Find the smallest subarray having atleast one duplicate
Last Updated :
26 Dec, 2022
Given an array arr of N elements, the task is to find the length of the smallest subarray of the given array that contains at least one duplicate element. A subarray is formed from consecutive elements of an array. If no such array exists, print “-1”.
Examples:
Input: arr = {1, 2, 3, 1, 5, 4, 5}
Output: 3
Explanation:
Input: arr = {4, 7, 11, 3, 1, 2, 4}
Output: 7
Explanation:
Naive Approach:
- The trick is to find all pairs of two elements with equal value. Since these two elements have equal value, the subarray enclosing them would have at least a single duplicate and will be one of the candidates for the answer.
- A simple solution is to use two nested loops to find every pair of elements.If the two elements are equal then update the maximum length obtained so far.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int subArrayLength( int arr[], int n)
{
int minLen = INT_MAX;
for ( int i = 1; i < n; i++) {
for ( int j = 0; j < i; j++) {
if (arr[i] == arr[j]) {
minLen = min(minLen, i - j + 1);
}
}
}
if (minLen == INT_MAX) {
return -1;
}
return minLen;
}
int main()
{
int n = 7;
int arr[] = { 1, 2, 3, 1, 5, 4, 5 };
int ans = subArrayLength(arr, n);
cout << ans << '\n' ;
return 0;
}
|
Java
class GFG
{
final static int INT_MAX = Integer.MAX_VALUE;
static int subArrayLength( int arr[], int n)
{
int minLen = INT_MAX;
for ( int i = 1 ; i < n; i++)
{
for ( int j = 0 ; j < i; j++)
{
if (arr[i] == arr[j])
{
minLen = Math.min(minLen, i - j + 1 );
}
}
}
if (minLen == INT_MAX)
{
return - 1 ;
}
return minLen;
}
public static void main(String[] args)
{
int n = 7 ;
int arr[] = { 1 , 2 , 3 , 1 , 5 , 4 , 5 };
int ans = subArrayLength(arr, n);
System.out.println(ans);
}
}
|
Python
n = 7
arr = [ 1 , 2 , 3 , 1 , 5 , 4 , 5 ]
minLen = n + 1
for i in range ( 1 , n):
for j in range ( 0 , i):
if arr[i] = = arr[j]:
minLen = min (minLen, i - j + 1 )
if minLen = = n + 1 :
print ( "-1" )
else :
print (minLen)
|
C#
using System;
class GFG
{
static int INT_MAX = int .MaxValue;
static int subArrayLength( int []arr, int n)
{
int minLen = INT_MAX;
for ( int i = 1; i < n; i++)
{
for ( int j = 0; j < i; j++)
{
if (arr[i] == arr[j])
{
minLen = Math.Min(minLen, i - j + 1);
}
}
}
if (minLen == INT_MAX)
{
return -1;
}
return minLen;
}
public static void Main()
{
int n = 7;
int []arr = { 1, 2, 3, 1, 5, 4, 5 };
int ans = subArrayLength(arr, n);
Console.WriteLine(ans);
}
}
|
Javascript
<script>
var INT_MAX = Number.MAX_VALUE;
function subArrayLength( arr , n) {
var minLen = INT_MAX;
for ( var i = 1; i < n; i++) {
for ( var j = 0; j < i; j++) {
if (arr[i] == arr[j]) {
minLen = Math.min(minLen, i - j + 1);
}
}
}
if (minLen == INT_MAX) {
return -1;
}
return minLen;
}
var n = 7;
var arr = [ 1, 2, 3, 1, 5, 4, 5 ];
var ans = subArrayLength(arr, n);
document.write(ans);
</script>
|
Time Complexity: O(N2)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach:
This problem can be solved in O(N) time and O(N) Auxiliary space using the idea of hashing technique. The idea is to iterate through each element of the array in a linear way and for each element, find its last occurrence using a hashmap and then update the value of min length using the difference of the last occurrence and the current index. Also, update the value of the last occurrence of the element by the value of the current index.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int subArrayLength( int arr[], int n)
{
int minLen = INT_MAX;
unordered_map< int , int > last;
for ( int i = 0; i < n; i++) {
if (last[arr[i]] != 0) {
minLen = min(minLen, i - last[arr[i]] + 2);
}
last[arr[i]] = i + 1;
}
if (minLen == INT_MAX) {
return -1;
}
return minLen;
}
int main()
{
int n = 7;
int arr[] = { 1, 2, 3, 1, 5, 4, 5 };
int ans = subArrayLength(arr, n);
cout << ans << '\n' ;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int subArrayLength( int arr[], int n)
{
int minLen = Integer.MAX_VALUE;
HashMap<Integer, Integer> last = new HashMap<Integer, Integer>();
for ( int i = 0 ; i < n; i++)
{
if (last.containsKey(arr[i]) && last.get(arr[i]) != 0 )
{
minLen = Math.min(minLen, i - last.get(arr[i]) + 2 );
}
last.put(arr[i], i + 1 );
}
if (minLen == Integer.MAX_VALUE)
{
return - 1 ;
}
return minLen;
}
public static void main(String[] args)
{
int n = 7 ;
int arr[] = { 1 , 2 , 3 , 1 , 5 , 4 , 5 };
int ans = subArrayLength(arr, n);
System.out.print(ans);
}
}
|
Python
n = 7
arr = [ 1 , 2 , 3 , 1 , 5 , 4 , 5 ]
last = dict ()
minLen = n + 1
for i in range ( 0 , n):
if arr[i] in last:
minLen = min (minLen, i - last[arr[i]] + 2 )
last[arr[i]] = i + 1
if minLen = = n + 1 :
print ( "-1" )
else :
print (minLen)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int subArrayLength( int []arr, int n)
{
int minLen = int .MaxValue;
Dictionary< int , int > last = new Dictionary< int , int >();
for ( int i = 0; i < n; i++)
{
if (last.ContainsKey(arr[i]) && last[arr[i]] != 0)
{
minLen = Math.Min(minLen, i - last[arr[i]] + 2);
}
if (last.ContainsKey(arr[i]))
last[arr[i]] = i + 1;
else
last.Add(arr[i], i + 1);
}
if (minLen == int .MaxValue)
{
return -1;
}
return minLen;
}
public static void Main(String[] args)
{
int n = 7;
int []arr = { 1, 2, 3, 1, 5, 4, 5 };
int ans = subArrayLength(arr, n);
Console.Write(ans);
}
}
|
Javascript
<script>
function subArrayLength(arr, n)
{
let minLen = Number.MAX_VALUE;
let last = new Map();
for (let i = 0; i < n; i++)
{
if (last.has(arr[i]) && last.get(arr[i]) != 0)
{
minLen =
Math.min(minLen, i - last.get(arr[i]) + 2);
}
last.set(arr[i], i + 1);
}
if (minLen == Number.MAX_VALUE)
{
return -1;
}
return minLen;
}
let n = 7;
let arr = [ 1, 2, 3, 1, 5, 4, 5 ];
let ans = subArrayLength(arr, n);
document.write(ans);
</script>
|
Time Complexity: O(N), where N is size of array
Auxiliary Space: O(N) because it is using unordered_map last
Share your thoughts in the comments
Please Login to comment...