N - K + 1):
if (indices[i][0] == indices[i + K - 1][0]):
mn = min(mn, indices[i + K - 1][1] - indices[i][1] + 1)
return -1 if(mn == 2**32) else mn
# Driver code
arr = [1, 2, 2, 2, 1]
N = len(arr)
K = 3
print(minLengthKRepetitions(arr, N, K))
# This code is contributed by Shubham Singh
// C# code to implement the above approach
using System;
public class GFG
{
// Function to return the minimum length of
// subarray having an element exactly K times
public static int minLengthKRepetitions(int[] arr, int N, int K)
{
int[,] indices = new int[N,2] ;
int mn = Int32.MaxValue, i;
for (i = 0; i < N; i++) {
indices[i, 0] = arr[i];
indices[i, 1] = i;
}
//Arrays.sort(indices);
for (i = 0; i <= N - K; i++) {
if (indices[i,0] == indices[i + K - 1,0])
mn = Math.Min(mn, indices[i + K - 1,1]
- indices[i,1] + 1);
}
return (mn == Int32.MaxValue) ? -1 : mn;
}
// Driver code
static public void Main ()
{
int[] arr = { 1, 2, 2, 2, 1 };
int N = arr.Length;
int K = 3;
Console.Write(minLengthKRepetitions(arr, N, K));
}
}
// This code is contributed by Shubham Singh
<script>
// JavaScript code for the above approach
// Function to return the minimum length of
// subarray having an element exactly K times
function minLengthKRepetitions(arr, N,
K) {
let indices = [];
let mn = Number.MAX_VALUE, i;
for (i = 0; i < N; i++) {
indices.push({ first: arr[i], second: i })
}
indices.sort(function (a, b) { return a.first - b.first; });
for (i = 0; i <= N - K; i++) {
if (indices[i].first == indices[i + K - 1].first)
mn = Math.min(mn, indices[i + K - 1].second
- indices[i].second + 1);
}
return (mn == Number.MAX_VALUE) ? -1 : mn;
}
// Driver code
let arr = [1, 2, 2, 2, 1];
let N = arr.length;
let K = 3;
document.write(minLengthKRepetitions(arr, N, K));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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